BASIC KNOWLEDGE 1. Two Important Terms (1.1). Permutations

50 AMC LECTURES (26) Counting

BASIC KNOWLEDGE 1. Two Important Terms (1.1). Permutations (1). Different elements, with no repetition. Take r elements each time from n distinct elements (1 ≤ r ≤ n).

Number of permutations )!(

!),(rn

nrnP−

= (1.1)

(2). n distinct objects can be permutated in n! permutations. We let n = r in (1) to get P(n, n) = n! (1.2) The symbol ! (factorial) is defined as follows:

0! = 1 (1.3) and for integers n ≥ 1,

n!= n · (n – 1) ···· 1 (1.4) 1! = 1, 2! = 2 · 1=2, 3! = 3 · 2 · 1=6, 4! = 4 · 3 · 2 · 1 = 24, 5! = 5 · 4 · 3 · 1 · 1 = 120, 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720. Proof of (1.2): The first object can be chosen in n ways, the second object in n −1 ways, the third in n − 2, etc. By the Fundamental Counting Principle, we have n(n − 1)(n − 2) · · · 2 · 1 = n! ways. A permutation is an arrangement or a listing of things in which order is important. Example 1: How many 5-digit positive integers can be formed by the digits of 0, 1, 2, 3, and 4? Solution: 12340 is different with 13240 so order is important and we use permutations.

1


Method 1: The number of permutations with 5 digits is 5! = 120. The number of permutations with 4 digits (0 is in the leftmost position) is 4! = 24. No positive 5-digit integers can be formed by these permutations. The answer is then 120 – 24 = 96. Method 2: The number of permutations with 0 as the units digit is 4! = 24. The number of permutations with 0 as the tens digit is 4! = 24. The number of permutations with 0 as the hundred digit is 4! = 24. The number of permutations with 0 as the thousand digit is 4! = 24. The answer is then 24 × 4 = 96. (1.2). Combinations

Definition Let n, r be non-negative integers such that 0 ≤ r ≤ n. The symbol (read “n

choose m”) is defined and denoted by

⎟⎟⎠

⎞rn⎜⎜⎝

⎛

)!(!!

),(),(

rnrn

rrPrnP

rn

−==⎟⎟

⎠

⎞⎜⎜⎝

⎛ (1.5)

Remember: , , and 1 nn

=⎟⎟⎠

⎞⎜⎜⎝

⎛10

=⎟⎟⎠

⎞⎜⎜⎝

⎛n1=⎟⎟

⎠

⎞⎜⎜⎝

⎛nn

Since n – (n – r) = r, we have (1.6) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛rn

nrn

Unlike Permutations, combinations are used when order does not matter. If we have n different elements, and it doesn’t matter which order we take the elements, the number of

ways to take m elements where 1 ≤ m ≤ n, is ⎟⎟⎠

⎞⎜⎜⎝

⎛mn

A combination is an arrangement or a listing of things in which order is not important. Example 2: (a). In how many parts at most do n lines cut a plane? (b). In how many parts at most do n planes cut a space? Solution:

2


(a). . ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛210nnn

(b). . ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛3210nnnn

2. Two Important Rules (2.1). The product rule (Fundamental Counting Principle) (Step Work) When a task consists of k separate steps, if the first step can be done in n1 ways, the second step can be done in n2 ways, and so on through the kth step, which can be done in nk ways, then total number of possible results for completing the task is given by the product: (2.1) knnnnN ××××= ...321

Example 3: (1998 North Carolina Math Contest) There are 8 girls and 6 boys in the Math Club at Central High School. The Club needs to form a delegation to send to a conference, and the delegation must contain exactly two girls and two boys. The number of possible delegations that can be formed from the membership of the Club is a) 48 b) 420 c) 576 d) 1680 e) 2304 Solution: (B). The task is to form a delegation. We need to go through the following two steps to complete the task.

Step 1: Selecting two girls: . ⎟⎟⎠

⎞⎜⎜⎝

⎛28

Step 2: Selecting two boys: . ⎟⎟⎠

⎞⎜⎜⎝

⎛26

By the product rule, we have the total number of distinct delegations:

⎟⎟⎠

⎞⎟⎟⎠

⎞⎜⎜⎝

⎛28

× = 28 × 15 = 420. ⎜⎜⎝

⎛26

(2.2). The sum rule (case work)

3


If an event E1 can happen in n1 ways, event E2 can happen in n2 ways, event Ek can happen in nk ways, and if any event E1, E2,.. or Ek happens, the job is done, then the total ways to do the job is N = n1 + n2 + ···+ nk (2.2) Example 4: Hope High School has three elective courses for social studies and four electives for science. How many ways are there for Alex to select three electives from them this semester? (A) 30 (B) 35 (C) 42 (D) 48 Solution: (B). Case I: Alex can select 1 social studies course and 2 science courses. By the product rule,

1824

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

Case II: Alex can select 2 social studies courses and 1 science course. By the product

rule, . 1214

23

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case III: Alex can select 3 social studies courses:

4

3

4

33

=⎟⎟⎠

⎞⎜⎜⎝

⎛.

Case III: Alex can select 3 science courses:

34

=⎟⎟⎠

⎞⎜⎜⎝

⎛.

By the sum rule, we have 18 + 12 + 1 + 4 = 35 ways. Example 5: How many two-digit numbers are there such that the units digit is greater than then ten’s digit? Solution:


The job is to count the two-digit number that has a larger units digit. We have the following cases. In any case, we finish the job of counting two-digit numbers that the units digit is greater than then ten’s digit. Case I: When units digit is 9, ten’s digit can be any digit of 1 to 8. So we have 8 such two-digit numbers. Case II: When units digit is 8, ten’s digit can be any of 1 to 7. So we have 7 such two-digit numbers. Case III: When units digit is 7, ten’s digit can be any of 1 to 6. So we have 6 such two-digit numbers. Similarly. When units digit is 2, ten’s digit can only be1. So we have 1 such two-digit number.

By The Sum Rule, we have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36942

8)81(=×=

×+ such

two-digit numbers. Example 6: (1989 AMC) Mr. and Mrs. Zeta want to name baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible? (A) 276 (B) 300 (C) 552 (D) 600 (E) 15600 Solution: (B). The last initial is fixed at Z. If the first initial is A, the second initial must be one of B, C, D, . . ., Y, so there are 24 choices for the second. If the first initial is B, there are 23 choices for the second initial: of C, D, E, . . ., Y. Continuing in this way we see that the number of monograms is 24 + 23 + 22 + · · · + 1. Use the formula

2

)1(21 +=+++

nnnL to obtain the answer .3002

2524=

⋅

Example 7: (2003 AMC 10B) How many distinct four-digit numbers are divisible by 3 and have 23 as their last two digits? Solution: First two digits should have a sum of 1, 4 , 7, 10, 13, or 16. If the sum of the first two digits is 1: There is only 1 such number: 1023;

5


If the sum of the first two digits is 4: There are 4 such numbers because 4 = 4 + 0 = 1 + 3 = 3 + 1 = 2 + 2;

If the sum of the first two digits is 7: There are 7 such numbers because 7 = 7 + 0 = 6 + 1 = 1 + 6 = 5 + 2 = 2 + 5 = 4 + 3 = 3 + 4;

If the sum of the first two digits is 10: There are 9 such numbers because 9 = 9 + 0 = 8 + 1 = 1 + 8 = 7 + 2 = 2 + 7 = 6 + 3 = 3 + 6 = 5 + 4= 4 + 5;

If the sum of the first two digits is 13: There are 6 such numbers because 13 = 9 + 4 = 4 + 9 = 8 + 5 =5 + 8 = 7 + 6 = 6 + 7;

If the sum of the first two digits is 16: There are 3 such numbers because 16 = 9 + 7 = 7 + 9 = 8 + 8.

By the sum rule, we get the answer: 1 + 4 + 7 + 9 + 6 + 3 = 30.

Example 8: (1989 AIME) Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices? (Polygons are distinct unless they have exactly the same vertices.) Solution: 968. The job is to form convex polygons. We have the following cases. In any case, we finish the job of forming convex polygons.

6

⎟⎟⎠

⎞Triangles : The number of triangles formed by selecting three points out of 10 points:

⎜⎜⎝

⎛310

Convex quadrilaterals: . ⎟⎟⎠

⎞⎜⎜⎝

⎛410

Convex pentagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛510

Convex hexagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛610

Convex heptagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛710

Convex octagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛810


Convex nonagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛910

Convex decagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛1010

By the sum rule, the number of convex polygons:

7

⎟⎟⎠

⎞⎟⎟⎠

⎞⎜⎜⎝

⎛

⎝N = + ⎜⎜ + + + + + + = 210 – ( + ⎜⎜

⎛+

) = 1024 – 56 = 968.

⎜⎜⎝

⎛310

⎟⎟⎠

⎞210

⎝

⎛410

⎟⎟⎠

⎞⎜⎜⎝

⎛510

⎟⎟⎠

⎞⎜⎜⎝

⎛610

⎟⎟⎠

⎞⎜⎜⎝

⎛710

⎟⎟⎠

⎞⎜⎜⎝

⎛810

⎟⎟⎠

⎞⎜⎜⎝

⎛910

⎟⎟⎠

⎞⎜⎜⎝

⎛1010

⎟⎟⎠

⎞010

⎟⎟⎠

⎞110

⎜⎜⎝

⎛

3. Three Important Theorems THEOREM 1: (Grouping) (a). Let the number of different objects be n. Divide n into r groups A1, A2, ..., Ar such that there are n1 objects in group A1, n2 objects in group A2, ..., nr objects in the group Ar, where n1 + n2 + · · · + nr = n. The number of ways to do so is

!

⎟⎟⎠

⎞

⎟⎟⎠

⎞

!!!

21 rnnnnN⋅⋅⋅

= (3.1)

Proof:

There are ways to take out n1 elements from n elements to form group A1. ⎜⎜⎝

⎛

1nn

There are ways to take out n2 elements from n – n1 elements to form group A2. ⎟⎟⎠

⎞⎜⎜⎝

⎛ −

2

1

nnn

Continue the process until there are nr elements left to form group Ar. The total number of ways, based on the Fundamental Counting Principle, is

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

2

1

nnn

⎜⎜⎝

⎛

1nn

.... = ⎟⎟⎠

⎞⎜⎜⎝

⎛

r

r

nn

!!!!

21 rnnnn⋅⋅⋅

(b). Let there be r types of objects: n1 of type 1, n2 of type 2; etc. The number of ways in which these n1 + n2 + · · · + nr = n objects can be rearranged is

!!!!

21 rnnnn⋅⋅⋅

(3.2)

The proof is the same as the proof for (3)


Example 9: A gardener plants eight trees out of three maple trees, two oak trees, and four birch trees in a row. How many ways are there? Solution: Case I: With two maple trees, two oak trees, and four birch trees, by (3.1), we have

420!4!2!2

!8=

Case II: With three maple trees, one oak tree, and four birch trees:

280!4!1!3

!8=

Case III: With three maple trees, two oak trees, and three birch trees:

560!3!2!3

!8=

By the sum rule, we know that the total number of ways: 420 + 280 + 560 = 1260. Example 10: Five number 1, 2, 3, 4, and 5 are arranged in a row like . How many arrangements are there such that a1 ≠ 1, a2 ≠ 2, a3 ≠ 3, a4 ≠ 4, a5 ≠ 5 ?

54321 aaaaa

Solution: We use 2 as an example and we have 11 arrangements: Since the leftmost digit can also be 3, 4, or 5, so the answer by the product rule will be 11 × 4 = 44. THEOREM 2: (Combinations with Repetitions) (a). n identical balls are put into r labeled boxes and the number of balls in each box is not limited. The number of ways is

8


9

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+n

rn

1 or (3.3) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Proof: Put r labeled boxes next to each other as shown in the figure below. Put n balls into these boxes (figure 1). Then we line these boxes up next to each other (figure 2). Now we take apart the top and bottom sides of the each box and the two sides of the two boxes at the end (figure 3). After that we have figure 4. The problem becomes finding the number of ways to permutate n identical balls with r – 1 identical partitions:

)!1(!1(

−−+

rnrn )! or or . ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

1⎟⎟⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Figure 1 Figure 2

Figure 3 Figure 4 (b) The number of terms in the expansion of , after the like terms combined, is

nrxxxx )....( 321 ++++

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛ −+n

rn

1 or (3.4) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

(c). Let n be a positive integer. The number of positive integer solutions to x1 + x2 + ⋅⋅⋅ + xr = n is

⎜⎜⎝

⎛−−

11

rn

. (3.5)


10

⎜⎜⎝

⎛

⎟⎟⎠

⎞

Proof. Write n as n =1+1+ ⋅⋅⋅ +1+1, where there are n 1’s and n − 1 plus signs. In order

to decompose n in r summands, we choose r − 1 plus signs from the n − 1, that is . ⎟⎟⎠

⎞−−

11

rn

(d). Let n be a positive integer. The number of non-negative integer solutions to y1 + y2 + ⋅⋅⋅ + yr = n is

⎜⎜⎝

⎛ −+n

rn

1 or (3.6) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Proof: Set xr − 1= yr. Then xr ≥ 1. The equation x1 + x2 + ⋅⋅⋅ + xr = n

is equivalent to x1 + x2 + ⋅⋅⋅ + xr = n + r, which has solutions. ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Example 11: A baking company produces four different cookies: Chocolate Chip Cookies, Peanut Butter Cookies, Oatmeal Cookies, and Blueberry Cookies. (a) If a package contains 8 cookies, how many different packages are possible? (b) If a package contains 8 cookies with at least one cookie of each kind , how many different packages are possible? Solution: (a). By (3.6), we write y1 + y2 + ⋅⋅⋅ + yr = n ⇒ y1 + y2 + y3 + y4 = 8

The answer is . 165311

3 148

1 1

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+

=r

rnN

(b). By (3.5), we write y1 + y2 + ⋅⋅⋅ + yr = n ⇒ y1 + y2 + y3 + y4 = 8


1418

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=rn

N

Example 12: (1998 AIME) Find the number of ordered quadruples (x1 , x2 , x3 , x4) of positive odd integers that satisfy x1 + x2 + x3 + x4 = 98. Solution: Since x1 , x2 , x3 , x4 are odd positive integers, we let


x1 = 2y1 − 1 x2 = 2y2 − 1 x3 = 2y3 − 1 x4 = 2y4 − 1 y1, y2, y3, and y4 are positive integers. The given equation becomes: 2(y1 + y2 + y3 + y4) = 98 + 4 ⇒ y1 + y2 + y3 + y4 = 51 (1) By (3.5), the number of positive integer solutions is

19600350

14151

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

.

Example 13: (a) How many ways to take 4 letters from a, b, b, c, c, c, d, d, d, d, d ?

(b) How many different 4-letter codes can be formed by using letters from a, b, b, c, c, c, d, d, d, d, d ?

Solution: (a) 20, (b) 152.

Method 1:

Case 1: 4 letters are the same: . 111

=⎟⎟⎠

⎞⎜⎜⎝

⎛


12

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

11

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛12

means that the number of ways we select the 3 letters (from either the letters c or d).

⎜⎜⎝

⎛13

means that we select the rest of the letter from a, b, or the letter (c or d) that is not

selected in the first time.

Case 3: 2 letters are the same and 2 other letters are the same: 32

12

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛


Case 4: 2 letters are the same, 2 other letters are different: 913

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 5: 4 letters are different: 144

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Total 1 + 6 + 3 + 9 + 1 = 20.

Method 2 (by listing): Case 1: 4 letters are the same: dddd. We have 1 way for case 1.

Case 2: 3 letters are the same: ccca 1 way. cccb 1 way cccd 1 way ddda 1 way dddb 1 way dddc 1 way We have 6 ways for case 2. Case 3: 2 letters are the same and 2 other letters are the same: bbcc 1 way bbdd 1 way ccdd 1 way We have 3ways for case 3. Case 4: 2 letters are the same, 2 other letters are different: bbac 1 way bbad 1 way bbcd 1 way ccab 1 way ccad 1 way ccbd 1 way ddab 1 way ddac 1 way ddbd 1 way

12


We have 9 ways for case 4. Case 5: 4 letters are different: abcd. We have 1 way for case 5. Total 1 + 6 + 3 + 9 + 1 = 20.

(b) 152.

Method 1:


=⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 2: 3 letters are the same: 24!3!4

13

12

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛12

means that the number of ways we select the 3 letters (from either the letters c or d).

⎜⎜⎝

⎛13

means that we select the rest of the letter from a, b, or the letter (c or d) that is not

selected in the first time.

!3!4 means that we arrange the 3 letters with fourth letter.

Case 3: 2 letters are the same and 2 other letters are the same: 18!2!2

!423

=×

×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 4: 2 letters are the same, 2 other letters are different: 108!2!4

23

13

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 5: 4 letters are different: 144

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Total 1 + 24 + 18 + 108 + 1 = 152.

Method 2 (by listing): Case 1: 4 letters are the same: dddd. We have 1 way for case 1.

Case 2: 3 letters are the same:

13


ccca 4!3!4= ways

cccb 4 ways cccd 4 ways ddda 4 ways dddb 4 ways dddc 4 ways We have 24 ways for case 2. Case 3: 2 letters are the same and 2 other letters are the same:

bbcc 16!2!2

!4=

×

bbdd 6 ways ccdd 6 ways We have 18 ways for case 3. Case 4: 2 letters are the same, 2 other letters are different:

bbac 12!2!4=

bbad 12 ways bbcd 12 ways cc ab 12 ways ccad 12 ways ccbd 12 ways ddab 12 ways ddac 12 ways ddbd 12 ways We have 108 ways for case 4. Case 5: 4 letters are different: abcd We have 1 way for case 5. Total 1 + 24 + 18 + 108 + 1 = 152.

14


THEOREM 3: (Circular Permutations) The number of circular permutations (arrangements in a circle) of n distinct objects is N = (n – 1)! (3.7) We can think of this as n people being seated at a round table. Since a rotation of the table does not change an arrangement, we can put person A in one fixed place and then consider the number of ways to seat all the others. Person B can be treated as the first person to seat and M the last person to seat. The number of ways to arrange persons A to M is the same as the number of ways to arrange persons B to M in a row. So the number of ways is N = (n – 1)!.

Example 14: In how many ways is it possible to seat seven people at a round table if Alex and Bob must not sit in adjacent seats? Solution: 480. By (3.7), the number of ways to sit 7 people at a round table is N = (n – 1)! = N = (7 – 1)! = 6! = 720. We find the number of ways Alex and Bob sit together by seeing them as a unit. There are (6 – 1)! = 5! ways. The result is multiplied by 2 if we alter Alex’s and Bob’s positions. The solution is then 720 – 2 × 5! = 720 – 2 × 120 = 480. Example 15: In how many ways can four married couples be seated at a round table if no two men, as well as no husband and wife are to be in adjacent seats? Solution: 12. We already know from the last problem that there are (4 – 1)! = 3! to seat four women. After the ladies are seated, person M4 (whose wife is not shown in the figure below) has two ways to sit. After he is seated in any one of the two possible seats, the other men have only one way to sit in the remaining seats. The solution is 3! × 2 = 12.

15


Example 16: Twelve student body members are seated at a round table electing president, vice president, and treasurer. How many possible ways are there such that at least two of the three elected had been sitting next to each other? Solution: Case I: Three had been sitting next to each other. We have 12 ways to select first person, say, A. Once A is selected, we have one way to select B and C. By the product rule, number of ways is the ways = 12 × 1 × 1 = 12.

Case II: Two of them had been sitting next to each other. We have 12 ways to select two that had been sitting next to each other. Let us say, A and B. For the third person, we are not allowed to select neither C nor D. So we have 12 – 4 ways to select the third person. By the product rule, number of ways is the ways = 12 × (12 – 4) = 12.

Finally by the sum rule, we have the answer: 12 + 96 = 108. 4. Typical Problems in Counting 4.1 Counting problems related to partitioning

16


17

ix students and dec etween two students. In

ow many ways can Professors Alpha, Beta and Gamma choose their chairs? 12 (B) 36 (C) 60 (D) 84 (E) 630

⎝aces at the ends do not count because each professor needs to sit between two students)

and there are 3! ways for them to rearrange themselves.

Example 17: (1994 AMC) Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the s

ide to choose their chairs so that each professor will be bh

(A) Solution:

We sit the students first. There are five spaces for three professors to choose ⎟⎟⎞

⎜⎜⎛5

(two ⎠3

sp

By the production rule, the answer is × 3! = 60.

eight students and two teachers lpha and Beta for a class picture. Two teachers decide not to sit next to each other.

B）520 （C）440 （D）720

teachers to sit so we get ⎜⎜⎝

⎛, and there

are 2! ways for them to rearrange themselves.

⎟⎞

⎜⎛5

⎟⎠

⎜⎝3

Example 18: Ten chairs in a row are to be occupied by AThere are n arrangements and the last two digits of n is A）040 （ Solution: (A). We sit the students first. There are 8! Ways.

After that, we see that there are 9 spaces for two ⎟⎟⎠

⎞29


By the production rule, the answer is !8 × ⎟⎟⎠

⎞⎜⎜⎝

⎛29

× 2! = 40320 × 72 = 2903040.

4.2. Counting problems related to coloring

18

s to be assigned a e color.

ore than once?

y th t ule, w 4 × 3 × 3 ×

n the figure, each of six regions ABCD is to be assigned a color. 4 colors to choose from, and no adjacent regions can be the same color. How

r is allowed to use more than once? 60 D．48

84 ways.

e sides of convex pentagon ABCDE are distinct. Each side is to be assigned a color. There are 3 colors (red, yellow and blue) to choose from, and no two sides sharing the same vertex must have different colors. How many different colorings are possible?

Example 19: As shown in the figure, each of five regions ABCDE icolor. There are 5 colors to choose from, and no adjacent regions can be the samHow many different ways are there if each color is allowed to be used m Solution: 540 ways. We color region A first. We have 5 ways. Then we color region B (we have 4 ways). Next we color region C (we have 3 ways). For region D, we have 3 ways (not the same color with A or C). Finally we have 3 ways to color E (not the same color with A or D). B

e produc r e have 5 × 3 = 540 ways.

Example 20: As shown iThere are many different ways are there if each colo

A．96 B．84 C． Solution: B. Solution: Case I: A and C have the same color. The number of ways to color: 4 × 3 × 1 × 3 = 36 Case II: A and C have different colors. The number of ways to color: 4 × 3 × 2 × 2 = 48.

36 + 48 =The answer is Example 21: The lengths of fiv


19

Solution: 30. We lable the sides as follows:

If a is colored red, b is colored yellow, then when c is colored red:

we have two arrangements:

If a is colored red, b is colored yellow, then when c is colored blue:

we have three arrangements:


By the sum rule, we have 2 + 3 = 5 ways. We see that (a, b) can also colored (red, blue), (yellow, blue), (yellow, red), (blue, red), and (blue, yellow). Therefore the total number of colorings is 5 × 6 = 30. Example 22: Each of four faces of a regular tetrahedron is colored one of 10 colors. How many distinct ways are there to color the tetrahedron? (Two colorings are considered distinct if they cannot be rotated to look like each other.) (A) 925 (B) 980 (C) 1024 (D) 1090 (E)1450 Solution: Case I: Four colors are used: We have two configurations as follows:

The number of ways to color is

Case II: Three colors are used: We have three configurations as follows:


Case III: Two colors are used: We have three configurations as follows:

20



Case IV: one color is used: We have only one configuration as follows:


Total ways: 420 + 360 + 135 + 10 = 925

Example 23: (2003 North Carolina Math Contest) The tips of a five-pointed star are to be painted red, white and blue. How many ways can this be done if no adjacent points can be the same color? Solution: 30 ways. Method 1 (official solution): Let the 5 points be ABCDE. Note that to paint the 5 points in the prescribed fashion, 2 colors must be used twice and one color just once. Also, note that if point A is a fixed color, then there are only two ways to paint the star. Consider one such set of colors, where red is the single color used. Then if A is painted red there are only two ways to paint the rest of the star. A similar situation occurs if the points B, C, D, or E are painted with the red. For this set of colors there are 10 ways to paint the star. Since there are three different sets of colors, there are a total of 30 ways to paint the star. Method 2 (our solution): If A and C have the same color, then the ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 2 = 12

21


If A and D have the same color, then the ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 2 = 12

If A and C have the different color, then the ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 1 = 6

Total ways should be 12 + 12 + 6 = 30. Method 3: There is a formula for this problem which can be derived by the recursion method. An = (m – 1)[(m – 1)(n-1) + (–1)n]. Where n is the number of points and m is the number of colors. n = 5 and m = 3 is the case for the NC Math contest problem. An = (m – 1)[(m – 1)(n-1) + (–1)n] = (3 – 1)[(3 – 1)(5 – 1) + (–1)5] = 30.

22


Example 24. There are five regions to be colored with four different colors. If no same color can be used for adjacent regions, how many ways are there to color? Solution: If regions 1 and 4 are colored by the same color, we have the following ways to do so: Regions 1 2 3 4 5

Ways to color 4 × 3 × 2 × 1 × 2 = 48 If regions 1 and 4 are colored by the different color, we have the following ways to do so: Regions 1 2 3 4 5

Ways to color 4 × 3 × 2 × 1 × 1 = 24 Total ways 48 + 24 = 72. Example 25. Each point of A, B, C, D, E, and F is to be assigned a color. There are 4 colors to choose from, and the ends of each line segment must have different colors. How many different colorings are possible? （A） 288 （B）264 （C） 240 （D）168 Solution: B. Case I：4 colors are used for B, D, E, and F. We have

ways. 44 1 1 2A × × = 4

2

Case II: 3 colors are used for B, D, E, and F. We have

ways. 34 2 2A × × + 3

4 2 1 2 19A × × × =

Case III: 2 colors are used for B, D, E, and F. We have 2

4 2 2 48A × × = ways. By the product rule, we have total 24 + 192 + 48 = 264 ways.

23


4.3. Counting problems related to geometry Example 26. (1993 AMC) How many triangles with positive area are there whose vertices are points in the xy-plane whose coordinates are integers (x, y) satisfying 1 ≤ x ≤ 4 and 1 ≤ y ≤ 4? Solution: 516.

There are = 560 sets of 3 points. We must exclude from our count those sets of

three points that are collinear.

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛316

There are 4 vertical and 4 horizontal lines with four points each. These 8 lines contain

8 ⎜⎜ = 32 sets of 3 collinear points. Similarly, there are 2 + 4 = 12 sets of 3

collinear points that determine lines of slope ±1. Because there are no other sets of 3 collinear points, the number of triangles is the total sets minus the degenerate triangles equals 560 – 32 – 12 = 516.

⎝

⎛34

⎟⎟⎠

⎞⎜⎜⎝

⎛34

⎟⎟⎠

⎞⎜⎜⎝

⎛33

Example 27. (1976 AMC) Lines L1, L2. . ., L100 are distinct. All lines L4n, n a positive integer, are parallel to each other. All lines L4n-3, n a positive integer, pass through a given point A. The maximum number of points of intersection of pairs of lines from the complete set {L1, L2, . . ., L100} is

(A) 4350 (B) 4351 (C) 4900 (D) 4901 (E) 9851 Solution: (B).

24


One hundred lines intersect at most at 49502

991002100

=⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛ points. But the 25 lines

L4, L8, . . ., L100 are parallel; hence intersections are lost. Also the 25 lines L1,

L5, . . ., L97 intersect only at point A, so that more

intersections are lost. The maximum number of points of intersection is 4950 – 300 – 299 = 4351.

300225

=⎟⎟⎠

⎞⎜⎜⎝

⎛

225⎜⎜⎝

⎛2991=−⎟⎟

⎠

⎞⎜⎜⎝

⎛2991

225

=−⎟⎟⎠

⎞

Example 28. There are six points in a plane. Sixteen triangles can be formed by connecting these points. How many lines are there that contain more 3 of the points? Solution:

If there are no three points collinear, we can form triangles. 2036

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Since we only get 16 triangles, we know that three or more points are collinear.

Since , we know that no 5 points collinear. 10102035

26

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

Suppose there are 4 points collinear, we have , which is true. 1642034

26

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

Suppose there are 3 points collinear, we have

, which is also true. 1642033

426

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

So we know that either one line containing 4 points or three lines each containing three points. 4.4. Counting problems related to numbers Example 29: Alex wants to select two different numbers from {2, 3, 4, 5, 6, 7, 8, 9}. How many ways are there such that these two numbers are not consecutive?

25


26

Solution: 21. Method 1 (Listing): Case I: First integer: 2 Second integer: 4, 5, 6, 7, 8, or 9 6 ways. Case II: First integer: 3 Second integer: 5, 6, 7, 8, or 9 5 ways. Case III: First integer: 4 Second integer: 6, 7, 8, or 9 4 ways. Case IV: First integer: 5 Second integer: 7, 8, or 9 3 ways. Case V: First integer: 6 Second integer: 8 or 9 2 ways. Case VI: First integer: 7 Second integer: 9 1 way. Total 6 + 5 + 4 + 3 + 2 + 1 = 21 ways. Method 2:


The formula: N = .

n is the total number of terms. k is the number of elements selected. n = 9 − 2 + 1 = 8. k = 2.

N = = .

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−kkn

)1(

⎟⎟⎠

⎞

)12(8⎜⎜⎝

⎛ −−k⎜⎜

⎝

⎛ −−kkn

)1(

2127

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞

Example 30: (2006 AMC 12 A) How many non-empty subsets S of {1, 2, 3,..., 15} have the following two properties? (1) No two consecutive integers belong to S. (2) If S contains k elements, then S contains no number less than k. (A) 277 (B) 311 (C) 376 (D) 377 (E) 405 Solution:

27

⎟⎟⎠

⎞N = ⎜⎜ .

⎝

⎛ −−kkn

)1(

If k = 1, n = 15, then N1 = . 151 15

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 2, n = 14, then N2 = . 782

13=⎟⎟

⎠

⎞⎜⎜⎝

⎛

If k = 3, n = 13, then N3 = . 1653

11=⎟⎟

⎠

⎞⎜⎜⎝

⎛

If k = 4, n = 12, then N4 = . 12649

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 5, n = 11, then N5 = . 2157

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 6, n = 10, then N6 = (not possible). ⎟⎟⎠

⎞⎜⎜⎝

⎛65


Total: 15 + 78 + 165 + 126 + 21 = 405. Note: this is the last problem in the test. Example 31: (1980 Bulgarian Mathematical Olympiad) Find the number of ways of choosing 6 among the first 49 positive integers, at least two of which are consecutive. Solution:

28

⎟⎟⎠

⎞There are ways to select 6 numbers among the first 49 positive integers. ⎜⎜

⎝

⎛6 49

There are = ways to select 6 numbers among the first 49 positive

integers such that no two of which are consecutive.

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−kkn

)1(

⎟⎟⎠

⎞⎜⎜⎝

⎛6 44

The answer will then be − . ⎟⎟⎠

⎞⎜⎜⎝

⎛6 49

⎟⎟⎠

⎞⎜⎜⎝

⎛6 44

Example 32. (1987 China Middle School Math Contest) Counting number n has the following property: if we take any 50 different numbers from 1, 2, 3, …, n, there always have two numbers with the difference of 7. Among many values of n, what is the largest value?

Solution: 98.

We take 7 numbers from 1 to 7 from S. Then we take 7 numbers again from 15 to 21 and we repeat the action until we get the last group: 85, 86, 87, 88, 89, 90, 91. Now we have 49 numbers. The difference between any two numbers in these 49 numbers is not 7.

Then if we take one more number (92 in our case), we get two numbers with a difference of 7 (92 – 85). However, 92 is not the largest one since we can have any one of 93 to 98. So 98 is the largest value of n.

Note that if we leave 92 to 98 but take 99, we can go on until we take 105 such that all these numbers do not have a difference of 7. So 99 is not the value for n.

When n = 98. we can divided all the numbers into the following 7 subsets:

},92,85,,15,8,1{ L


},93,86,,16,9,2{ L

},94,87,,17,10,3{ L

},95,88,,18,11,4{ L

},96,89,,19,12,5{ L

},97,90,,20,13,6{ L

},98,91,,21,14,7{ L

Each subset has 14 numbers and any two neighboring numbers have a difference of 7. If we take 50 numbers from these 7 sets (50 = 7 × 7 + 1) from the pigeonhole principle, we are sure that there must be one subset containing 8 numbers. There must be two consecutive numbers among these 8 numbers and there must be two numbers with a difference of 7.

So the largest value of n is 98.

Example 33: A subset of integers 1, 2,…, 100 has the property that none of its members is either the sum of other two or two times of another. What is the largest number of members such a subset can have?

Solution: 50.

All 50 odd numbers can be selected. If we have 51 numbers, without lose generality, we assume a1, a2, a3, .. a51, with a51 is the greatest of these numbers. Then we have a51 – a1, a51 – a2, …. a51 – a50 and all these numbers are within 100. Then we got 101 numbers. Among these 101 numbers, there must be two numbers that are equal. i.e. there must be ai = a51 – ai, or 2ai = a51. Contradiction. So the largest number of members is 50.

Example 34: A subset of integers 1, 2,…, 10 has the property that the sum of its members is odd. How many such subsets are there?

Solution: 526.

Let A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 10}.

29


If we form a subset by taking out 1, 3, or 5 numbers from set A and 0, 1, 2, 3, 4, or 5 numbers from set B, the sum of the members in this subset is odd. So the number of such subsets is:

526255

45

35

25

15

05

55

35

15 9 ==⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛×⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

Note: A subset of integers 1, 2,…, n has the property that the sum of its members is odd. The number of such subsets is 2n – 1.

A subset of integers 1, 2,…, n has the property that the sum of its members is even. The number of such subsets is 2n – 1– 1.

30


31

PROBLEMS

Problem 1: (1986 AMC) The 120 permutations of AHSME are arranged in dictionary order, as if each were an ordinary five-letter word. The last letter of the 86th word in this list is

(A) A (B) H (C) S (D) M (E) E Problem 2: (1990 AMC) At one of George Washington’s parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If 13 married couples attended, how many handshakes were there among these 26 people?

(A) 78 (B) 185 (C) 234 (D) 321 (E) 325 Problem 3: Six cards each labeled with one of the digits 1 to 6 are put into 3 different envelopes. Each envelope will have two cards and cards 1 and 2 will be in the same envelope. How many ways are there to do so?

（A） 12 (B) 18 (C) 36 (D) 54

Problem 4: Hope High School has three elective courses for social studies and four electives for science. How many ways are there for Alex to select three electives from them this semester with at least one from each subject? (A) 30 (B) 35 (C) 42 (D) 48 Problem 5: A palindrome number is a number that is the same when written forwards or backwards. How many palindrome numbers less than 1000 are there? Problem 6: How many 3-digit even numbers can be formed by using the digit 0, 2, 3, 4, and 5? Problem 7: You have nine line segments with the lengths of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. How many ways are there to form a square by connecting the ends of some of these line segments? No overlapping of the line segments is allowed.


Problem 8: (1985 AMC) How many distinguishable rearrangements of the letters in CONTEST have both the vowels first? (For instance, OETCNST is one such arrangement, but OTETSNC is not.)

(A) 60 (B) 120 (C) 240 (D) 720 (E) 2520 Problem 9: In how many ways can three married couples be seated in a row if no husband and wife are to be in adjacent seats? Problem 10: A baking company produces five different cookies: Chocolate Chip Cookies, Peanut Butter Cookies, Oatmeal Cookies, Sugar Cookies, and Blueberry Cookies. If a package contains 8 cookies with at least one each kind, how many different packages are possible? Problem 12: In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats? Problem 13: In how many ways can a family of six people be seated at a round table if the youngest kid must sit between the parents? Problem 14: A gardener plants three maple trees, four oak trees, and five birch trees in a row. How many ways are there such that no two birch trees are next to one another? Problem 15: Seven identical chairs in a row are to be seated by four students. How many arrangements are there such that the only two of the three empty chairs are next to each other? Problem 16: As shown in the figure, each of six regions ABCDEF is to be assigned a color. There are 4 colors to choose from, and no adjacent regions can be the same color. How many different ways are there if each color is allowed to use more than once?

32


Problem 17: There are five regions that need to be colored by six different colors as shown in the figure. Each region can only be colored with one color. How many different ways to do the coloring?

Problem 18: (2011 AMC 10 A) Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

Problem 19: (2007 AMC 12B ) Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

(A) 15 (B) 18 (C) 27 (D) 54 (E) 81

Problem 20: Nine squares of a 3 × 3 board are painted using three colors: black, red and yellow with the following restrictions: (1) each square is pained with one color, (2) each color is used exactly three times, (3) each column is painted with three colors, and each row is colored with three colors. How many ways can this be done?

Problem 21. How many ways to color the 4 regions using 3 different colors, if no two neighboring regions can have the same color?

33


Problem 22. Using red, yellow, blue, and black colors to color the figure as shown. Each region is colored with one color and no adjacent region can have the same color. If each color is allowed to use more than once, how many ways are there to color? (A) 48 (B) 36 (C) 30 (D) 24 Problem 23. What is the size of the largest subset, S, of {1, 2,…, 2013} such that no pair of distinct elements of S has a sum divisible by 3? Problem 24. Line a is parallel to line b. There are 10 points on line a and 9 points on line b. At most how many points of intersection of line segments obtained by connect all these points on line a to all the points on b? Problem 25. In how many ways can two squares be selected from an 8-by-8 chessboard so that they are not in the same row or the same column? Problem 26: Alex wants to select three different numbers from {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. How many ways are there such that no two numbers are consecutive? Problem 27: Alex wants to select four different numbers from {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. How many ways are there such that no two numbers are consecutive? Problem 28. How many ways are there to take two different numbers from 1, 2,…, 32 such that the sum of them is divisible by 4?

Problem 29. At least how many numbers need to be removed from the list of 1, 2, 3,…, 2013 such that among the remaining numbers, any number is not a product of other two numbers?

34


35

Problem 30: A subset of integers 1, 2,…, 100 has the property that the sum of its two members is always divisible by 10. What is the largest number of members such a subset can have?

Problem 31: Taking 7 numbers from 1 to 12 such that none of the numbers taking is two times of another. How many ways to do this? Problem 33. How many ways are there to take three different numbers from 1, 2,…, 30 such that the sum of them is divisible by 3? Problem 34. What is the size of the largest subset, S, of {1, 2,…, 2013} such that no pair of distinct elements of S has a sum divisible by 3?


SOLUTIONS:

Problem 1: Solution: (E). The first 24 = 4! words begin with A, the next 24 begin with E and the next 24 begin with H. So the 86th begins with M, and it is the 86 – 72 = 14th such word. The first six words that begin with M begin with MA and the nest six with ME. So the desired word begins with MH and it is the second such word. The first word that begins with MH is MHAES, the second is MHASE. Thus E is the letter we seek. Problem 2: Solution: No matter Alex shakes hand with Bob or Bob shakes hand with Alex, it only counts one time. So order does not matter and we use combinations to solve this problem.

We have 26 people so at most we have hand shakes. Among them, there are ⎜⎜⎛

hand shakes between women, and 13 hand shakes between spouses. Therefore the final

answer is ⎜⎜ − − 13 = 234.

⎟⎟⎠

⎞⎜⎜⎝

⎛226

⎟⎟⎠

⎞213

⎝

⎟⎟⎠

⎞

⎝

⎛226

⎟⎟⎠

⎞⎜⎜⎝

⎛213

Problem 3: Solution: (B). We deal with special situation first. We have 3 ways to put cards 1 and 2 into the same

envelope. After that, we have ways to select two cards from the four cards left

and put them into one envelope. The two left will go to the last envelope.

624

=⎟⎟⎠

⎞⎜⎜⎝

⎛

By the product rule, we have 3 × 6 = 18 ways. Problem 4: Solution: Case I: Alex can select 1 social studies course and 2 science courses. By the product rule,

1824

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

Case II: Alex can select 2 social studies courses and 1 science course. By the product

rule, . 1214

23

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

By the sum rule, we have 18 + 12 = 30 ways. Method 2:

36


3 3 37 3 4 30C C C− − = .

Problem 5: Solution: 109. The job is to count palindrome numbers. We have the following three cases. In any case, we finish the job of counting palindrome numbers. Case I: E1: to count one-digit palindrome numbers. There are 10 one-digit palindromes: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We finished E1 in n1= 9 ways. Case II: E2 to count two-digit palindrome numbers, and There are 9 two-digit palindromes: 11, 22, 33, 44, 55, 66, 77, 88, 99. We finished E2 in n2 = 9 ways. Case III: E3 to count three-digit palindrome numbers. There are 90 3-digit palindromes: 101, 111, 121, ...999. We finished E3 in n3= 90 ways. N = n1+ n2 + n3 = 10 + 9 + 90 = 109. Problem 6: Solution: The job is to form even numbers. We have the following two cases. In any case, we finish the job of forming even numbers.

Case I: When 0 is in the ones position, we have four digits left. By )!(

!),(rn

nrnP−

= , we

have 12)!24(

!4)2,4( =−

=P such numbers.

Case II: When 0 is not in the ones position. By )!(

!),(rn

nrnP−

= , we have ways

to select the units digit (only digits 2 and 4 available), ways to select the hundred digit (0 is excluded, one of the two digits 2 or 4 is also excluded since one of them has been used already as the units digit), ways for the middle digit. By the production rule, we get:

)1,2(P

)1,3(P

)1,3(P18)1,3()1,3()1,2( =×× PPP such numbers.

By the sum rule, we get the final answer 12 + 18 = 30.

37


Problem 7: Solution: 9.

We see that 124454)9321( <=÷++++ L . So the length of the side of the square is

less than 12. Case I: The side length is 11. We have one way: 9 + 2 = 8 + 3 = 7 + 4 = 6 + 5 Case II: The side length is 10. We have one way: 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 Case III: The side length is 9. We have 5 ways: 9 = 8 + 1 = 7 + 2 = 6 + 3 = 5 + 4 Case IV: The side length is 8. We have 1 way: 8 = 7 + 1 = 6 + 2 = 5 + 3 Case V: The side length is 7. We have 1 way: 7 = 6 + 1 = 5 + 2 = 4 + 3 Case VI: The side length is 6 or less. We have no way: Total we have 1 + 1 + 5 + 1 + 1 = 9 different ways. Problem 8: Solution: (B). There are two ways to order the vowels. By (3.1), we have the following ways to rearrange the rest of the letters C, N, T, T, S:

60!1!1!1!2

!5!!!

!

21

=×××

=⋅⋅⋅

=rnnn

nN .

By the production rule, we have the answer: 2 × 60 = 120. Problem 9: Solution: 240. Let three couples be (A, A), (B, B), (C, C). If A is seated first, we have the following 5 arrangements:

38


If we switch the genders for A, B, and C, we have the factors 2× 2 ×2. If A is followed by C, we get another factor of 2. The number of arrangements is 5 × 2 × 2 × 2 × 2 = 80. Similarly, B or C can also be seated first, then the final answer will be 3 × 80 = 240. Problem 10: Solution: By (3.5), we have to x1 + x2 + ⋅⋅⋅ + xr = n ⇒ to x1 + x2 + x3 + x4 + x5 = 8


1518

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=rn

N

Problem 11: Solution: 174. We focus on 2x1 since this term is special. We know that x1 ≥ 0. So we have two cases: Case I: x1 = 0. The given equation becomes: x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 3. The number of nonnegative integer solutions is given by (3.6):

1653

193

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

Case II: x1 = 1. The given equation becomes: x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 1. The number of nonnegative integer solutions is given by (3.6):

91

191

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

We do not have case III since x1 is less than 2. The answer is 165 + 9 = 174.

39


Problem 12: Solution: 144. We seat four women first. There are (4 – 1)! = 3! ways to sit them. After the ladies are seated, we have 4! ways to seat four men in the small rectangles as shown in the figure below. 4! × 3! = 144.

Problem 13: Solution: 12. We link two parents and the youngest kid together to form a unit. There are (4 – 1)! ways to seat them at the table. The result must be multiplied by 2 since we can switch the positions of the two parents. The solution is (4 – 1)! × 2 = 12. Problem 14: Solution: We plant the non-birch trees first. There are 7! ways to do so. There are eight spaces for

three birch trees to choose and there are 5! ways for them to rearrange themselves. ⎟⎟⎠

⎞⎜⎜⎝

⎛58

By the production rule, the answer is 7! × × 5! = 33868800. ⎟⎟⎠

⎞⎜⎜⎝

⎛58

Problem 15: Solution: 480. We tie the two empty chairs next to each other together and treat them as on unit, namely A. the other empty chair is B.

We have 4! = 24 ways to sit four students. There are ways to insert A and B.

and we multiply the result by 2 because A and B can switch their positions.

1025

=⎟⎟⎠

⎞⎜⎜⎝

⎛

40


By the product rule, the answer will be 24 × 10 × 2 = 480. Problem 16: Solution: Case I: A, C, and E have the same color. The number of ways to color: 4 × 3 × 3 × 3 = 108 Case II: A, C, and E have two different colors. The number of ways to color: 3 × 4 × 3 × 3 × 2 × 2 = 432 . Case III: A, C, and E have the three different colors. The number of ways to color: P(4,3) × 2 × 2 × 2 = 192. Total 108 + 432 + 192 = 732 ways. Problem 17: Solution: We color region A first. We have 6 ways. Then we color region B (we have 5 ways), region C (4 ways). For region D, we have two cases: Case I: Different color from B. We have 3 ways to color D (because we have three colors left) and 3 ways to color E (either from 2 colors left or the same color as C). Case II: Same color as B. We have 1 way to color D and 4 ways to color E. The answer will be 6 × 5 × 4 × (3 × 3 + 1 × 4) = 1560 ways.

Problem 18: Solution: 3120. Method 1:

41


Case I: 5 colors are used:

720!556

=×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case II: 4 colors are used:

18005!31446

=××××⎟⎟⎠

⎞⎜⎜⎝

⎛

We consider the vertex A first. We color point A and point B with the same color. 5 means we have five vertices (5 sub cases).

Case III: 3 colors are used: ⎜⎜⎛

6005112336

=×××××⎟⎟⎠

⎞

⎝We consider the vertex A first. We color points A and B with the same color. We multiply the result by 5 since we have 5 vertices. No other cases exist. So the answer is 720 + 1800 + 600 = 3120.

Method 2: There is a formula for this problem which can be derived by the recursion method. An = (m – 1)[(m – 1)(n-1) + (–1)n]. Where n is the number of points and m is the number of colors. n = 5 and m = 6. An = (m – 1)[(m – 1)(n-1) + (–1)n] = (6 – 1)[(6 – 1)(5 – 1) + (–1)5] = 3120.


Let r, w, and b be the number of red, white, and blue faces, respectively. Case 1: One color is used. rrrr, wwww, bbbb. We have 3 ways to color.

42


Case 2: Two colors are used. rrww, rrbb, wwbb; rrrb, rrrw; wwwr, wwwb, bbbr, bbbw. We have 9 ways to color. Case 3: Three colors are used. rrwb, wwrb, bbrw. R We have 3 ways to color. Total we have 3 + 9 + 3 = 15 ways. Problem 20: Solution: We have 3 ways to color square 8. Let us say we use black color.

In the region that contains the squares 5, 9, 2, and 7, we have two ways to use the black color (color 5, 2 black or 7, 9 black). After that, we have exactly two ways to color other squares. By the product rule, we have 3 × 2 × 2 = 12 ways. Problem 21. Solution: If regions 1 and 3 are painted the same color: Regions 1 2 3 4 Ways to color 3 × 2 × 1 × 2 If regions 1 and 3 are the different color: Regions 1 2 3 4 Ways to color 3 × 1 × 2 × 1 Total ways 12 + 6 = 18. Problem 22. (D) 24 Solution: We have two cases: Case I: Region A and region B have the same color. We have 4 ways to color region A and 1 way to color region B.

43


Then we can color region C in 3 ways and region D in 2 ways. That is 4 × 3 × 2 = 24 ways. Case II: Region A and region B have different color. We have 4 ways to color region A, 3 ways for region B, 2ways for region C, and 1 way for region D.

That is 4 × 3 × 2 × 1 = 24 ways.

The answer is 24 + 24 = 48.

Problem 23. Solution: 672. Let A = {3, 6, 9, …, 2013}, B = {1, 4, 7,…, 2011}, and C = {2, 5, 8,…, 2012}. At most we can select one element from A to add to B or C to from S. Since B and C each has 671 elements, so S has at most 671 + 1 = 672 elements.

Problem 24. Solution: 1620. We need two points on line a and two points on line b to form 1 point of intersection. The maximum number of points of intersection is obtained by considering no three line segments will intersect at one point:

162029

210

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Problem 25. Solution: 1568. Consider the coordinate system (a, b), which designates that a piece is in row a, column b. In order to ensure that two pieces are in different rows and different columns, both

their coordinates must be different. There are ways to choose 2 different row

coordinates and ways to choose 2 different column coordinates. Given the row

2828

=⎟⎟⎠

⎞⎜⎜⎝

⎛

2828

=⎟⎟⎠

⎞⎜⎜⎝

⎛

44


coordinates and column coordinates, there are 2 different ways to pair them into two ordered pairs. Therefore, the answer is 2 × 28 × 28 = 1568 . Problem 26: Solution:

45

⎟⎟⎠

⎞

⎟⎟⎠

⎞

N = ⎜⎜ = .

Problem 27: Solution:

⎝

⎛ −−kkn

)1(

220312

3 )13(14

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

N= ⎜⎜ = .

⎝

⎛ −−kkn

)1(

330411

4 )14(14

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

Problem 28. Solution: 120. Let A = {4, 8, 12, 16, 20, 24, 28, 32}, B = {1, 5, 9, 13, 17, 21, 25, 29}, and C = {2, 6, 10, 14, 18, 22, 26, 30}, and D = {3, 7, 11, 15, 19, 23, 27, 31}. If we take two numbers from A, or two numbers from C, the sum of them must also be divisible by 3. If we take one number from B and one number from D, the sum of them must be divisible by 3.

So the number of ways is: . 12018

18

28

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

Problem 29. Solution: 43.

We know that 44 × 45 = 1980 and 45 × 46 = 2070. So after we remove 43 numbers (2, 3,…, 44), the remaining numbers satisfy the given condition.

In the other hand, if we only remove 42 numbers from the list, at least there is one set with all three numbers remaining.

{k, 89 – k, k(89 – k)}, k = 2, 3, …, 44.

So one number is a product of other two numbers.

Therefore at least we need to remove 43 numbers.


46


If the sum of any two members is divisible by 10, each of them must be a multiple of 10, or each has a remainder of 5 when divided by 10.

We can select two numbers in the form of 10k + 5 (k = 0, 1, 2, …, 9) or 10k (k = 1，2，…，10). But we can not take the numbers at the same time from both groups. So we at most can take 10 numbers. Problem 31: Solution: 47. First we classify these numbers by 2 times: 1, 2, 4, 8， 3, 6, 12， 5, 10， 7， 9， 11， At most we can select 2 numbers each from (1, 2, 4, 8) and (3, 6, 12). Take two numbers from (1, 2, 4, 8), we have three ways and one way from (3, 6, 12). Other 3 numbers are to be selected from the rest of the numbers. We need also pay attention to these 3 numbers that if we selected one of （5, 10). The ways to take 7 numbers are 3 × 1 ×（1 + 2 × 3）= 21； Second, we take two numbers from (1, 2, 4, 8), one number from (3, 6, 12), there are total ways: 3 × 3 × 2 = 18；

Third, we take one number from (1, 2, 4, 8), and 2 numbers from (3, 6, 12), total ways: 4 × 1 × 2 = 8.

Final solution will then be: 21 + 18 + 8 = 47. Problem 32. Solution: 110. Let A = {7, 14, 21 28} and B = {1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 29, 30}. If we take two numbers from A, or we take one number from A and one number from B, the product of them will be divisible by 7.


47

110

1030


14

24

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

Problem 33. Solution: 1030. Let A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}, B = {1, 4, 7, 10, 13, 16, 19, 22, 25, 28}, and C = {2, 5, 8, 11, 14, 17, 20, 23, 26, 29}. If we take one number from A, one number from B, and one number from C, the sum of them must be divisible by 3. If we take three numbers from A, or three numbers from B, or three numbers from C, the sum of them must also be divisible by 3.


3110

110

110

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛

Problem 34. Solution: 672. Let A = {3, 6, 9, …, 2013}, B = {1, 4, 7,…, 2011}, and C = {2, 5, 8,…, 2012}. At most we can select one element from A to add to B or C to from S. Since B and C each has 671 elements, so S has at most 671 + 1 = 672 elements.

Date post:	17-Feb-2022
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

BASIC KNOWLEDGE 1. Two Important Terms (1.1). Permutations

Documents