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Veritas Combinatorics - Flush in Poker

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by dellaboemia » Sat Mar 31, 2012 12:09 pm
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
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by five.pointer » Sat Mar 31, 2012 1:59 pm
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
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by GMATGuruNY » Sat Mar 31, 2012 2:04 pm
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.
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by dellaboemia » Sat Mar 31, 2012 5:33 pm
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?
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by Bill@VeritasPrep » Sat Mar 31, 2012 5:38 pm
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.
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by dellaboemia » Sun Apr 01, 2012 3:34 pm
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?
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by Bill@VeritasPrep » Sun Apr 01, 2012 3:42 pm
Yes, that is a good summation.
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by GMATGuruNY » Mon Apr 02, 2012 5:09 am
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.
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