Conditional probability on a poker hand Question (from Hoel&Stone, Chapter 2):
Find the probability that a poker hand of 5 cards will contain no card smaller than 7, given that it contains at least 1 card over 10 (i.e. J, Q, K, A).
There is a solution for this problem which I fully understand:
$\frac{\binom{32}{5}-\binom{16}{5}}{\binom{52}{5} - \binom{36}{5}}$
There are $\binom{36}{5}$ ways of drawing a hand that contains only 2,3,...,10; the same is true for the relationship between $\binom{32}{5}$ and 7,...,A and between $\binom{16}{5}$ and 7,8,9,10.
My solution, on the other hand, approaches this problem from another angle, and goes like this: $$\frac{4^2\binom{32}{4}}{4^2 \binom{51}{4}}$$
Concerning the denominator: There are 4 ranks (J,Q,K,A) in 4 suits for the first card. Since the task is to provide "at least" one such card, I should be able to draw the remaining 4 from the remaining 51 cards.
Concerning the numerator: Same for the first card. The remaining 4 can be 7,8,9,...,A which makes $4*8 = 32$ in total.
I'd like to know why my solution is wrong. Also, can you recommend some ressource on combinatorics which explains this topic in a more intuitive manner?

Think for example of the hand that contain the $\heartsuit$ Queen and the $\spadesuit$ Jack, and $3$ other specific cards $\le 10$. In the denominator, you have counted this hand twice, for the "first" card could have been the $\heartsuit$ Queen, and one of the remaining $51$ could have been the $\spadesuit$ Jack, or the other way around. Some other hands have been counted more times, and some only once.
Remark: The main way to avoid such mistakes is to solve many problems, make mistakes, and become sensitized to the common sources of error. "Multiple-counting" is a common problem. 
