SOLUTION: Let {{{N = N[1] + N[2]}}} , and {{{1 <= k <= N[1]}}} and {{{1 <= k <= N[2]}}}.
Prove that
{{{C(N,k) = C(N[1],k)*C(N[2],0) + C(N[1],k-1)*C(N[2],1) + C(N[1],k-2)*C(N[2],2)}}}+.
Algebra ->
Permutations
-> SOLUTION: Let {{{N = N[1] + N[2]}}} , and {{{1 <= k <= N[1]}}} and {{{1 <= k <= N[2]}}}.
Prove that
{{{C(N,k) = C(N[1],k)*C(N[2],0) + C(N[1],k-1)*C(N[2],1) + C(N[1],k-2)*C(N[2],2)}}}+.
Log On
You can put this solution on YOUR website! Use a combinatorial approach.
Suppose a group of N people consists of two groups, with group A having people and group B having people, and that .
The number of ways of selecting 0 people from group A and selecting k people from group B is .
The number of ways of selecting 1 person from group A and selecting k-1 people from group B is .
The number of ways of selecting 2 people from group A and selecting k-2 people from group B is .
This goes on until we come to the number of ways of selecting k people from group A and selecting 0 people from group B which is .
The total number of ways of selecting k people from N people, based on membership on group A or B, is then
+ + +...+
But the total number of ways of selecting k people from N people, without any restriction (or regardless of membership), is .
Therefore,
= + + +...+.
