Binomial IdentitiesBinomial Identities
Expansion of (a + x)Expansion of (a + x)nn
• (a + x) = a + x = 1C0a + 1C1x
• (a + x)(a + x) = aa + ax + xa + xx = x2 + 2ax + a2 =
2C0x2 + 2C1ax + 2C2a
2
• The 4 red terms are the “formal” expansion of (a+x)2
• The 3 blue terms are the “simplified” expansion of (a+x)2
• (a + x)(a + x)(a + x) = aaa + aax + axa + axx + xaa + xax + xxa + xxx = x3 + 3a2x + 3ax2 + a3 =
3C0x3 + 3C1a2x + 3C2ax2 + 3C3a3
Generalizing . . .Generalizing . . .
• In (a + x)4, how many terms does the:
• formal expansion have?
• simplified expansion have?
• In (a + x)n, how many terms does the:
• formal expansion have?
• simplified expansion have?
The Coefficient on aThe Coefficient on akkxxn-kn-k
• The coefficient on akxn-k is the number of
terms in the formal expansion that have
exactly k as (and hence exactly n-k xs).
• It is equal to the number of ways of
choosing an a from exactly k of the n
binomial factors: nCk.
Binomial TheoremBinomial Theorem
• (1 + x)n = nC0x0 + nC1x1 + nC2x2 + . . . nCnxn
• In addition to the combinatorial argument
that the coefficient of xi is nCi, we can prove
this theorem by induction on n.
Binomial IdentitiesBinomial Identities
• nCk = n!/[k!(n-k)!] = nCn-k
The number of ways to pick k objects from n = the ways to pick not pick k (i.e., to pick n-k).
• Pascal’s identity: nCk = n-1Ck + n-1Ck-1
• The number of ways to pick k objects from n can be partitioned into 2 parts:
• Those that exclude a particular object i: n-1Ck
• Those that include object i: n-1Ck-1
• Give an algebraic proof of this identity.
nnCCkk kkCCmm = = nnCCmm n-mn-mCCk-mk-m
• Each side of the equation counts the number of k-subsets with an m-subsubset.
• The LHS counts:1. Pick k objects from n: nCk
2. Pick m special objects from the k: kCm
• The RHS counts:1. Pick m special objects that will be part of the k-
subset: nCm
2. Pick the k-m non-special objects: n-mCk-m
Pascal’s TrianglePascal’s Triangle
• kth number in row n is nCk:
1
1 1
1 2 1
1 3 3 1
n = 4
n = 3
n = 2
n = 1
n = 0
1 4 6 4 1
k = 0
k = 1
k = 2
k = 3
k = 4
Displaying Pascal’s IdentityDisplaying Pascal’s Identity
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
Block-walking InterpretationBlock-walking Interpretation
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
nCk = # ways toget to corner n,kstarting from 0,0
nCk = # strings of n Ls & Rs with k Rs.
Pascal’s Identity via Block-Pascal’s Identity via Block-walkingwalking
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
# routes to corner n,k = # routes thru corner n-1,k+ # routes thru corner n-1,k-1
nnCC00 + + nnCC11 + + nnCC22 + . . . + + . . . + nnCCnn = 2 = 2nn
• LHS counts # subsets of n elements using the sum rule: partitioning the subsets according to their size (k value).
• RHS counts # subsets of n elements using the product rule: • Is element 1 in subset? (2 choices)• Is element 2 in subset? (2 choices) … • Is element n in subset? (2 choices)
rrCCrr + + r+1r+1CCrr + + r+2r+2CCrr + . . . + + . . . + nnCCrr = = n+1n+1CCr+1r+1
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
rrCCrr + + r+1r+1CCrr + + r+2r+2CCrr + . . . + + . . . + nnCCrr = = n+1n+1CCr+1r+1
• RHS = routes to corner 4,2• LHS: Partition the routes to 4,2 into those:
• whose last right branch is at corner 1,1: 1C1
• whose last right branch is at corner 2,1: 2C1
• whose last right branch is at corner 3,1: 3C1
• For each subset of routes, there is only 1 way to complete the route from that corner to 4,2: RLL, RL, & R respectively.
• The identity generalizes this argument.
nnCC002 2 + + nnCC11
2 2 + + nnCC222 2 + … + + … + nnCCnn
2 2 = = 2n2nCCnn
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
nnCC002 2 + + nnCC11
2 2 + + nnCC222 2 + … + + … + nnCCnn
2 2 = = 2n2nCCnn
• RHS = all routes to corner 4,2
• LHS partitions routes to 4,2 into those that:
• go thru corner 2,0: 2C0 2C2
• go thru corner 2,1: 2C1 2C1
• go thru corner 2,2: 2C2 2C0
• The identity generalizes this argument:
• # routes to 2n, n that go thru n,k = nCk nCn-k
• Sum over k = 0 to n
11223 + 23 + 2334 + 34 + 3445 +…+ (n-2)5 +…+ (n-2)(n-2)(n-2)n = ?n = ?
• The general term = (k-2)(k-1)k
= P(k,3)
= k!/(k-3)!
= 3! kC3
• Sum = 3!3C3 + 3!4C3 + 3!5C3 +...+ 3!nC3
= 3! [3C3 + 4C3 + 5C3 +...+ nC3 ]
= 3! n+1C4
A StrategyA Strategy
• When the general term of a sum is not a
binomial coefficient:
• break it into a sum of P(n, k) terms, if possible;
• rewrite these terms using binomial coefficients
• 12 + 22 + 32 +. . . + n2 = ?
• General term:
= k2
= k(k-1) + k
= P(k, 2) + k
= 2! kC2 + k
• Sum
Σk=1,n (2! kC2 + k )
= 2! Σk=1,n kC2 + Σk=1,n k
= 2! n+1C3 + n+1C2
Another Strategy: Manipulate the Another Strategy: Manipulate the Binomial TheoremBinomial Theorem
• (1 + 1)n = 2n = nC0 + nC1 + . . . + nCn
• (1 - 1)n = 0 = nC0 - nC1 + nC2 - . . . +(-1)n nCn
or
nC0 + nC2 . . . = nC1 + nC3 + . . . = 2n-1
• Differentiate the Binomial theorem,
• n(1 + x)n-1 = 1nC1x0 + 2nC2x1 + 3nC3x2 + … + nnCnxn-1
• n(1 + 1)n-1 = 1nC1 + 2nC2
+ 3nC3 + … + nnCn