0

Suppose you have a deck of 30 cards, where 13 cards each have one duplicate, leaving 4 cards to be unique. So, for the sake of simplicity, if we were to assign letters as card values, we would have 2 cards with the letter A, 2 cards with letter B, ....2 cards with letter L (so 13 cards x 2), and 4 remaining cards lettered R, S, T, U.

How do you compute the number of possible combinations when drawing n cards. For example, say I draw 3 cards, how many possible combinations of 3 cards can I get?

(We don't care about order, so AAB is the same as ABA or BAA)

Jürgen
  • 121
  • What did you try? – gamma Oct 05 '15 at 18:22
  • Nothing, since I have absolutely no idea how to even approach this problem. I am aware of how to compute the combinations if the deck would have been unique (30 choose 3, binomial coefficient), but for this problem, I am completely in the dark. I tried googling, but I dont even know what to search for. – Jürgen Oct 05 '15 at 18:24

1 Answers1

1

There are seventeen types, so firstly we have ${17 \choose n}$ for all distinct case.
Next, we have one duplicate case: we choose $1$ type from the $13$ types and choose $(n-2)$ from the remaining types, ${13\choose1}\times{16\choose{n-2}}$.
Next we have $2$ duplicate case, ${13\choose2}\times{15\choose{n-4}}$
And so on, until you cannot substract from $n$ any more. Then you add all these cases up to get the final anser.

When $n=3$, it is simply ${17\choose3}$+${13\choose1}\times{16\choose{1}}$.

cr001
  • 12,598
  • I don't understand why you are adding up and then multiplying. Can you elaborate please? – Jürgen Oct 05 '15 at 19:03
  • Multiplication always happens first in any expression. It is ${17\choose3}+({13\choose1}\times{16\choose{1}})$ – cr001 Oct 05 '15 at 19:07
  • I ment, why are you adding it up to start with. Also, I just computed your example, and the result would be that in a deck of 30 cards, where 13 cards are duplicated, and I draw 3 cards, I can get 141.440 different combinations? That just does not seem right? – Jürgen Oct 05 '15 at 19:11
  • 1
    The answer is 888, not 141440. You can google "17 choose 3" to get the value of ${17\choose3}$. For the adding up part, consider the following small example: There are two types of cards, A and B, and you can have 1 card out of them. How many ways you can have? One way of thinking of the answer is there is $1$ way for A case and $1$ way for B case, so totally $1+1=2$. For similar reason, when you split into different cases you add them up. – cr001 Oct 05 '15 at 19:17
  • Oh right, I am sorry, I muliplicated instead of adding up the first two. I know how to compute the binomial coefficient. Thank you very much for enlightning me, it is truly much appreciated. – Jürgen Oct 05 '15 at 19:23