3
$\begingroup$

I'm trying to solve the following problem:

In a fraternity house, three boys share a room with a single closet. Each boy can wear each of the other boys items of clothing, and they share freely. The closet contains 3 pairs of shoes, 7shirts, 5 pairs of pants, 8 pairs of socks and 4 coats. If each boy dresses in shoes, shirt, pants, socks, and coat, in how many combinations of clothing may the boys appear together?

This is from Hays, 1973.

In the end it gives the solution: 3*7!8! or (7!)(8!)/ (3!)ˆ2*2!

I've spent the last 15 minutes trying to figure out where all the other items disappeared. I thought the answer would be 3!*7!*5!*8!.

Where am I going wrong?

  • 0
    I don't understand the *or* answer which is $(7!)(8!)/ (3!)^2 \times2!$ as the first one is correct but $ 3 \times 7! \times 8! \neq (7!)(8!)/ (3!)^2 \times2$.2011-09-22

1 Answers 1

0

The combinations for each article of clothing are independent, so it suffices to count the combinations for each independently and then multiply these numbers together.

  1. Given $x$ articles, how many ways are there of picking out $3$ for the boys?
  2. Given $3$ articles picked out, how many ways can they permuted among the boys?

If you answered these two correctly and multiply them together, you should find a simple factorial/factorial expression for each article combination. If you put all of these in a product, there will be some cancellation...

  • 0
    @anon yes I noticed that thanks :) thanks for all the help I think I understand the problem now!2011-09-22