Hi lehder,

if "a" and "b" are in the selection of p balls, chosen from n,

then there remain p-2 balls to choose from the remaining n-2.

Hence there are selections containing "a", "b" and p-2 other balls.

If the p balls contain "a" but not "b",

then we have n-2 balls to choose from and we require p-1 new additional balls to join with "a".

Hence there are selections of "a" with p-1 others, none of which are "b".