1. ## permutations and combinations

I ran into this problem:

You are choosing 5 teams out of a group of 23 people. You do this by choosing a team lead for each team, and then letting the team leads choose their teams in round-robin fashion. You choose the order in which the round-robin selection goes at random. Each team lead is guaranteed to pick the most-desirable unpicked person each time, and it turns out everyone has the same priorities as to who they would pick, so the competition is likely to be stiff. How many possible outcomes are there? Outcomes are the "same" only if they have exactly the same lead+team arrangements.

I know it has to do with permutations, but I have no idea how to solve it.

Any help will be appreciated

2. Originally Posted by andyandy26
I ran into this problem:

You are choosing 5 teams out of a group of 23 people. You do this by choosing a team lead for each team, and then letting the team leads choose their teams in round-robin fashion. You choose the order in which the round-robin selection goes at random. Each team lead is guaranteed to pick the most-desirable unpicked person each time, and it turns out everyone has the same priorities as to who they would pick, so the competition is likely to be stiff. How many possible outcomes are there? Outcomes are the "same" only if they have exactly the same lead+team arrangements.

I know it has to do with permutations, but I have no idea how to solve it.

Any help will be appreciated
Once the team leads have been chosen which member is chosen in each round is fixed, only their order varies.

so there are:

$TL=\frac{23!}{18! 5!}$

ways of choosing the team leads (ignoring order)

Then for any selection of leads there are $5!$ arrangements of persons in each round except the last where there are $60$ arrangement

Then the required answer is the product of all these.

CB