# Combinations or Perms

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• Dec 9th 2009, 08:24 AM
dave1022
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?
• Dec 9th 2009, 08:56 AM
Plato
Quote:

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 $\{a,a,b,b,b,c,c,c,c\}$ .
Then the answer would be $\frac{9!}{2!\cdot 3! \cdot 4!}$.
We divide to account for the repetitions.

What is the answer to your original question?
• Dec 9th 2009, 09:09 AM
dave1022
Surely if I only have two discrete elements recurring a number of times it's just $total!/a!*b!$ which is just $totalC(a or b)$
• Dec 9th 2009, 09:13 AM
Plato
Quote:

Originally Posted by dave1022
Surely if I only have two discrete elements recurring a number of times it's just $total!/a!*b!$ which is just $totalC(a or b)$

Yes. But don't you want to know the general solution?