# simple probability question using counting rule

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?

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How can we show you where you're going wrong if you don't explain how you got your answer? – anon Sep 22 '11 at 10:01
@anon I thought that there are 4 types of clothing and we want to know all the possible was in which each clothing can be selected meaning that there are 3! ways to select shoes, 7! to select shirts etc... – dbr Sep 22 '11 at 10:04
@dbr: 3 boys can only wear 3 shirts, not 7. Also you seem to have forgotten the coats. – heijp06 Sep 22 '11 at 10:07
dbr: But you're not selecting all of the shirts, pants, socks, or coats, because there are only three boys. Do you see what you have to do now? – anon Sep 22 '11 at 10:08
I don't understand the or answer which is $(7!)(8!)/ (3!)^2 \times2!$ as the first one is correct but $3 \times 7! \times 8! \neq (7!)(8!)/ (3!)^2 \times2$. – Quixotic Sep 22 '11 at 18:54

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...

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so basically for your first point: I want to find how many ways 3 can be selected out of 3, out of 5, out of 7 and out of 8 which is (3 3) (5 3) (7 3) (8 3) which for each is: 3!/3!*0! ; 5!/3!*2!; 7!/3!*4!, 8!/3!*5! and I need to multiply these together? – dbr Sep 22 '11 at 10:16
and for your second point its x * x * x * x the number of wazs thez can be permuted among the boys – dbr Sep 22 '11 at 10:20
@dbr: You've misunderstood my second point. Given precisely three (say) shirts picked out for the boys, how many ways can they be distributed to them? There's more than one way they can wear them. (But yes, your calculations in the first part are correct, just don't go multiplying them yet because you haven't finished counting each article's combination total.) – anon Sep 22 '11 at 10:23
thanks for all your help, then I guess the number of ways it can be distributed to them is 3 so I have to multiply each term by 3 then I get the number of ways in which it can be distributed to them – dbr Sep 22 '11 at 10:30
@dbr: If the boys choose from n items, the first boy gets to choose from n, the second from n - 1 and the third from n - 2. How many ways to do that are there? – heijp06 Sep 22 '11 at 10:32