# Day 10 Binomial coefficient identities .pdf

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Combinatorial Analysis

Binomial coefficients identities

The binomial coefficient nk counts the number of k-subsets of a set of

n elements. These numbers have many fascinating properties and satisfy a

number of interesting identities. Recall that

n

n

=

.

k

n−k

Next we prove some more identities. Use combinatorial proofs.

Exercise 1 Show that, for all n and k positive integers such that 0 ≤ k ≤ n

(except n = k = 0)

n

n−1

n−1

=

+

.

k

k

k−1

This identity is called Pascal’s formula.

Exercise 2 For n ≥ 1,

X n

k≥0

2k

= 2n−1 .

Exercise 3 Show that, for all n and k positive integers such that 0 ≤ k ≤ n,

n

n−1

k

=n

.

k

k−1

Exercise 4 For n ≥ 1,

n

X

n

k

= n2n−1 .

k

k=0

Exercise 5 Dividing both sides of the last identity by 2n allows us to give a

different combinatorial proof of the equivalent identity:

Pn

n

n

k=0 k k

= .

n

2

2

The next identity is called Vandermonde’s identity.

Exercise 6 For m ≥ 0, n ≥ 0,

X

k

m+n

m

n

=

.

k

j

k−j

j=0

The next identity has an interesting application to number theory.

Exercise 7 For 0 ≤ m ≤ k ≤ n,

n

k

n

n−m

=

.

k

m

m k−m

As a simple consequence of this last identity, Erdos and Szekeres proved

the following simple fact about binomial coefficients. It seems that this was

not known prior to 1978.

n

Exercise 8 For 0 < m ≤k < n, m

and nk have a nontrivial common

n

factor, that is, gcd m

, nk > 1.

Next we examine identities involving the quantity nk , spoken “n multichoose k”, which counts the ways to select k objects from a set of n elements,

where order is not important, but repetition is allowed.

Exercise 9 nk = n+k−1

.

k

Exercise 10 For 0 ≤ n ≤ m,

n

m−n

m−1

=

.

n−1

Not surprisingly, there are many multichoose identities that resemble earlier

binomial identities. We begin with the Pascal-like identity.

Exercise 11 For n ≥ 0, k ≥ 0 (except n = k = 0),

n

n

n−1

=

+

.

k

k−1

k

Exercise 12

n

n+1

k

=n

.

k

k−1

Exercise 13 For k ≥ 1,

X

n

n

m

=

.

k

k

−

1

m=1

Although there is no closed form for

Pm

k=0

n

k

, m 6= n, we have

Exercise 14 For n ≥ 0,

m

X

n

k=0

k

=

2

n+1

m

.

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