# HW Probability Problem

How many different bridge hands are possible containing five spades, three diamonds, threee clubs, and two hearts?

Please provide some helpful hints as I am unfamiliar with most card games.

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All you need to know about bridge for this is that in a standard deck there are four suits, each of thirteen cards, and a bridge hand is thirteen cards. So select five of thirteen spades, etc. – Ross Millikan Mar 18 '13 at 23:55
Normal deck of card used: 4 suits, 13 cards of each suit. Bridge hands have 13 cards. So your required number of cards of each suit totals 13 the number of cards in a hand. – amWhy Mar 18 '13 at 23:55

Not much card knowledge is needed here, only that a bridge hand is any collection of $13$ cards. These are chosen from the standard $52$-card deck, which has $13$ spades, $13$ hearts, $13$ diamonds, and $13$ clubs. The $13$ spades are all different, as are the $13$ hearts, and so on.

For counting the hands of the kind described, note that the $5$ spades can be chosen in $\binom{13}{5}$ ways. For every way of choosing the spades, there are $???$ ways of choosing the $3$ diamonds from the $13$ available. And for every way of choosing the spades and diamonds, there are $???$ $\dots$.

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• Since a bridge hand has 13 cards, and every deck of cards has $13$ spades, $13$ diamonds, $13$ clubs, and $13$ hearts $= 52$ cards in all, and
• since a "hand" is a "hand" (that is, the order in which the cards in a hand doesn't matter,

we thus have

$\displaystyle \binom{13}{5}$ ways of choosing five spades, and for each way of choosing spades, there are

$\quad\times$

$\displaystyle \binom{13}{3}$ ways of choosing 3 diamonds. Then for each $8$-card hand thus far chosen, there are

$\quad \times$

$\quad\vdots\quad$ ways of choosing 3 clubs...

$\quad\times$

$\quad \vdots \quad$ ways of choosing 2 hearts for each partial hand thus far determined. $=======================$

And we are almost done.

Now we simply calculate the product of the binomial coefficients (why the product?) to determine the total number of hands bridge hands that can be formed subject to the given restrictions : $$\binom{13}{5}\times \binom{13}{3} \times \binom{13}{3}\times \binom{13}{2}$$ Recall $$\binom nr = \dfrac{n!}{r!(n-r)}$$

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This is one of the problems where you can check the answer at the back of the book, and the solution seems to be 8, 211, 173, and 256. – JuanSancen Mar 19 '13 at 0:20
Are you sure you are checking the solution to the right problem? – amWhy Mar 19 '13 at 0:24
Yes. This is the primary source of confusion – JuanSancen Mar 19 '13 at 0:47
It does; the professor had explained something in class from chapter 1 and I had to go back and do one of the beginning problems using a the bars and stars method for distinguishable and indistinguishable elements, so I was thinking that I to use this method. But I see not, thanks for the help. – JuanSancen Mar 19 '13 at 0:55
You're very welcome! Yes, it can get confusing distinguishing between methods of constructing combinations. It helped me (for study purposes) to create a chart/table listing the various methods and the conditions required/circumstances appropriate for each + picking an informative example of when/how each is used/computed. – amWhy Mar 19 '13 at 0:58

amWhy and Andre have great answers, but I'd like to expand on amWhy's post if your still having trouble with why you're supposed to take the product:

In combinatorics we multiply when the problem can be broken down in stages. This means, in the first stage there are $x_1$ different outcomes, in the second stage there are $x_2$ many outcomes, and continuing until the last stage (call it stage $n$), we have $x_n$ many outcomes. Then, if the outcomes are independent and the outcomes of each stage is distinct, then the problem has $x_1\cdot x_2 \cdots x_n$ many combined outcomes.

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