I am not understanding this problem:
In a deck of 52 cards, of 13 ranks, and 4 suits, how many different 5 card hand can we get such that, there is always exactly one pair.
There is a similar example here . But I don't understand the logic behind it.
My thoughts are:
assume we have the hand ${a_{1},a_{2},b,c,d}$ where $a_{1}$ and $a_{2}$ are two cards of the same rank and $b,c,d$ are distinct cards from each other and $a$.
So my logic:
there are $\binom{52}{1}$ ways to pick $a_{1}$ and $ \binom{3}{1}$ ways to pick $a_{2}$.
and then $ \binom{50}{1}$ for $b$, $\binom{49}{1}$ for $c$, and $ \binom{48}{1}$ for $d$
which gives us $\binom{52}{1}\binom{3}{1}\binom{50}{1}\binom{49}{1}\binom{48}{1}$ = $18345600$ ways to get a pair…
…which is the wrong answer.
Where have I gone wrong?
Best Answer
There are $\binom{13}{1}$ ways to pick the kind we have a pair in. For each such way, we have $\binom{4}{2}$ ways to pick the actual cards.
For each of these ways, there are $\binom{12}{3}$ ways to pick the kinds we will have one each of. For each of these kinds, there are $\binom{4}{1}$ ways to pick the actual cards, for a total of $\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$ ways.
Remark: One of common problems in counting is inadvertent double (or multiple) counting. For example, your $\binom{52}{1}\binom{3}{1}$ includes the Ace of $\spadesuit$ together with the Ace of $\heartsuit$. But it also counts the Ace of $\heartsuit$ together with the Ace of $\spadesuit$. These give the same one-pair hand.