Find the probability of getting one diamond and one spade in a five-card hand, using binomial coefficients. 


*A five card hand is dealt at random from a standard $52$ card deck. Let $X = \text{# spades}$ and $Y = \text{# diamonds}$. Find $P(X = 1\text{ and }Y =1)$. Leave your answer as a ratio of products of binomial coefficients.


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Please help me with this question. I don't know how to apply the binomial coefficient in to this question. My attempt would be:
$ ^{52}C_5 \times^{13}C_1\times^{13}C_1$
But I know for sure that this is not the right answer. Could you please help me out?
 A: There are $\binom{52}{5}$ equally likely $5$-card hands. 
Now we count the "favourables," the hands that have precisely $1$ $\spadesuit$ and $1$ $\diamondsuit$.
The $\spadesuit$ can be chosen in $\binom{13}{1}$ ways. For each of these ways, the $\diamondsuit$ can be chosen in $\binom{13}{1}$ ways. And for each of these ways the remaining $3$ cards can be chosen from the $26$ $\heartsuit$ and/or $\clubsuit$ in $\binom{26}{3}$ ways, for a total of $\binom{13}{1}\binom{13}{1}\binom{26}{3}$ ways.
Finally, divide by $\binom{52}{5}$ for the probability.
A: We may assume that the sample space we are working in is the one where order of cards does not matter (it will not change the final answer compared to if we had decided to work in the space where order does in fact matter).  Let us count how many hands exist where there is exactly one spade and exactly one diamond.  Approach via multiplication principle:


*

*Pick the one spade used



 $\binom{13}{1}$ choices



*

*Pick the one diamond used



 $\binom{13}{1}$ choices



*

*Pick the remaining three non-spade/non-diamond cards used



 $\binom{26}{3}$ choices

There are then a total of $\underline{~~~~}$ possible hands with exactly one spade and exactly one diamond:

 $\binom{13}{1}\binom{13}{1}\binom{26}{3}$

The probability is then the ratio of the number of "good" possibilities to the number of elements in the sample space (assuming every element in the sample space is equally likely as is the case in this scenario).

 $\binom{13}{1}\binom{13}{1}\binom{26}{3} / \binom{52}{5}$


Remember that when using an equiprobable sample space $S$, (one in which all outcomes in the sample space are equally likely to occur), for any event $A\subseteq S$ we have $Pr(A) = \frac{|A|}{|S|}$
In our case, the sample space being the set of all possible five card hands from a fifty-two card deck where order doesn't matter, we have $|S|=\binom{52}{5}$
