Prove:
 n  n 
   

k n  k 
[n choose k is equal to n choose n minus k]
Proof:
1. Combinatorial Proof:
One must prove that in a set of n elements, there are the same number of combinations
(subsets whose order is irrelevant) of the k and n – k elements. Thus, one constructs a set A
of n elements and must prove that for each subset of k elements, there is exactly one
corresponding subset of n – k elements.
A = {a1, a2, a3, a4, …, an-1, an}
If k elements are chosen out of A, there are n – k elements of A that are not chosen.
Thus, every subset of k elements of A corresponds to exactly one unordered set of n – k
elements. Therefore, n choose k has the same number of combinations as n choose n minus
k, and thus the two quantities are equal.
2. Algebraic Proof:
1. By definition of n choose k,
 n
n!
  
 k  k!( n  k )! .
2. Since
k  n  (n  k ) ,
 n
n!
  
 k  ( n  (n  k ))!(n  k )! .
3. By the commutative property of multiplication,
 n
n!
  
 k  (n  k )!(n  ( n  k )! .
4. Substituting n-k for k in the definition of n choose k leads to
 n  n 
   

k n  k  .