I'm having some trouble with combinatorics and would like some help resolving this confusion. The problem is as follows.
You are drawing five cards and get the following 3 in your hand: 4h, 4c, Qs. What is the probability that you will draw exactly one pair?
To solve this problem, I reasoned it out like this: There are ${49 \choose 2} $ total ways of drawing your last two cards. However, you can only pick cards that won't pair up your hand; removing the two remaining '4's and the three remaining 'Q's, you are left with 44 cards. After choosing the next card, you must remove the other three of the rank you just chose from the remaining deck, resulting in 40 cards. Therefore, you get ${44 \choose 1}{40 \choose 1} $ ways to draw into a pair. Probability will be ${44 \choose 1}{40 \choose 1}/{49 \choose 2} $.
The given answer says that there are 11 remaining ranks in the deck to choose from, and out of each rank, there are 4 suits. So the number of ways to draw a pair will be $4^2{11 \choose 2} $.
These two answers are not equivalent, and are in fact off by a factor of 2. That is, ${44 \choose 1}{40 \choose 1} = 2\cdot 16{11\choose 2} $. What mistake did I make originally and what are some tips to avoid making such errors? I don't understand why these two answers are not the same.
Thanks!