# Disc. math: permutations

1. Apr 14, 2010

### n00neimp0rtnt

How many permutations of the letters ABCDEFGH contain the string ABC?

This is an example problem in my book, and the answer is 6! = 720. Could someone please explain to me the reasoning behind this (my book does a poor job explaining)? And would this reasoning apply if the string to be found was, say, just AB?

2. Apr 15, 2010

### Filip Larsen

Since ABC has appear just like that you can treat ABC as an indivisible block just like the other letters so that you have ABC, D, E, F, G and H, i.e. 6 blocks in all to find permutations for, which is 6! = 720.

3. Apr 15, 2010

### n00neimp0rtnt

OK, I see now. Thank you. Here's another problem I'm stuck on:

A department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have more women than men?

What I figure out so far is that this would be a combination, not a permutation. If gender was not a concern, it could just simply be C(25, 6). Since there must be MORE women than men, there are gender possibilities: WWWWWW, WWWWWM, WWWWMM. At first glance, it looks like you might be able to just half the amount of answers, but the amount of women and men are not equal. I tried an approach like this:

C(10, 2) + C(10, 1) + C(13, 4) + C(13, 5) + C(13, 6)

but I'm not sure if this is correct.

*C(n, r) = n!/(r!*(n-r)!)
this is the combination theorem for number of r-combinations of a set with n elements.

4. Apr 15, 2010

### Staff: Mentor

Given the restrictions, the committee must fall into one of three types:
4 women, 2 men
5 women, 1 man
6 women.
Find the number of combinations for each of the three committee arrangements and add them.