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.
 A: 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 $???$ $\dots$.
A: *

*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 $$\binom nr = \dfrac{n!}{r!(n-r)}$$
A: 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.
