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-1)(n-1)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