permutation with repeated identical elements 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

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 $7$.
