Combinations type question...

**rodders** · April 20th 2012, 01:23 AM

Ok, stuck on this problem..

Suppose you have 6 letters: A, B, C, D, E , F

And you are allowed to pick 3 but with the following rules:

1. ABC is the same as ACB, BCA etc so they are not counted as unique
2. repetitions are allowed so AAB is allowed or CCC

How many would there be in total...after playing around i think there are 56, but i would like a general approach and a generalisation... So n letters and you are allowed to pick q with the rules above.

Many thanks

**princeps** · April 20th 2012, 01:38 AM

You can obtain a solution using formula for combinations with repetition :

$N=\frac{(n+k-1)!}{k! \cdot (n-1)!}$

In this particular case :

$N=\frac{(6+3-1)!}{3! \cdot (6-1)!}=56$