The directions for the problem say: Find the number of possible 5-card hands that contain the cards specified.
28. 1 ace and 4 other cards (none of which are aces)
1 ace AND 4 non-aces.
There are 4 aces, so there are "4 choose 1"
ways to get the ace.
For each of the 4C1 ways we can pick the ace,
there are 48 non-aces, so there are "48 choose 4"
ways to get the 4 non-aces.
"AND" usually means to multiply, so the number of
ways that can be had is
(4C1)(48C4) = 778320
Edwin
