# Combinations or Perms

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

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