# Thread: Combinations or Perms

1. ## Combinations or Perms

If I have a set of 6 elements {a,a,a,b,b,b} and I want to find the number of ways to arrange these in different orders, is it perms or combs that I use?

2. Originally Posted by dave1022
If I have a set of 6 elements {a,a,a,b,b,b} and I want to find the number of ways to arrange these in different orders, is it perms or combs that I use?
Strictly speaking it is neither. It is arrangements with repetitions.
Say the problem was the same instructions but with $\displaystyle \{a,a,b,b,b,c,c,c,c\}$ .
Then the answer would be $\displaystyle \frac{9!}{2!\cdot 3! \cdot 4!}$.
We divide to account for the repetitions.

What is the answer to your original question?

3. Surely if I only have two discrete elements recurring a number of times it's just $\displaystyle total!/a!*b!$ which is just $\displaystyle totalC(a or b)$

4. Originally Posted by dave1022
Surely if I only have two discrete elements recurring a number of times it's just $\displaystyle total!/a!*b!$ which is just $\displaystyle totalC(a or b)$
Yes. But don't you want to know the general solution?