How many $7$ letter sequences (formed from the $26$ letters in the alphabet, with repetition allowed) How many $7$ letter sequences (formed from the $26$ letters in the alphabet,  with repetition allowed)  contain exactly one A and exactly two Bs? 
Soln:
My constraint is the one A and two Bs,  thinking about this conatraint I am not concerned with order,  but I want to know how many ways this set of 3 elements can be arranged in a string of length 7, tbis would be $$\binom{7}{3}$$
now the other thing I am concerned about is the number of ways the other four letters can be selected: $$24^4$$
now to find the number of sequences it would be the product of these two values: $$ \binom{7}{3} 24^4$$
Questions: 
1) Is this the right way to decompose the problem?
2) the way I accounted for the A and Bs,  why would I not have to do the same for the other set of letters? 
 A: 1) You are close but you forgot to arrange $A$ and $B$ which you need to multiply again by ${3\choose1}$.
2) You could choose the other four letters instead and ${7\choose3}={7\choose4}$ so the result will be the same. You need not to do both because when you choose one set the other set is automatically chosen as well.
A: 2)
You knew you had 1 A and 2 Bs and needed to know where to put them.  If you know what the remaining 4 letters were, you could (and would have to) do this but don't so you can't and don't need to do this.
Now you could say that there are 24 ways the remaining 4 letters are the same and ${4 \choose 4}$ ways to place them, 24*23 ways for 3 of the remaining letters to be the same and ${4 \choose 3}$ ways to place them 24*23 ways for the remaining letter to be two different pairs and ${4 \choose 2}{2 \choose 2}$ ways to place them, 24*23*22 ways for the remaining numbers to have just one pair and ${4 \choose 2}{2 \choose 1}{1 \choose 1}$ and there are 24*23*22*21 ways they can all be different and ${4 \choose 1}{3 \choose 1}{2 \choose 1}{1 \choose 1}$ ways to place them.  And then you can add them all up.
But that's way too much unnecessary work.  Only need to worry about choosing if you know the values or have restrictions on the values.  If you don't then you can simply consider the options for each remaining position.  There are 4 and the values are each 24 options.  So there are $24^4$ possible options. 
This way even if the remaining letters were ... say....F,F,S,L all orders were already taken into account when we first tried to determine all possible letters in all possible orders.
