Combinatorics Derivations
The definition of the combination function is:
C(n, r) = n! / (r! *
(n – r)!)
Today I am going to mathematically verify three equivalents
in combinatorics.
Newton’s Identity
C(n,r) * C(r,k) = C(n,k) * C(n – k, r – k)
Here I am going to assume that n > r > k.
Hence:
C(n, r) * C(r, k)
= n! / (r! * (n – r)!) * r! / (k! * (r – k)!)
= n! / (n – r)! * 1 /
(k! * (r – k)!)
Rearrange:
= n! / k! * 1 / ((n – r)! * (r – k)!)
Multiply by (n – k)!/(n- k)!:
= n! / (k! * (n – k)!) * (n – k)! / ((n – r)! * (r – k)!)
Observe that (n – k) – (r – k) = n – k – r + k = n – r. Hence,
= C(n,k) * C(n – k, r – k)
QED
Pascal’s Identity
C(n,r) = C(n-1, r) + C(n – 1, r – 1)
I’m going to start with C(n-1, r) + C(n – 1, r – 1)
C(n-1, r) + C(n – 1, r – 1)
= (n – 1)! / (r! * (n – 1 – r)!) + (n – 1)! / ((r – 1)! * (n
– r)!)
= (n – 1)! / (r! * (n – 1- r)!) + (r * (n – 1)!) / (r! *(n –
r)!)
= ((n – 1)! * (n – r)) / (r! * (n – r)!) + (r * (n – 1)!) /
(r! * (n – r)!)
= (n * (n – 1)! – r * (n – 1)! + r * (n – 1)!) / (r! * (n –
r)!))
= (n * (n - 1)!) /
(r! * (n – r)!)
= n! / (r! * (n – r)!)
= C(n, r) QED
Combinatorial Proof
C(m + n, 2) – C(m, 2) – C(n, 2) = m * n
C(m + n, 2) – C(m, 2) – C(n, 2)
Note that 2! = 2
= (m + n)! / (2 * (m + n – 2)!) – m! / (2 * (m – 2)!) – n! /
(2 * (n – 2)!)
= ( (m + n)! * (m – 2)! * (n – 2)! – m! * (m + n – 2)! * (n –
2)! - n! * (m – 2)! * (m + n – 2)!) / (2
* (m + n – 2)! * (m – 2)! * (n – 2)!)
= ( (m + n) * (m + n – 1) *(m + n – 2)! * (m – 2)! * (n – 2)!
– m * (m – 1) *(m – 2)! * (n – 2)! * (m + n – 2)! – n * (n – 1) * (n – 2)! * (m
– 2)! * (m + n – 2)! )
/ (2 * (m + n – 2)! *
(m – 2)! * (n – 2)!)
= ( (m + n)*(m + n – 1) – m * (m – 1) – n * (n – 1) ) / 2
= ( m^2 + m*n – m + m*n + n^2 – n – m^2 + m – n^2 + n) / 2
= (2 * m * n) / 2
= m * n QED
Eddie
Source where I got the identities from:
V.K. Balakrishnan Schaum’s Theory and Problems:
Combinatorics including concepts of Graph Theory McGraw-Hill, Inc. New York: 1995 ISBN 0-07-003575-X
The derivation and proof details are my work.
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