Binomial Identities

1
Binomial Identities
2
Expansion of (a x)n

• (a x) a x 1C0a 1C1x
• (a x)(a x) aa ax xa xx x2 2ax
  a2 2C0x2 2C1ax 2C2a2
• The 4 red terms are the formal expansion of
  (ax)2
• The 3 blue terms are the simplified expansion
  of (ax)2
• (a x)(a x)(a x) aaa aax axa axx
  xaa xax xxa xxx x3 3a2x 3ax2 a3
• 3C0x3 3C1a2x 3C2ax2 3C3a3

3
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?

4
The Coefficient on akxn-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.

5
Binomial 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.

6
Binomial 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).
• Pascals 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.

7
nCk kCm nCm n-mCk-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

8
Pascals Triangle

• kth number in row n is nCk

k 0
1
n 0
k 1
k 2
1
1
n 1
k 3
1
2
1
n 2
k 4
1
3
3
1
n 3
n 4
1
4
6
4
1
9
Displaying Pascals Identity
k 0
0C0
n 0
k 1
k 2
1C0
1C1


n 1
k 3
2C0
2C1
2C2
n 2
k 4
3C0
3C1
3C2
3C3
n 3
4C0
4C1
4C2
4C3
4C4
n 4
10
Block-walking Interpretation
nCk strings of n Ls Rs with k Rs.
nCk ways to get to corner n,k starting from
0,0
k 0
0C0
n 0
k 1
k 2
1C0
1C1


n 1
k 3
2C0
2C1
2C2
n 2
k 4
3C0
3C1
3C2
3C3
n 3
4C0
4C1
4C2
4C3
4C4
n 4
11
Pascals Identity via Block-walking
routes to corner n,k routes thru corner
n-1,k routes thru corner n-1,k-1
k 0
0C0
n 0
k 1
k 2
1C0
1C1


n 1
k 3
2C0
2C1
2C2
n 2
k 4
3C0
3C1
3C2
3C3
n 3
4C0
4C1
4C2
4C3
4C4
n 4
12
nC0 nC1 nC2 . . . nCn 2n

• 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)

13
rCr r1Cr r2Cr . . . nCr n1Cr1
k 0
0C0
n 0
k 1
k 2
1C0
1C1


n 1
k 3
2C0
2C1
2C2
n 2
k 4
3C0
3C1
3C2
3C3
n 3
4C0
4C1
4C2
4C3
4C4
n 4
14
rCr r1Cr r2Cr . . . nCr n1Cr1

• 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.

15
nC02 nC12 nC22 nCn2 2nCn
k 0
0C0
n 0
k 1
k 2
1C0
1C1


n 1
k 3
2C0
2C1
2C2
n 2
k 4
3C0
3C1
3C2
3C3
n 3
4C0
4C1
4C2
4C3
4C4
n 4
16
nC02 nC12 nC22 nCn2 2nCn

• 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

17
1?2?3 2?3?4 3?4?5 (n-2)?(n-2)?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! ? n1C4

18
A 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

19

• 12 22 32 . . . n2 ?
• General term
• k2
• k(k-1) k
• P(k, 2) k
• 2! kC2 k

20

• Sum
• Sk1,n (2! kC2 k )
• 2! Sk1,n kC2 Sk1,n k
• 2! n1C3 n1C2

21
Another Strategy Manipulate the Binomial 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