Question on word combinations with exclusivity "How many 4 letter words on the alphabet {a,b,c} in which 'a' occurs exactly twice are there?"
My answer is incorrect as I answered 3*3*2*2 4 letter words. 
However, this doesn't necessarily remove 'a' from the letter choices. What would be the correct way to solve this?
 A: Approach via multiplication principle.
$$\underline{~~~}~\underline{~~~}~\underline{~~~}~\underline{~~~}$$


*

*Step 1: Since we know that $a$ must occur exactly twice, let us pick which two positions within our word are occupied by $a$'s.  For example we might have chosen the second and third positions like below:


$$\underline{~~~}~\underline{~a~}~\underline{~a~}~\underline{~~~}$$


*

*Step 2: Pick what non-a letter appears in the furthest left unoccupied space.  For example we could have chosen $b$ like below:


$$\underline{~b~}~\underline{~a~}~\underline{~a~}~\underline{~~~}$$


*

*Step 3: Pick what non-a letter appears in the final remaining unoccupied space.  For example, we could again have chosen $b$ like below:


$$\underline{~b~}~\underline{~a~}~\underline{~a~}~\underline{~b~}$$
Multiply the number of options available to complete each step to get the final count.
A: Instead of proceedng in order i.e. choosing first letter, then second etc. , choose which positions you want $a$ to occur in , and then choose what should come in the other positions.
For example, we have four positions $1,2,3,4$. Two of these positions can be chosen in $\binom 42 = 6$ ways. We place $a$ in these places.
Now, in the rest of the places, we have a choice of either $b$ or $c$ going in each place. Two places, two choices for each place gives a total of $2^2 = 4$ choices.
Multiplying these gives us $6 \times 4 = 24$ choices.
To actually write down all the words, pick each possible combination of two places, put $a$ there, and vary $b,c$ across the other places.
For example, fixing positions $2$ and $4$ for $a$ gives $baba,baca,caba,caca$, four of the desired words.

As for your answer, reducing the option pool cannot be done by position, since the positions of $a$ don't occur in order, or say within the first two positions, or in any particular pattern. However, treating them independently of the other positions does the job.
