Number of length-five words How many length-five words can be written using two A's, two T's and one E? 
Why is it not $\binom5 2 \times\binom 5 2 \times \binom 5 1$? 
Is it $ \binom 5 3 = 10$?
 A: Once you’ve chosen $2$ positions for the A’s, there are only $3$ positions remaining from which to choose positions for the T’s, not $5$, so the second factor should be $\binom32$. And once the positions for the A’s and the T’s have been chosen, there is only one position left for the E, so there is only $\binom11=1$ way to ‘choose’ it.
Added: Here’s a problem for which your first calculation would be correct. If I have $5$ pieces of paper, and I write A on two of them, T on two of them, and E on two of them, allowing more than one letter on a piece of paper, there are $\binom52\binom52\binom51$ possible outcomes: there are $\binom52$ ways to choose which two pieces of paper get an A, $\binom52$ ways to choose which two get a T, and $\binom51$ ways to choose which one gets the E. But note that I might have as many as $3$ blank pieces of paper, if I happened to write the A’s and T’s on the same two pieces and then wrote the E on one of those pieces.
A: One way of seeing this is to consider letters $A_1 A_2 T_1 T_2 E_1$, those you can arrange in $5!$ ways. For any arrangement the $A$s are over-arranged in $2!$ ways, etc. So in the end you have:
$$
\frac{5!}{2! \, 2! \, 1!} = \binom{5}{2 \; 2 \; 1}
$$
A multinomial coefficient.
Turn subindices on, turn subindices off.
