This problem asks to explain why the method used is overcounting or undercounting cards in a 7-card poker hand that contains at least two 3-of-a-kind (3-of-a-kind means three cards of the same rank) and calculate the correct probability.For example, this would be a valid hand: ace of hearts, ace of diamonds, ace of spaces, 7 of clubs, 7 of spades, 7 of hearts and queen of clubs. (Note that a hand consisting of all 4 aces and three of the 7s is also valid.). The $P_1(E)$ answer is given and basically we have to refute it. My answer is that the correct answer is $P_2(E)$ but cannot really grasp why the first one is wrong and if it is overcounting or undercounting cards.
$P_1(E) = \frac{{13\choose 2}{4\choose 3}{4\choose 3}{46\choose 1}}{{52\choose 7}}$ or $P_2(E) = \frac{{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 3}{46\choose 1}}{{52\choose 7}}$ .
Any explanations on this would be really helpful.