Question 1046042
Use a combinatorial approach.

Suppose a group of N people consists of two groups, with group A having {{{N[1]}}} people and group B having {{{N[2]}}} people, and that {{{N= N[1] + N[2]}}} .

The number of ways of selecting 0 people from group A and selecting k people from group B is {{{(matrix(2,1, N[1],0))*(matrix(2,1, N[2],k))}}}.

The number of ways of selecting 1 person from group A and selecting k-1 people from group B is {{{(matrix(2,1, N[1],1))*(matrix(2,1, N[2],k-1))}}}.

The number of ways of selecting 2 people from group A and selecting k-2 people from group B is {{{(matrix(2,1, N[1],2))*(matrix(2,1, N[2],k-2))}}}.

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 {{{(matrix(2,1, N[1],k))*(matrix(2,1, N[2],0))}}}.

The total number of ways of selecting k people from N people, based on membership on group A or B, is then

{{{(matrix(2,1, N[1],0))*(matrix(2,1, N[2],k))}}} + {{{(matrix(2,1, N[1],1))*(matrix(2,1, N[2],k-1))}}} + {{{(matrix(2,1, N[1],2))*(matrix(2,1, N[2],k-2))}}}+...+{{{(matrix(2,1, N[1],k))*(matrix(2,1, N[2],0))}}}

But the total number of ways of selecting k people from N people, without any restriction (or regardless of membership), is {{{(matrix(2,1, N,k))}}}.

Therefore,

{{{(matrix(2,1, N,k))}}} = {{{(matrix(2,1, N[1],0))*(matrix(2,1, N[2],k))}}} + {{{(matrix(2,1, N[1],1))*(matrix(2,1, N[2],k-1))}}} + {{{(matrix(2,1, N[1],2))*(matrix(2,1, N[2],k-2))}}}+...+{{{(matrix(2,1, N[1],k))*(matrix(2,1, N[2],0))}}}.