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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?2011-09-22
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    @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...2011-09-22
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    @dbr: 3 boys can only wear 3 shirts, not 7. Also you seem to have forgotten the coats.2011-09-22
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    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?2011-09-22
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    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

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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?2011-09-22
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    and for your second point its x * x * x * x the number of wazs thez can be permuted among the boys2011-09-22
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    @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.)2011-09-22
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    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 them2011-09-22
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    @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?2011-09-22
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    @dbr: Not 3, you want $3!$ - this is the number of ways 3 of the same type of article can be permuted between the boys.2011-09-22
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    so yes sorry its 3! and there are 5 items so 5*3! is the number of ways 3 of each can be permuted is that correct?2011-09-22
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    dbr: Nope. Like I said, **split the problem** so that you're *only* looking at one article of clothing at a time. Say, shirts. There are 5 shirts, so there are (5 choose 3) ways of selecting 3 shirts for the boys. Multiply that by 3! (the number of ways to permute them among the boys) and you get 5!/2!. Another way to get this is through @Peter's method: pick a shirt for the boys one at a time; there are 5 choices, 4 choices, and then 3 choices for the boys respectively, so there are 5*4*3 total combinations. Can you apply this reasoning to the other articles?2011-09-22
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    so there are 3*2*1; 7*6*5; 5*4*3; 8*7*6; 4*3*2 ways each.2011-09-22
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    @dbr: (Edit:) Yes! You can also write them as 3!, 7!/4!, 5!/2!, 8!/5!, and 4!/1!. Now what happens when you multiply all these together? (And do you understand why you should multiply them together?)2011-09-22
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    @anon We need to multiply these together to get all the possible combinations of each 3 item selected. right?2011-09-22
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    @dbr: **I** made the mistake this time. ;) I edited my comment right afterwards though.2011-09-22
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    @anon yes I noticed that thanks :) thanks for all the help I think I understand the problem now!2011-09-22