Reason for difference in number of five-card hands.

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 :$$\binom{12}{1}, \binom{11}{1}, \binom{10}{1}$$.

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.