How do I count the number of $5$-letter words out of $\{a,b,c,d\}$ with 2 letters occurring exactly twice?
My current attempt to a solution looks like this.
- Select two different letters out of $\{a,b,c,d\}$: $4 \cdot 3$ possibilities.
- Assign two occurrences of the first of these two letters to two out of the $5$ places: $\binom{5}{2}$ possibilities.
- Assign two ocurrences of the second of these two to the remaining $3$ places: $\binom{3}{2}$
- Choose one out of $2$ letters for the remaining space: $2$
Can I conclude that the amount of words is equal to the following?
$$ 12\binom{5}{2} \binom{3}{2} 2 $$
A simple way to tackle this type of question is as follows:-
The letters can be chosen in $\binom{4}{2} \binom{2}{1}=12$ ways.
We now have letters $xxyyz$ say. These can be arranged in $\frac {5!}{2!2!1!}=30$ ways.
Answer $360$.