# counting: permutation/combination?

• Apr 16th 2010, 09:18 PM
swimmergirl
counting: permutation/combination?
I don't understand how to do this problem.
I need to figure out how many different 14 letter words I can form with 5 A's, 3 B's, and 6 C's. I understand that since some letters are the same the words would be indistinguishable if it was just 14 factorial. But why would I only divide that number by (5 x 3 x 6)? Can you help me understand this concept of combinations vs. permutations?
• Apr 16th 2010, 09:33 PM
Debsta
Quote:

Originally Posted by swimmergirl
I don't understand how to do this problem.
I need to figure out how many different 14 letter words I can form with 5 A's, 3 B's, and 6 C's. I understand that since some letters are the same the words would be indistinguishable if it was just 14 factorial. But why would I only divide that number by (5 x 3 x 6)? Can you help me understand this concept of combinations vs. permutations?

Imagine if the letters are distinguishable eg A1, A2, ...A5 etc.
Then the solution would be 14 factorial.
But, as you said they are not distinguishable, so eg
A1,A2,A3,A4,A5 is the same as A2, A1, A3, A4, A5 etc etc.
Thus each AAAAA can be formed in 5! ways.
Similar argument for the Bs and Cs.
So you need to divide 14! by (5!x3!x6!) not (5x3x6) as you suggested.