I have a problem as such:

How many ways are there to choose nine cards out of a standard deck of 52 cards in such a way that every suit is represented in the selection at least twice?

Here's my solution:

Partition the set into the four suits - then pick two cards from each suit: there are ${13 \choose 2}^4$ ways to do this. Then there are $44$ cards left over, of which one has to be picked. ${44 \choose 1} = 1$. So the total number of ways the goal can be achieved is $44 \cdot {13 \choose 2}^4$.

The trouble is that this gives me something like $1.6$ billion ways, which seems unrealistically large. Have I made a mistake?

  • $\begingroup$ This isn't exactly right, because you're counting the case where you pick, say, the Ace and Jack of spades in the first round and then the 10 of spades as the leftover, as distinct from {10s, Js} + {As}. $\endgroup$ – mjqxxxx Oct 28 '13 at 15:33
  • $\begingroup$ Yes, you are right. How should I fix this mistake? $\endgroup$ – Newb Oct 28 '13 at 15:35
  • $\begingroup$ You need to decide by what factor that is causing you to overcount, then divide by it. (Alternatively, choose the three-card suit first, then choose the determined number of cards from each suit.) $\endgroup$ – mjqxxxx Oct 28 '13 at 15:35
  • $\begingroup$ Sure - I could just arbitrarily pick the three-card suit first. So a solution would be ${13 \choose 3} \cdot {13 \choose 2}^3$? $\endgroup$ – Newb Oct 28 '13 at 15:38
  • $\begingroup$ and also * 4 because, there are 4 ways to choose the suit from which you take 3 cards :) $\endgroup$ – Gintas K Oct 28 '13 at 15:45

If you need that "every suit is represented in the selection at least twice" in the selection of 9 cards, so you have 3 cards of one suit and 2 cards of each other suit. you can pick 2 cards out of 13 in ${13 \choose 2}$ different ways and you can pick 3 cards out of 13 in ${13 \choose 3}$ different ways. And if you have 4 different suits, so you need to multiply the answer by 4. There are $4 \cdot {13 \choose 2}^3 \cdot {13 \choose 3}$ different ways :) 542887488 :)

  • $\begingroup$ Thanks for your answer! By the way, in the future, you might be able to make your answers easier to read by using LaTeX formatting: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Newb Oct 28 '13 at 15:49
  • $\begingroup$ Thanks, I'm new here, so didn't got used to it yet :/ I rewrite my answers on my free-time, or wait for someone to edit it :D $\endgroup$ – Gintas K Oct 28 '13 at 15:50

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