Question 1073142
{{{(matrix(2,1,n,k-1))}}}{{{"="}}}{{{n!/((k-1)!(n-(k-1))!)}}}{{{"="}}}{{{n!/((k-1)!(n-k+1)!)}}} is n choose k-1.
{{{(matrix(2,1,n,k))}}}{{{"="}}}{{{n!/(k!(n-k)!)}}} is n choose k.
The key to the proof is understanding that for any positive integer, {{{h}}} ,
{{{(h+1)!}}}{{{"="}}}{{{(h+1)*h!}}} .
Let us add those two fractions.
{{{(matrix(2,1,n,k-1))}}}{{{"+"}}}{{{(matrix(2,1,n,k))}}}{{{"="}}}{{{n!/((k-1)!(n-k+1)!)+n!/(k!(n-k)!)}}}
Taking out the common factors, we get
{{{(matrix(2,1,n,k-1))}}}{{{"+"}}}{{{(matrix(2,1,n,k))}}}{{{"="}}}{{{(n!/((k-1)!(n-k)!))*(1/(n-k+1)+1/k)}}}
Now we just have to deal with that simpler sum,
using as common denominator the product
{{{(n-k+1)k}}} .
So, we get
{{{(matrix(2,1,n,k-1))}}}{{{"+"}}}{{{(matrix(2,1,n,k))}}}{{{"="}}}{{{(n!/((k-1)!(n-k+1)!))((k+n-k+1)/((n-k+1)k))}}}
  {{{"="}}}{{{(n!(n+1))/(((k-1)!k)((n-k+1)(n-k)!))}}}
  {{{"="}}}{{{(n+1)!/(k!(n-k+1)!)}}}{{{"="}}}{{{(n+1)!/(k!((n+1)-k)!)}}}{{{"="}}}{{{(matrix(2,1,n+1,k))}}}