I'm not sure about how to solve the following problem
Randomly draw 5 cards from a deck of cards. What is the probability of getting two pairs and one of the 5 cards is Heart-7 (♥7)?
I solved it this way:
Case 1 (7 is a pair): (3C1)(12C1)(4C2)(11C1)
Case 2 (7 is not a pair): (12C2)(4C2)^2
So the probability is: ((3C1)(12C1)(4C2)(11C1) +(12C2)(4C2)^2)/(52C5)
But I think I'm missing something
On
Case 1 (7 is a pair): (3C1)(12C1)(4C2)(11C1)
Select $7\heartsuit$ and one $7$ from three other suits, one from twelve other kinds in two from four suits (the other pair), and one from eleven remaining kinds $\color{red}{\textit{in one from four suits}}$ (the singleton).
$$\def\binom#1#2{\mathop{^{#1}\mathsf C_{#2}}} {\binom 31\,\binom{12}{1}\binom 42\,\binom {11}1\binom 41}$$
Case 2 (7 is not a pair): (12C2)(4C2)^2
Select $7\heartsuit$ as the singleton, and two from twelve kinds each in two from four suits.$$\color{silver}{\binom{1}{1}}\,\binom{12}2{(\binom 42)}^2$$
Not quite. In the first case there are actually $44$ possibilities for the odd card, not $11$: it can be any of the $44$ cards of denominations different from those of the two pairs. If you prefer, it’s $\binom{11}1\binom41$: $11$ ways to choose the odd denomination, and $4$ ways to choose one card of that denomination.