# Thread: 2 part permutation calcuations

1. ## 2 part permutation calcuations

A club has 8 members, namely A, B, C, D, E, F, G, and H.

i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?

ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?

2. ## Re: 2 part permutation calcuations

Originally Posted by azollner95
A club has 8 members, namely A, B, C, D, E, F, G, and H.

i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?

ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?
You must show us all some effort on your part.

3. ## Re: 2 part permutation calcuations

Well for i) I got 48, because I assumed that any of the 8 members can be chosen, then it has to be 6 options left, because G and H cannot be together. So 8x6=48.

for ii) I only got 6 because there are only 3 men (FGH) and if it is only men then the next option only leaves the other two guys so 3x2=6 ways.

4. ## Re: 2 part permutation calcuations

Originally Posted by azollner95
A club has 8 members, namely A, B, C, D, E, F, G, and H.
i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?
ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?
i) There are three possible ways to do this:
1. neither G nor H is chosen: $6\cdot 5$
2. G is chosen but not H: $6\cdot 2$
3. H is chosen but not G: $6\cdot 2$
This should give us the same answer as $8\cdot 7-2$ WHY?