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?

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- Dec 9th 2009, 08:24 AMdave1022Combinations 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 AMPlato
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? - Dec 9th 2009, 09:09 AMdave1022
Surely if I only have two discrete elements recurring a number of times it's just $\displaystyle total!/a!*b!$ which is just $\displaystyle totalC(a or b)$

- Dec 9th 2009, 09:13 AMPlato