Number of choices in a 52-set of cards Given a set of 52 playing cards (4 suits, 13 cards of each type numbered 1-13).
In how many sets of 5 cards there is a pair with the same number and a different suit
    (considering that the rest 3 cards are of different number each)
I thought about the following (which is wrong):

We choose 1 card from 52.
From the rest 3 cards with the same number, we choose 1.
Then we put away the 2 leftover cards with the same number.
Now out of 48 cards we choose one and remove the rest 3 with the same number.
Now out of the 44 cards we do the same,
      and the same goes to 40 cards.

Thus: $52 \times 3 \times 48 \times 44 \times 40$
    which is totally wrong.
 A: That's not totally wrong.
In fact, it is almost right, except you end up counting each hand with a pair 12 times, according to which order you pick the two cards in the pair (2 possible orders) and which order you pick the three non-pair cards ($3!=6$ possible orders).
So the total number of one-pair hands is
$$ \frac{52\times 3 \times 48 \times 44 \times 40}{2\times 6} $$
A: Consider a hand of the specified type, say $\heartsuit2,\spadesuit2,\heartsuit7,\diamondsuit8,\clubsuit 9$. You counted it $12$ times, one for each of the following orders of choosing the five cards:
$$\begin{align*}
&\heartsuit2,\spadesuit2,\heartsuit7,\diamondsuit8,\clubsuit 9\\
&\heartsuit2,\spadesuit2,\heartsuit7,\clubsuit 9,\diamondsuit8\\
&\heartsuit2,\spadesuit2,\diamondsuit8,\heartsuit7,\clubsuit 9\\
&\heartsuit2,\spadesuit2,\diamondsuit8,\clubsuit 9,\heartsuit7\\
&\heartsuit2,\spadesuit2,\clubsuit 9,\heartsuit7,\diamondsuit8\\
&\heartsuit2,\spadesuit2,\clubsuit 9,\diamondsuit8,\heartsuit7\\
&\spadesuit2,\heartsuit2,\heartsuit7,\diamondsuit8,\clubsuit 9\\
&\spadesuit2,\heartsuit2,\heartsuit7,\clubsuit 9,\diamondsuit8\\
&\spadesuit2,\heartsuit2,\diamondsuit8,\heartsuit7,\clubsuit 9\\
&\spadesuit2,\heartsuit2,\diamondsuit8,\clubsuit 9,\heartsuit7\\
&\spadesuit2,\heartsuit2,\clubsuit 9,\heartsuit7,\diamondsuit8\\
&\spadesuit2,\heartsuit2,\clubsuit 9,\diamondsuit8,\heartsuit7\\
\end{align*}$$
In more detail, your first step of choosing the two $2$’s can be carried out in two ways: choose $\heartsuit2$ first and then $\spadesuit2$, or choose $\spadesuit2$ first and then $\heartsuit2$. Similarly, the other three cards can be chosen in any of $3!=6$ different orders.
Since you’ve counted each hand $12$ times, you can get the correct answer by dividing your answer by $12$.
You can get the answer more directly by approaching the problem in a slightly different way. First, there are $13$ possible values for the pair. Once you’ve chosen the value, there are $\binom42$ ways to pick two of the four cards having that value. Thus, there are $13\binom42$ ways to choose the pair. Now you must choose three cards of different values from the remaining $12$ values. First choose the three values that will be represented; in my example they’re $7,8,$ and $9$. There are $\binom{12}3$ ways to do this. For each of these three values you have a choice of $4$ suits, so you must make a $4$-way choice $3$ times; this can be done in $4^3$ ways. Thus, you can choose the three odd cards in altogether $4^3\binom{12}3$ ways.
Finally, you can choose the pair and the three odd cards in 
$$13\binom42\cdot4^3\binom{12}3=13\cdot6\cdot64\cdot220=1,098,240$$
ways.
A: You can also choose


*

*Common number of pair: $13$

*Colors of pair: ${4 \cdot 3} \over 2$

*Numbers of three other cards: ${12 \cdot 11 \cdot 10}\over6$

*Colors of these three cards: $4^3$


Thus ${{13 \cdot 12 \cdot 11 \cdot 10 \cdot 4 \cdot 3 \cdot 4^3}\over{2 \cdot 6}}=1098240$ hands.
