permutation with repeated identical elements

Question

First of all I do know the solution to below problem I'm asking different way!!

The problem is like this:

consider the word $AABBB$ how many 3 letter words can be written using the given word?

clearly this is a permutation problem, my problem is can we find the answer only using $nPr$ equation?

this is the way I'll do it (if you have different way please answer thanks.)

(In my answer I used $nCr$, I'm asking a way to do this using $nPr$)

when 3 words equal permutations- $1C1\times\frac{3!}{3!}=1$

when only 2 words equal permutations- $2C2\times\frac{3!}{2!}=6$

numbers of words 7

Answer 1

In a case like this, it may be easier to start from all words that can be made using the letters A and B, then subtract those which don't work. So there are $2^3=8$ three-letter words using only As and Bs (since each letter independently has $2$ choices), and the only one which doesn't fit into AABBB is AAA, leaving you with $7$.

Answer 2

While I feel that the optimal solution to this problem is the one suggested by Especially Lime, you can do this problem using combinations by choosing which positions in the word are filled by which letter.

There are two cases:

1. Three identical letters are used.
2. Two different letters are used.

This seems to be the approach you took.

Three identical letters are used: Since there are only two $A$s available, this can only be done by using the three $B$s. If you want to write this in terms of combinations, choose the only letter you can use three times and then choose all three positions to be filled with that letter. $$\binom{1}{1}\binom{3}{3}$$

Two different letters are used: This means one letter must be used twice and the other letter must be used once. Choose which of the two available letters will be used twice, which can be done in $\binom{2}{1}$ ways. Choose which two of the three positions it will occupy, which can be done in $\binom{3}{2}$ ways. The other letter must occupy the remaining open position. Hence, there are $$\binom{2}{1}\binom{3}{2}\binom{1}{1}$$ such arrangements.

Since the two cases are mutually exclusive and exhaustive, the number of three-letter words that may be formed with the letters $A, A, B, B, B$ without repetition is $$\binom{1}{1}\binom{3}{3} + \binom{2}{1}\binom{3}{2}\binom{1}{1} = 1 + 6 = 7$$ as you found.