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Math Help - Combination Problem

  1. #1
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    Combination Problem

    I thought this would be a simple problem, but I can still not come up with the correct answer.
    The problem is:
    12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

    Since order is not important it is a combination problem and not a permuntation problem.

    So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

    Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

    I get 200 for the answer but according to the quiz this is not the answer.

    Could someone tell me what I am doing wrong?
    Thanks in advance
    James
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jamesinsc
    I thought this would be a simple problem, but I can still not come up with the correct answer.
    The problem is:
    12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

    Since order is not important it is a combination problem and not a permuntation problem.

    So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

    Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

    I get 200 for the answer but according to the quiz this is not the answer.

    Could someone tell me what I am doing wrong?
    Thanks in advance
    James
    The first room can be filled in 12\times 11\times 10\times 9 \times 8=12!/7! ways (counting each
    permutation as a distinct way of filling the room. Thus for each combination
    we have counted 5! permutations, so the number of ways of filling the first
    room is 12!/(7! 5!).

    Similaly for the next room except that we have only the 7 left overs to use,
    so there are now 7!/(3!4!).

    There are no options for who is allocated to room three once the first
    two rooms are filled.

    So our answer is:

    \frac{12!}{7!5!} \frac{7!}{3!4!}=\frac{12!}{5!4!3!}=27720 <br />

    RonL
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  3. #3
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    Thats it!!

    Thanks Ron!!!
    I was going about that one all wrong. It's hard for an old man to go back to school.
    Thanks again
    James
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  4. #4
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    'tis maybe is an indistinguishable permutation

    Quote Originally Posted by jamesinsc View Post
    I thought this would be a simple problem, but I can still not come up with the correct answer.
    The problem is:
    12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

    Since order is not important it is a combination problem and not a permuntation problem.

    So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

    Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

    I get 200 for the answer but according to the quiz this is not the answer.

    Could someone tell me what I am doing wrong?
    Thanks in advance
    James

    .........'TIS COULD BE AN INDISTINGUISHABLE PERMUTATION

    use the formula: n!/p!q!r!...

    12!/5!4!3!
    =479001600/17280
    =27,720

    MAYBE YOU COULD TRY THIS ONE...
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  5. #5
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    Lexington, MA (USA)
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    Hello, James!

    Another approach . . .


    12 Students are in a class.
    Five can go to room A, Four to room B, and Three to room C.
    How many ways can this happen?

    Assign 5 students to room A.
    . . There are: . _{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792 ways.

    From the remaining 7 students, assign 4 students to room B.
    . . There are: . _7C_4 \:=\:\frac{7!}{4!3!} \:=\:35 ways.

    From the remaining 3 students, assign 3 students to room C.
    . . Of course, there is: . _3C_3 \:=\:1 way.


    Therefore, there are: . 792 \times 35 \times 1 \:=\:27,\!720 ways.

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