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Basics of Counting Techniques

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Before learning the basics of Probability Theory, one must have a good command of Counting Techniques
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COUNTING TECHNIQUES Efren A. Medallo
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Page 1: Basics of Counting Techniques

COUNTING TECHNIQUES

Efren A. Medallo

Page 2: Basics of Counting Techniques

Why learn counting?

Counting techniques are the very bases of being able to find the different probabilities of events in any kind of situation.

Page 3: Basics of Counting Techniques

Why learn counting?

This is not counting one-to-one but this is collectively counting all possible ways of a given instance. (Ex. Counting all 4-digit numbers whose digits are different among one another)

Page 4: Basics of Counting Techniques

COUNTING TECHNIQUES

FCP Factorial Notation Permutation Combination

Page 5: Basics of Counting Techniques

Fundamental Counting Principle

PROBLEM Liza brought 3 different pairs of pants and 4 shirts

in a camp. How many combinations of a shirt and a pair of pants can she choose from to wear?

Page 6: Basics of Counting Techniques

Fundamental Counting Principle

Let’s name the shirts of Liza as Shirts A, B, C, and D. Let’s also name her pants as Pants 1, 2, and 3.

Let us now enumerate the possible combinations that Liza can wear.

Page 7: Basics of Counting Techniques

Fundamental Counting Principle

The answer is there are 12 possible combinations.

We found the answer by drawing a diagram. But what if there are large numbers given in the problem (say, 179 pairs of pants and 83 shirts)?

Page 8: Basics of Counting Techniques

Fundamental Counting Principle

If an independent event can occur in m ways, another independent event can occur in n ways, and another in p ways, then the total number of ways that all events can occur simultaneously is

n(E) = m ∙n∙p ways

Page 9: Basics of Counting Techniques

Fundamental Counting Principle

In the given problem,

n(E) = 4 x 3 = 12 ways

Page 10: Basics of Counting Techniques

Drill

How many 4 digit numbers are there that have no single digit repeated in it?

Page 11: Basics of Counting Techniques

Solution

Let us analyze.

How many numbers (from 0-9) can fit in the first digit?

Page 12: Basics of Counting Techniques

Solution

There are only 9. This is because 0 cannot be the first digit. If it were, then it would not be a 4-digit number anymore.

Page 13: Basics of Counting Techniques

Solution

How about the second digit?

Here, zero can already be included, which makes 10 possibilities.

HOWEVER, in the problem, it is stated that digits must not be repeated.

Page 14: Basics of Counting Techniques

Solution

Therefore, whatever was used on the first digit cannot be used on the second digit anymore, which gives us 9 possibilities.

Page 15: Basics of Counting Techniques

Solution

How about the third digit?

Again here, we have ten possibilities, but we have to put into account what was used in the previous digits, so that makes 8.

Page 16: Basics of Counting Techniques

Solution

In the last digit, since 3 digits have been used already, it is safe to say that there are only 7 possibilities.

Page 17: Basics of Counting Techniques

Solution

All in all, we have

Digit 1st 2nd 3rd 4th

9 x 9 x 8 x 7

Answer: 4536 numbers

Page 18: Basics of Counting Techniques

Tip

For every independent event, count carefully the number of possibilities before multiplying them.

Page 19: Basics of Counting Techniques

Drill

How many three letter combinations can you make with the following conditions (separate)? a. Repetition is allowed b. Repetition is not allowed c. The first letter must be a vowel with no repetition

of letters d. There must be no vowel in the first two letters,

repetition is not allowed and the last letter must be a vowel.

Page 20: Basics of Counting Techniques

Factorial Notation

The factorial of an integer k is the product of all integers from 1 to k.

This is usually denoted as k! read as “k factorial”

Page 21: Basics of Counting Techniques

Factorial Notation

Examples: 7! = 7∙6∙5∙4∙3∙2∙1 = 5040 5! = 5∙4∙3∙2∙1 = 120

Page 22: Basics of Counting Techniques

Properties of Factorial Notation

The factorial operation is NOT distributive. Ex. (5-3)! ≠ 5! – 3!

The factorial operation cannot be performed on non-integer numbers. Ex. (2.5)!, (√6)!

Page 23: Basics of Counting Techniques

Properties of Factorial Notation

The factorial notation precedes multiplication and/or division. Ex. 5!3! ≠ (5∙3)! 8! / 2! ≠ (8/2)!

0! = 1

Page 24: Basics of Counting Techniques

Properties of Factorial Notation

n! = n(n-1)!= n(n-1)(n-2)!= n(n-1)(n-2)(n-3)…(n-k+1)(n-k)!

This property is very crucial in making cancellations in factorial expressions.

Page 25: Basics of Counting Techniques

Properties of Factorial Notation

Example

= 4896

Page 26: Basics of Counting Techniques

Permutation

A permutation is an ordered arrangement of objects.

It tells us how many possible orders there can be given a number of objects.

In permutation, if all objects are distinct, then they cannot be repeated.

Page 27: Basics of Counting Techniques

Permutation

Example How many ways can three books be arranged in a

shelf?

Possible ArrangementsABC BCA

ACB CAB

BAC CBA

Page 28: Basics of Counting Techniques

Permutation

Take Note: Since order is essential, ABC is different from

ACB, and all other similar instances.

Page 29: Basics of Counting Techniques

Permutation

Permutation is usually denoted by

nPr

read as “the permutation of n objects taken r at a time”

Page 30: Basics of Counting Techniques

Permutation

The general formula for permutation is

nPr = n(n-1)(n-2)…(n-r+1)

or

Page 31: Basics of Counting Techniques

Permutation

n is the total number of objects r is the number of objects in consideration

In the problem given, r=n

If r=n, the equation becomes

Page 32: Basics of Counting Techniques

Drill

How many ways can you choose your top 2 senators from a list of 35 candidates?

Page 33: Basics of Counting Techniques

Drill

Remember Liza and her clothes? Find the number of ways she can wear her clothes if the camp lasts for three days and she can only wear each garment once.

Page 34: Basics of Counting Techniques

Permutation with Repetitions

If we were asked how many permutations we can get if we rearrange the letters of the word “cat”, we get 3!

What if we rearrange the letters of the word “dad”?

Page 35: Basics of Counting Techniques

Permutation with Repetitions

Permutations of the word “dad” dad dda add

Page 36: Basics of Counting Techniques

Permutation with Repetitions

Actually, there are still 3! permutations for the word “dad”. However, since we cannot distinguish the first ‘d’ from the second, then “dad” with the first ‘d’ first will be the same as “dad” with the second ‘d’ first.

This leaves us with 3 distinguishable permutations.

Page 37: Basics of Counting Techniques

Permutation with Repetitions

The formula for finding the distinguishable permutations from n set of objects is

where a, b, and c are the number of times a particular object exists on n.

Page 38: Basics of Counting Techniques

Drill

Example Find the number of distinguishable permutations of

the word “technicalities”

Page 39: Basics of Counting Techniques

Circular Permutation

How many ways can 4 people line up?

The answer is 4P4 or 4! = 24 ways

Let us enumerate them.

Page 40: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Page 41: Basics of Counting Techniques

Circular Permutation

Now, how many ways can these 4 people sit on a round table?

Page 42: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 43: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 44: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 45: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 46: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 47: Basics of Counting Techniques

Circular Permutation

ABCD BACD CABD DABCABDC BADC CADB DACBACBD BCAD CBAD DBACACDB BCDA CBDA DBCAADBC BDAC CDAB DCABADCB BDCA CDBA DCBA

Look closely at the bold italicized letters. If theyare arranged in a circular manner, they allresemble the same pattern.

Page 48: Basics of Counting Techniques

Circular Permutation

The number of ways n objects can be arranged in a circular manner is given by

Page 49: Basics of Counting Techniques

Circular Permutation

In the given problem,

Page 50: Basics of Counting Techniques

Combination

A combination is an arrangement of objects with no respect to order.

Example. Find the possible permutations and combinations

of the word “cat”

Page 51: Basics of Counting Techniques

Combination

For permutations:CAT CTAACT ATCTAC TCA

There are 6 permutations for the word “cat”.

Page 52: Basics of Counting Techniques

Combination

For combinations,

Since order is regardless, thenCAT = CTA = ACT = ATC = TAC = TCA

which gives us only 1 combination.

Page 53: Basics of Counting Techniques

Combination

What if we take only two letters from the word “cat”? How many permutations and combinations are there?

Page 54: Basics of Counting Techniques

Combination

For permutations, According to the formula, there are 3P2 = 6

permutations, which are

AC AT

CA CT

TA TC

Page 55: Basics of Counting Techniques

Combination

For combinations,

Again, order is not essential, so AC=CA, TC=CT, and AT=TA.

This gives us only 3 combinations.

Page 56: Basics of Counting Techniques

Combination

The general formula for a combination is

and it is read “the combination of n objects taken r at a time”

Page 57: Basics of Counting Techniques

Drill

Find how many possible combinations there are of 5 cards when randomly selected from a standard deck of 52 cards.

Page 58: Basics of Counting Techniques

Solution

Since in getting 5 cards, order is not essential (i.e getting K♦ 3♠ K♣ 3♦ K♥ is the same as getting K♥ K♦ 3♠ 3♦ K♣, and all other orders), then we use combination.

Page 59: Basics of Counting Techniques

Drill

Find the number of ways of getting just a pair when randomly getting three cards from a standard deck.

Page 60: Basics of Counting Techniques

Drill

Find the number of ways of getting at least a pair when randomly getting three cards from a standard deck.

Page 61: Basics of Counting Techniques

Drill

In a party, there are 53 guests who shook hands with one another exactly once. How many handshakes took place?


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