Card combinatorics - two answers A hand of $10$ cards is dealt from a standard pack of $52$ cards.
How many ways can the hand contain exactly $3$ cards of the same value and the remaining $7$ cards from the remaining suit?  
I have two answers and one is correct. If so, why is the other incorrect?  
Answer 1:  

Choose any possible card in $52$ ways. Fix this card value.
  There are now 3 remaining cards in that value, and we choose 2 in $\binom{3}2$ ways.
  The remaining suit is chosen in $\binom{1}1$ ways and we choose $7$ out of this suit in $\binom{12}7$ ways.
  Total is 123552.

Answer 2:  

Choose a card value in $\binom{13}1$ ways. Fix this card value.
  We choose 3 out of the 4 in $\binom{4}3$ ways.
  The remaining suit is now fixed and we choose the remaining 7 cards in $\binom{12}7$ ways.
  Total is 41184.

 A: As Peter stated in the comments, your second answer is correct.
Why is your first answer wrong?
You count each hand three times, once for each way you could designate one card of the three of a kind as the card of that rank.  For instance, if you have the hand $$\color{red}{7\heartsuit, 7\diamondsuit}, 7\clubsuit, 3\spadesuit, 4\spadesuit, 6\spadesuit, 8\spadesuit, 9\spadesuit, Q\spadesuit, A\spadesuit$$
your first method counts this in three ways:
$$\color{red}{7\heartsuit}, \qquad \color{red}{7\diamondsuit}, 7\clubsuit, \qquad 3\spadesuit, 4\spadesuit, 6\spadesuit, 8\spadesuit, 9\spadesuit, Q\spadesuit, A\spadesuit$$
$$\color{red}{7\diamondsuit}, \qquad \color{red}{7\heartsuit}, 7\clubsuit, \qquad 3\spadesuit, 4\spadesuit, 6\spadesuit, 8\spadesuit, 9\spadesuit, Q\spadesuit, A\spadesuit$$
$$7\clubsuit, \qquad \color{red}{7\heartsuit, 7\diamondsuit}, \qquad 3\spadesuit, 4\spadesuit, 6\spadesuit, 8\spadesuit, 9\spadesuit, Q\spadesuit, A\spadesuit$$
Notice that $$3\binom{13}{1}\binom{4}{3}\binom{12}{7} = 52\binom{3}{2}\binom{12}{7}$$
A: The first one is incorrect because you are overcounting. Take for instance, picking Jc as one of the possible 52 cards. Then choosing two other jacks. Say you have JsJd. The remaining suit is hearts, and so you pick 7 hearts. Say 2,3,4,5,6,7,8 hearts.
Now say instead the first card that you picked was the Js. Then the two other jacks that get chosen are JcJd, and then you chose the 2,3,4,5,6,7,8 of hears.
In the first solution these both get counted as different hands, yet as you can see they are the same hand.
A: The second answer is the right one. As for the first, consider the case where your first card is the 2 of spades. Now let's say the other two suits you choose are diamonds and hearts. Let's then choose the 7 remaining cards as the 1,2,3,4,5,6 and 7 of clubs. We can now observe that, had we choosen the 2 of hearts at the beginning, we could have picked the rest of the cards to form the exact same hand as before. In short, the first answer is wrong since you consider the same hands more than once.
