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Math Help - combinations

  1. #1
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    combinations

    A commitee of 6 is to be selected from 10 people of whom A and B are two. HOw many commitees can be formed excluding A if B is included

    i have already solved the first part of this qs which was if A and B were both included and that came out to be 70, using combination formula
    i really need help for the second part. the answer is 140 but i dont know how to get there ><""

    and can someone help me with second half of question too?
    In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women? What proportion of these committees contain a particular man and 2 particular women?
    ive worked out the first part which is 700 but i dont know the second half which is 1/7 T.T

    thanks guyz for any help given !
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  2. #2
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    Quote Originally Posted by flyinhigh123 View Post
    A commitee of 6 is to be selected from 10 people of whom A and B are two. HOw many commitees can be formed excluding A if B is included

    i have already solved the first part of this qs which was if A and B were both included and that came out to be 70, using combination formula
    i really need help for the second part. the answer is 140 but i dont know how to get there ><""

    [snip]
    A committee either contains B or it doesn't. If it does, then you're not allowed to have A. If it doesn't, then you can any of the remaining nine:

    (Number of committees with B and not A) + (number of committees without B) =  {8 \choose 5} + {9 \choose 6}.

    Quote Originally Posted by flyinhigh123 View Post
    [snip]
    In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women? What proportion of these committees contain a particular man and 2 particular women?
    ive worked out the first part which is 700 but i dont know the second half which is 1/7 T.T

    thanks guyz for any help given !
    Similar to the other question. You need another 2 men from the remaining 5 and another 2 women from the remaining 5. So the total number is {5 \choose 2} \cdot {5 \choose 2}. Now divide this number by 700 to get the required proportion.
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  3. #3
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    Hello, flyinhigh123!

    A commitee of 6 is to be selected from 10 people of whom A and B are two.
    How many commitees can be formed excluding A if B is included?
    There are ten people: . \{A,B,C,D,E,F,G,H,I,J\}

    With no restrictions, there are: . {10\choose6} \:=\:{\color{blue}210} possible committees.

    But A and B will not serve together.
    We will count the committees in which A and B are together.

    Since A and B are already on the committee,
    . . we must choose 4 more people from the remaining 8 people.
    There are: . {8\choose4} \:=\:{\color{blue}70} committees with both A and B.


    Therefore, there are: . 210 - 70 \:=\:140 committees
    . . which do not have both A and B.


    This agrees with Mr. F's solution.




    In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women?
    What proportion of these committees contain a particular man and 2 particular women?
    This is Mr. F's solution . . . with baby-talk.


    You are right: there are 700 possible committees with 3 men and 4 women.


    There are six men: . \{A,B,C,D,E,F\}
    and seven women: . \{T,U,V,W,X,Y,Z\}

    Suppose man A and women T, U must be on the committee.

    We must select 2 more men from \{B,C,D,E,F\}
    . . . and 2 more women from \{V,W,X,Y,Z\}

    There are: . {5\choose2}{5\choose2} \:=\:10\cdot10 \:=\:100 ways.


    So the proportion is: . \frac{100}{700} \:=\:\frac{1}{7}

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  4. #4
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    thank you so much for your help !
    special thanks to Soroban for your clear and step by step explanation, you really helped me =D
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