Probability on cards involving two conditions . $\text{Problem (i) :}$

When $4$ cards are drawn out of $52$ cards, what is the probability of only $2$ cards  being  same value and the rest being different  ?

MY WORK:
The probability I found is :
$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \binom{4}{1}\times \binom{4}{1}}{\binom{52}{4}}$$
$\text{Problem (ii) :}$

When $4$ cards are drawn out of $52$ cards,  what is the probability of $2$ cards to be of same value of same colour and the other two cards to be of same value ( different from the previous one ) of different colours? 

MY WORK:
The probability I found is :
$$\frac{\binom{2}{1}\times \binom{13}{1}\times \binom{2}{2}\times \binom{12}{1}\times \binom{2}{1}\times \binom{2}{1}}{\binom{52}{4}}$$
Here,  I at first select a particular colour $\binom{2}{1}$ and then a particular value $\binom{13}{1}$  .  Then,  there is only single way of first condition. After,  I select a different value $\binom{12}{2}$ . For different colour,  I take $\binom{2}{1}\times \binom{2}{1}$ .
AM I CORRECT  ?
N. B.  Value means ace,  king,  jack,  queen and the numbers from $2$-$10$  .


$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \binom{4}{1}\times \binom{4}{1}}{\binom{52}{4}}$$

$\checkmark$ Select a rank and two suits for it, and select two other ranks, and a suit for each.

$$\frac{\binom{2}{1}\times \binom{13}{1}\times \binom{2}{2}\times \binom{12}{1}\times \binom{2}{1}\times \binom{2}{1}}{\binom{52}{4}}$$

$\checkmark$ Select a colour, a rank, and two suits of that colour for that rank, and select another rank, and a suit of each colour.
