Given a 50 card deck with cards numbered from 1 through 10 in each of 5 suits, how many Given a 50 card deck with cards numbered from 1 through 10 in each of 5 suits, how many 5 card hands are there that include exactly one pair of two cards that have the same numeric value?
 A: Hint : You have $10$ possibilities for the value of the pair and $\frac{9\times 8\times 7}{6}$ possibilities for the value of the other cards. For the pair, you have $10$ different suit combinations and for the other cards, you have $5$.
A: [Choose a rank for the pair, and 2 suits]$\times$ [choose 3 other ranks and suits for each of them]
$${10\choose1}\cdot{5\choose2}\times {9\choose 3}\cdot{5^3}$$
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A: My counting is rusty, but I believe the way to approach this is the following: first we need to be clear about what we want. Two cards of the same value, that is two of the same card from different suits. The other three cards must not be the same number as the other two, nor have any matching amongst themselves.
Firstly, we must choose the suit of the two same-numbered cards: we have $5$ suits, and want to choose $2$ suits; $\binom{5}{2}$. We also want to choose one number for the cards: $\binom{10}{1}$.
Next, we must choose the remaining cards. For the third card, $5$ possibilities are eliminated, the ones matching with the first two cards in the remaining suits, and the first cards themselves. So, we have $50-5 = 45$ options for one card. $\binom{45}{1}$. 
Now, we lost $5$ more possible cards to choose from (including the one we just chose). So, we have $40$ cards to choose from, for one card: $\binom{40}{1}$.
With the same logic, we subtract another $5$ cards, and are left with $35$ options for one final card: $\binom{35}{1}$.
Our final answer is as follows: $\frac{\binom{5}{2}\binom{10}{1}\binom{45}{1}\binom{40}{1}\binom{35}{1}}{5!}$. The $5!$ accounts for the over-counting, that is, considering the ordering of the cards, when the ordering does not matter.
