Every suite has 13 cards. you can choose 5 cards from the same suite in 13C5 ways.

The cards can be from any suite out of the 4 suits.

so total no of combinations for a flush =13C5 * 4=1287*4=5148 = Option D

dellaboemia wrote:A flush in poker is defined as all 5 cards being of the same suit. There are four suits: spades, diamonds, clubs and hearts - all of which there are 13 of in a standard deck of 52 cards. How many ways are there to draw a flush?

(A) 4
(B) 676
(C) 1287
(D) 5148
(E) 154,440

Correct Answer D

There are 13 cards of each suit.
A flush is a combination of 5 cards of the same suit.
Number of combinations of 5 that can be formed from 13 options = 13C5 = (13*12*11*10*9)/(5*4*3*2*1) = 1287.

Since there are 4 suits, the result above must be multiplied by 4:
4*1287 = 5148.

The correct answer is D.

Thanks guys. I actually tried this two ways. The first attempt, I used 13C5 formula and got the right answer. My second attempt, I set it up as
52 ways to pick the first card times
13 ways to pick the second card, since you know have a suit, times
12 ways to pick the third card, times
11 ways to pick the third.

Didn't work.

Then I tried
4 ways to pick a specific suite
13 ways to pick the second card of that suit,
12
11.

That doesn't work either and I can't really figure out why? Comment?

The problem with your second two methods is that the "fill in the blank" method only works for permutations, where order matters. For a flush, A-5-3-J-2 is the same as J-5-A-2-3; we end up with the same five cards and the same flush.

By the way, on your first attempt (52*13*12...), your second number should be 12. The first card takes away 1 card from that suit, so you have 12 possibilities remaining for your second card.

Ahh got it. Thanks Bill. The "fill in the blanks" approach is really about arranging things, not selecting x things from a set containing y things where y > x. Would that be a fair takeaway?

Yes, that is a good summation.

dellaboemia wrote:Ahh got it. Thanks Bill. The "fill in the blanks" approach is really about arranging things, not selecting x things from a set containing y things where y > x. Would that be a fair takeaway?

The slot method -- your second attempt -- works perfectly fine here.

Number of options for the first card = 52.
Number of options for the second card = 12. (There are 12 remaining cards of the same suit as the first card.)
Number of options for the third card = 11.
Number of options for the fourth card = 10.
Number of options for the fifth card = 9.

To combine these options, we multiply:
52*12*11*10*9.

Since the order of the cards doesn't matter, we divide by the number of ways that the 5 cards can be arranged:
(52*12*11*10*9)/(5*4*3*2*1) = 5148.