# Probability ( counting principles)

I just need an explaination to a question I saw in a statistics book. I understand the concepts but I don't really understand what the question is asking. This question asks '' Find the probability of being dealt at random and without replacement a thirteen card bridge hand consisting of a)thirteen cards of the same suit b) 2 clubs, 3 diamonds, 5 hearts and 3 spades. A thorough explaination will assist me.

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Part a) is easy: There is a total of $52\choose 13$ hands that can be dealt.Out of these there are $4$ of the same suit, hence the answer is $4\over{52\choose13}$.

For b) the number of "good" possibilities is counted by first selecting 2 out of 13 clubs, then 3 out of 13 diamonds, then 5 out of 13 heatrs, finally 3 out of 13 spades. Thus the answer is ${{13\choose2}{13\choose3}{13\choose5}{13\choose3}}\over{52\choose13}$. Do we have to worry about mixing the suits to accomodate for the fact that these cards might be dealt e.g. in order spades, hearts, clubs, hearts, hearts, diamonds, spades, ... or the like? No! Since we count unordered hands with $52\choose13$ in the denominator, we msut do the same in the numerator - which means that we may assume any specific order we wish, as if after dealing we sort our hand by suits.

EDIT: Adpated solution to fixed typo in problem statement.

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@ Hagen, thank you for your assistance. for the b) I did the same way but wasn't that sure. Thank you... –  Eugene Mettle Sep 3 '12 at 5:08
Hagen, I think its suppose to be 13 cards consiting of 2 clubs, 3 diamonds 5 hearts and 3 spades. It was a typo but i get it now.. –  Eugene Mettle Sep 3 '12 at 5:24
For a: There are four hands that are all the same suit. There are $52 \choose 13$ total hands.