How many ways are there to make a “one pair” in poker

I'm confused at the general method of solving this type of problem. The wikipedia page says that there are:

${13 \choose 1} {4 \choose 2} {12 \choose 3} {4 \choose 1}^{3}$ ways to select a pair when 5 cards are dealt. Can someone outline what each calculation means? For example, is $13 \choose 1$ the process of selecting 1 card in the beginning?

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Each card has a rank (e.g.: king, jack, $2$, $6$) and a suit (e.g.: heart, club).
The $\binom{13}1$ is the number of ways to pick the rank of the single pair, and the $\binom42$ is the number of ways to choose the suits of the cards in the pair. To make sure that it's the only pair, then we have to choose other ranks for each of the three other cards. There are only $12$ ranks left to choose from, so $\binom{12}3$ is the number of ways to choose our other three ranks. Each of the other three ranks will have an associated suit, and $\binom41$ is the number of ways to choose the suit of one of the other three ranks. Thus, the number of ways to choose the suits of the other three ranks is $\binom41^3$.
$13\choose 1$ is the number of ways of choosing the denomination of the pair, whether it is a pair of kings, or a pair of threes, or whatever. Then there are $4\choose 2$ ways to choose suits for the two cards of the pair.
Then there are $12\choose 3$ ways to choose the three different denominations of the remaining three cards from the twelve that are different from the denomination of the pair. Each of those other three cards can be of ${4\choose 1}=4$ different suits, for ${4\choose 1}^3$ choices of three suits in all.
 Why does $13 \choose 1$ represent both cards, though? Why wouldn't we do ${13 \choose 1}^{2}$, say? – user1038665 Oct 29 '12 at 6:39 Becausr it's a pair. After you choose the denomination for the first card, you don't get to choose the denomination for the other half of the pair, because it has to be the same, or you won't have a pair. – MJD Oct 29 '12 at 12:37