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Math Help - Combinations

  1. #1
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    Combinations

    I am just seeking confirmation to my answers. Thanks in advance!

    Here is the question.

    A company employs 8 persons in department A, 5 in department B and 3 in department C. A project team comprising of 6 people is required. How many ways can this team be assembled if:

    a) From every department, 2 persons are picked.
    b) A minimum of 2 are picked from department A and
    c) A representative for every department is picked.

    My attempts:

    a) \left(^{8}\mathcal{C}_2\right)\left(^{5}\mathcal{C  }_2\right)\left(^{3}\mathcal{C}_2\right)

    b) \left(^{8}\mathcal{C}_2\right)\left(^{14}\mathcal{  C}_4\right)

    c) \left(^{16}\mathcal{C}_6\right)

    Am I on the right track or I veered off somewhere along the way?
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  2. #2
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    Your answer to part a) is correct.

    This is part b) \sum\limits_{k = 2}^6 {\left( {^8 C_k } \right)\left( {^9 C_{6 - k} }\right)}

    Now you do part c). Your answer is not correct.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Your answer to part a) is correct.

    This is part b) \sum\limits_{k = 2}^6 {\left( {^8 C_k } \right)\left( {^9 C_{6 - k} }\right)}

    Now you do part c). Your answer is not correct.
    I did not understand part b. The first part where it is \left(^{8}\mathcal{C}_2\right) is clear, choosing 2 from department A. The second part is \left(^{9}\mathcal{C}_4\right). Here also I understand the subscript number four, which is the number of representatives remaining since we already chose 2. But why is the superscript 9?

    Question c is a bit tricky for me because there is no number given. You are just told picking a representative for every department. It could be picking 3 for one department and 1 for the remaining departments or 2 for each or or ... I thought adding all the numbers (8+5+3) and choosing 6 from the total will be right.
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  4. #4
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    Quote Originally Posted by Keep View Post
    I did not understand part b. The first part where it is \left(^{8}\mathcal{C}_2\right) is clear, choosing 2 from department A. The second part is \left(^{9}\mathcal{C}_4\right).

    Question c is a bit tricky for me because there is no number given.
    For part b) we can choose 2, 3, 4, 5 or 6 from department A
    There are seventeen employees in all, eight in department A and then nine not in department A.
    If we choose only two from department A, \left(^{8}\mathcal{C}_2\right), then we must choose four not in department A, \left(^{9}\mathcal{C}_4\right).
    You do that for each of k=3,4,5,6,  \left(^{8}\mathcal{C}_k\right)\left(^{9}\mathcal{C  }_{6-k}\right)
    Now you add them up.

    Part c) is difficult. There are nine combinations you have to calculate.
    \begin{array}{ccccc}<br />
   A &\vline &  B &\vline &  C  \\<br />
\hline<br />
   4 &\vline &  1 &\vline &  1  \\<br />
   3 &\vline &  2 &\vline &  1  \\<br />
   3 &\vline &  1 &\vline &  2  \\<br />
   2 &\vline &  2 &\vline &  2  \\<br />
   2 &\vline &  3 &\vline &  1  \\<br />
   2 &\vline &  1 &\vline &  3  \\<br />
   1 &\vline &  2 &\vline &  3  \\<br />
   1 &\vline &  3 &\vline &  2  \\<br />
   1 &\vline &  4 &\vline &  1  \\<br />
 \end{array}
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  5. #5
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    Quote Originally Posted by Plato View Post
    For part b) we can choose 2, 3, 4, 5 or 6 from department A
    There are seventeen employees in all, eight in department A and then nine not in department A.
    If we choose only two from department A, \left(^{8}\mathcal{C}_2\right), then we must choose four not in department A, \left(^{9}\mathcal{C}_4\right).
    You do that for each of k=3,4,5,6,  \left(^{8}\mathcal{C}_k\right)\left(^{9}\mathcal{C  }_{6-k}\right)
    Now you add them up.

    Part c) is difficult. There are nine combinations you have to calculate.
    \begin{array}{ccccc}<br />
   A &\vline &  B &\vline &  C  \\<br />
\hline<br />
   4 &\vline &  1 &\vline &  1  \\<br />
   3 &\vline &  2 &\vline &  1  \\<br />
   3 &\vline &  1 &\vline &  2  \\<br />
   2 &\vline &  2 &\vline &  2  \\<br />
   2 &\vline &  3 &\vline &  1  \\<br />
   2 &\vline &  1 &\vline &  3  \\<br />
   1 &\vline &  2 &\vline &  3  \\<br />
   1 &\vline &  3 &\vline &  2  \\<br />
   1 &\vline &  4 &\vline &  1  \\<br />
 \end{array}

    That is really a lot of work for part c. You said we have 17 employees but we have 16. 8 in dept A, 5 in dept B and 3 in dpt. C which is 16 in total. So if we have 8 employees in A, there is another 8, not 9, employees in B and C.
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  6. #6
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    Quote Originally Posted by Keep View Post
    That is really a lot of work for part c. You said we have 17 employees but we have 16. 8 in dept A, 5 in dept B and 3 in dpt. C which is 16 in total. So if we have 8 employees in A, there is another 8, not 9, employees in B and C.
    You are right. I cannot do addition.
    So adjust the numbers above. Change 9 to 8.

    There is no change in part c).
    Yes that is a lot of work.
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  7. #7
    Super Member malaygoel's Avatar
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    I want to say that the coefficient of x^6 in
    [(1+x)^8-1][(1+x)^5-1][(1+x)^3-1] gives the answer.

    Explanation:

    (1+x)^8-1=\sum{x}^r...r varies from 1 to 8
    hence this expression contains
    C_{1}^8x,C_{2}^8x^2,C_{3}^8x^3,C_{4}^8x^4,....

    similarly for other terms of the product given above.

    Now consider the first case of the table given by Plato
    It has the no. of ways
    C_{4}^8.C_{1}^5.C_{1}^3
    now this case is covered in the product by the multiplication x^4 of first term, x of second and third term.

    Hence, all the cases in the table are summed up in the coefficient of x^6 in the product.
    Last edited by malaygoel; June 20th 2009 at 06:49 PM.
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by malaygoel View Post
    I want to say that the coefficient of x^6 in
    [(1+x)^8-1][(1+x)^5-1][(1+x)^3-1] gives the answer.

    Explanation:

    (1+x)^8-1=\sum{x}^r...r varies from 1 to 8
    hence this expression contains
    C_{1}^8x,C_{2}^8x^2,C_{3}^8x^3,C_{4}^8x^4,....

    similarly for other terms of the product given above.

    Now consider the first case of the table given by Plato
    It has the no. of ways
    C_{4}^8.C_{1}^5.C_{1}^3
    now this case is covered in the product by the multiplication x^4 of first term, x of second and term.

    Hence, all the cases in the table are summed up in the coefficient of x^6 in the product.
    Say you want the number ways to form a team if at least 2 people must come from department A and at least one person must come from department B.

    Would you then be looking for the coefficient of the x^{6} term in the exapansion of

     \big( (1+x)^{8}-1-8x\big)\big((1+x)^{5}-1\big)\big(1+x\big)^{3} ?
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Random Variable View Post
    Say you want the number ways to form a team if at least 2 people must come from department A and at least one person must come from department B.

    Would you then be looking for the coefficient of the x^{6} term in the exapansion of

     \big( (1+x)^{8}-1-8x\big)\big((1+x)^{5}-1\big)\big(1+x\big)^{3} ?
    Yes...that would work. That is the real advantage here. You have to form the product only and it will take care of all the cases.
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