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Find the number of five-card hands dealt from a deck of $52$ cards, s.t. there is one pair (two cards of one denomination), a third card of a different denomination, a fourth card of a third different denomination, and a fifth card of a fourth different denomination.

My approach is:
There is application of product principle for all the sub-cases (& in the sub-cases as well).

(i) There are $\binom{13}{1}$ ways to get one denomination, then choose two cards out of four suits by $\binom{4}{2}$.
(ii) Further, the third card can be chosen in $\binom{12}{1}$ ways, with a particular card chosen in $\binom{4}{1}$ ways.
(iii) The fourth card can be chosen in $\binom{11}{1}$ ways, with a particular card chosen in $\binom{4}{1}$ ways.
(iv) Further, the fifth card can be chosen in $\binom{10}{1}$ ways, with a particular card chosen in $\binom{4}{1}$ ways.

The answer is : $(\binom{13}{1}*\binom{4}{2})*(\binom{12}{1}*4)*(\binom{11}{1}*4)*(\binom{10}{1}*4)$

But the answer is given by :
$\binom{13}{1}* \binom{4}{2}* \binom{12}{3}* \binom{4}{1}^3$

The both approaches given differ by a factor of $3$.

My approach yields higher value by effectively yielding a permutation of $3$ cards from $12$; while the answer yields a combination of them. But, am unclear how my approach is wrong in taking individual choices of :$\binom{12}{1}, \binom{11}{1}, \binom{10}{1}$.

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They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.

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