From a deck of $52$-playing cards, $5$ cards are drawn. How many $5$-card hands are there consisting of $3$ face cards, $1$ king, and $1$ spade?
Solution:
This can be shown in two cases:
Case 1: If king is not drawn,we have
$\displaystyle \binom{8}{3} \times \binom{4}{1} \times \binom{4}{1}= 896 $ ways
Case 2:
If king is drawn,
$\displaystyle \binom{4}{3} \times \binom{3}{1} \times \binom{4}{1}= 48 $ ways
Hence, the total number of ways is $896+48= 944$ ways.
Is the solution presented above is conclusively correct?
Best Answer
I’m assuming that you mean that the five cards can be divided into three face cards, a king and a spade (and not just that they contain three cards, contain a king and contain a spade, possibly overlapping).
Let’s count separately depending on whether the spade is also a face card.
If the spade is a face card, we have $5$ face cards, of which at least one is a spade and at least one is a king. There are $\binom{12}5$ hands with $5$ face cards, of which $\binom95$ have no spade, $\binom85$ have no king and $\binom65$ have neither, so by inclusion–exclusion that makes $\binom{12}5-\binom95-\binom85+\binom65$. But now we’ve counted hands with only the king of spades and no other king and no other spade, of which there are $\binom64$, so we have to subtract those.
If the spade is not a face card, we have one of $10$ non-face-card spades, and $4$ face cards, of which one is a king. There are $\binom{12}4$ sets of $4$ face cards, of which $\binom84$ have no king, so that makes $10\left(\binom{12}4-\binom84\right)$.
Thus, in total, there are $\binom{12}5-\binom95-\binom85+\binom65-\binom64+10\left(\binom{12}4-\binom84\right)=4851$ such hands.
Since this doesn’t agree with another answer that’s been posted, I wrote Java code to check it by enumeration.