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Math Help - A permutation/ race problem

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    Member zzizi's Avatar
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    A permutation/ race problem

    Permutation questions please help?

    There are 12 children in a race wearing different coloured hats. 4 Red, 4 blue and 4 white hats.


    How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?

    I would really appreciate an explanation too. Thank you!
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    Re: A permutation/ race problem

    Start by looking just at the first five racers. Imagine that the first three are wearing red hats. There are four possibilities for the first child, then 3 for the second, two for the third. There are 4(3)(2)= 24 ways that can happen. Now, there are 8-non-red hats so 8(7)= 56 ways the fourth and fifth students can be "non-red hats". That would give a total of (56)(24)= 1344 ways the first 5 children can finish "red, red, red, non-red, non-red". But there are also 5!/(3!2!)= 10 different orders for "RRRNN" so, in fact, 13440 ways the first five children could be "3 red hat, 2 non-red hat". Now, we have 7 children remaining. There are 7! ways they can finish.

    (This is assuming that we are talking about the ways individual children, not just colors of hats, can finish. That is, "Jimmy first, Sarah second" is different from "Sarah first, Jimmy second" even though both are wearing red hats.)
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    Re: A permutation/ race problem

    Quote Originally Posted by zzizi View Post
    There are 12 children in a race wearing different coloured hats. 4 Red, 4 blue and 4 white hats.
    How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?
    I assume that you are to count the number of different ways the children can finish the race in which three red hats are among the first five places.
    \left[\binom{4}{3}\binom{8}{2}(5!)\right]\left[7!\right]
    We select the three red hats and two non-red hats and then arrange the first five; then arrange the other seven.
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    Re: A permutation/ race problem

    Thank you very much for your responses.


    \left[\binom{4}{3}\binom{8}{2}(5!)\right]\left[7!\right]
    Can you just verify, what is the calculation for this? Are you multiplying these together?


    I received another answer from someone else I posed the question to and it differs from the answers given here, so am alittle confused.

    He stated the following:

    if we number the children as 1 2 3 4 || 5 6 7 8 || 9 10 11 12
    we want combos in which 3 out of 5 are from the 1st group
    possible color combos are RRRBB RRRWW & RRRBW
    considering distinct #s, combos*perms for RRRBB = 4c3*4c2*5!/(3!*2!) = 240
    ditto for RRRWW -> 240
    for RRRBW -> 4c3*4c1*4c1*5!/3! = 1280

    total possible outcomes = 2*240 + 1280 = 1760
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    Re: A permutation/ race problem

    Quote Originally Posted by zzizi View Post
    Can you just verify, what is the calculation for this? Are you multiplying these together?
    Here is the OP "How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?"

    Now that is poorly posed.
    I took it quite literally. The order of the children is all important here. Say Bob has a red hat and Jim has a red hat. If Bob is first and Jim is third is totally different from Jim first and Bob second. Although in both cases those two are among the first five to finish.

    That is the basis of my reply. If the order of the finishers makes no difference then we must ask. Do you simply mean how many ways can the first five be wearing three red hats and two non-red hats? And that is regardless of order of any kind.

    Or what exactly does the question mean?
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    Re: A permutation/ race problem

    Quote Originally Posted by Plato View Post
    Here is the OP "How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?"

    Now that is poorly posed.
    I took it quite literally. The order of the children is all important here. Say Bob has a red hat and Jim has a red hat. If Bob is first and Jim is third is totally different from Jim first and Bob second. Although in both cases those two are among the first five to finish.

    That is the basis of my reply. If the order of the finishers makes no difference then we must ask. Do you simply mean how many ways can the first five be wearing three red hats and two non-red hats? And that is regardless of order of any kind.

    Or what exactly does the question mean?
    Apologies if it was poorly posed, that's how it was asked of me. That's why I put the question here, so that I could make better sense of what is meant and how to solve. Thank you for your input though, it's appreciated.
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    Re: A permutation/ race problem

    Quote Originally Posted by zzizi View Post
    Apologies if it was poorly posed, that's how it was asked of me. That's why I put the question here, so that I could make better sense of what is meant
    If we assume that only the order of the colors is all that matters then the answer is
    \frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350
    Last edited by Plato; November 17th 2012 at 01:11 PM.
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    Re: A permutation/ race problem

    Quote Originally Posted by Plato View Post
    If we assume that only the order of the colors is all that matters then the answer is
    \frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350
    That's great thank you
    Last edited by Plato; November 17th 2012 at 01:13 PM.
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    Re: A permutation/ race problem

    Quote Originally Posted by Plato View Post
    If we assume that only the order of the colors is all that matters then the answer is
    \frac{5!}{3!\cdot 2!}\cdot\frac{7!}{(4!)(3!)}=350
    I'd just like to ask a clarifying question:

    when you are multiplying by: \ffrac{8!}{(4!)^2}=700
    Can you explain to me how you derived the 4!^2?
    Last edited by Plato; November 17th 2012 at 01:14 PM.
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    Re: A permutation/ race problem

    Quote Originally Posted by zzizi View Post
    I'd just like to ask a clarifying question:
    when you are multiplying by: \ffrac{8!}{(4!)^2}=700
    Can you explain to me how you derived the 4!^2?
    Please note the edits I have just made.
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    Re: A permutation/ race problem

    Thank you !
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