a)all diamonds
That's 13 diamonds, Choose 5. That's C(13,5) = 1287
b) exactly 3 face cards
Choose the face cards. That's 12 face cards, Choose 3. That's C(12,3)
For each of those choices of the 3 face cards, we
choose the non-face cards. That's 40 non-face cards, Choose 2.
That's C(40,2).
Answer: C(12,3)×C(40,2) = 220×780 = 171,600.
c) at most 2 Queens
There are three cases. No queens, 1 queen, and 2 queens
1. No queens. That's 48 non-queens, Choose 5. That's C(48,5)=1,712,304.
2. 1 queen. That's 4 queens, choose 1. That's C(4,1)=4.
For each of those choices of the 1 queen, we choose the 4 non-queens.
That's 48 non-queens, Choose 4. That's C(48,4). So for case 2, the answer
is C(4,1)×C(48,4) = 4×194580 = 778,320
3. 2 queens. That's 4 queens, choose 2. That's C(4,2)=6.
For each of those choices of the 2 queens, we choose the 3 non-queens.
That's 48 non-queens, Choose 3. That's C(48,3). So for case 3, the answer
is C(4,2)×C(48,3) = 6×17296 = 103,776.
Total = 1,712,304 + 778,320 + 103,776 = 2,594,400
d) exactly 2 pairs (e.g., 2 Queens, 2 kings)
We choose the two denominations of the pairs.
That's 13 denominations Choose 2, or C(13,2).
We choose the suits for the lower pair. That's 4 suits Choose 2, or C(4,2).
We choose the suits for the higher pair. That's 4 suits Choose 2, or C(4,2).
We choose the 5th card that doesn't have the same denomination as either
of the two pair. (We must be sure we don't have a full house!)
That's 11 denominations that the pairs aren't of, choose 1. That's C(11,1).
We choose the suit of the 5th card. That's 4 suits choose 1. That's C(4,1).
Answer: C(13,2)×C(4,2)×C(4,2)×C(11,1)×C(4,1) = 78×6×6×11×4 = 123,552.
Edwin
