# [SOLVED] Permutations and Combinations problem

• Feb 13th 2009, 02:47 AM
struck
[SOLVED] Permutations and Combinations problem
In a game of bridge the pack of 52 cards is shared equally between all four players. What is the probability that one particular player has no hearts?

So there are (52C13) hearts AFAIK (Worried) but I have no idea how would the distribution occur. Normally I would do a (n+1 Object A and n-1 Object B) or (n-1 Object A and n+1) if there were only 2 objects involved and the difference was very small. But in this case that way would go on forever (Shake)

I have to solve it using permutations and combinations so please take that into account.
• Feb 13th 2009, 05:13 AM
mr fantastic
Quote:

Originally Posted by struck
In a game of bridge the pack of 52 cards is shared equally between all four players. What is the probability that one particular player has no hearts?

So there are (52C13) hearts AFAIK (Worried) but I have no idea how would the distribution occur. Normally I would do a (n+1 Object A and n-1 Object B) or (n-1 Object A and n+1) if there were only 2 objects involved and the difference was very small. But in this case that way would go on forever (Shake)

I have to solve it using permutations and combinations so please take that into account.

There are $^{52}C_{13} = \frac{52!}{13! 39!} = 635,013,559,600$ different possible bridge hands.

There are 39 non-heart cards. The number of ways to get a bridge hand with 13 of those 39 cards (that is, a hand with no hearts) is $^{39}C_{13} = \frac{39!}{13! 26!} = 8,122,425,444$.

Now divide and get $\frac{8,122,425,444}{635,013,559,600} \approx 0.01279$.