Combinations trouble











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A class consists of 14 males and 12 females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of 6 men and 4 women be chosen from the class?



My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?



so (13,5) demonstrates that male.a in on the committee therefore (11,4) is female.a not being allowed on the committee. (13,5) male.a not allowed on committee and (11,4) is female.a on the committee.










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    A class consists of 14 males and 12 females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of 6 men and 4 women be chosen from the class?



    My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?



    so (13,5) demonstrates that male.a in on the committee therefore (11,4) is female.a not being allowed on the committee. (13,5) male.a not allowed on committee and (11,4) is female.a on the committee.










    share|cite|improve this question


























      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      A class consists of 14 males and 12 females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of 6 men and 4 women be chosen from the class?



      My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?



      so (13,5) demonstrates that male.a in on the committee therefore (11,4) is female.a not being allowed on the committee. (13,5) male.a not allowed on committee and (11,4) is female.a on the committee.










      share|cite|improve this question















      A class consists of 14 males and 12 females. If one male A and one female B cannot be in the same committee, how many ways can a committee consisting of 6 men and 4 women be chosen from the class?



      My thought process so far is to find total combinations subtract the constraint scenarios. Therefore ending with c(14,6)* c(12,4) - (c(13,5)*(11,4) + c(13,6)*c(11,3)) can someone tell me if im on the right track?



      so (13,5) demonstrates that male.a in on the committee therefore (11,4) is female.a not being allowed on the committee. (13,5) male.a not allowed on committee and (11,4) is female.a on the committee.







      combinatorics combinations






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      edited Nov 16 at 10:31









      N. F. Taussig

      42.7k93254




      42.7k93254










      asked Nov 16 at 8:40









      Georges Shimmon

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          My thought process so far is to find total combinations subtract the constraint scenarios.




          Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.



          You are allowed to select commitees containing only one from the twain (so don't subtract these).






          share|cite|improve this answer




























            up vote
            1
            down vote













            You may construct the number of "forbidden" committees as follows:





            • $A$ and $5$ other males and $B$ and 3 other females: $color{blue}{binom{13}{5}cdotbinom{11}{3}}$


            Subtracting from the total number of committees you get the "allowed" committees:
            $$binom{14}{6}cdotbinom{12}{4} - color{blue}{binom{13}{5}cdotbinom{11}{3}}$$






            share|cite|improve this answer





















            • ahhh i see, thank you for your help
              – Georges Shimmon
              Nov 16 at 9:26











            Your Answer





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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

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            up vote
            1
            down vote














            My thought process so far is to find total combinations subtract the constraint scenarios.




            Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.



            You are allowed to select commitees containing only one from the twain (so don't subtract these).






            share|cite|improve this answer

























              up vote
              1
              down vote














              My thought process so far is to find total combinations subtract the constraint scenarios.




              Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.



              You are allowed to select commitees containing only one from the twain (so don't subtract these).






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote










                My thought process so far is to find total combinations subtract the constraint scenarios.




                Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.



                You are allowed to select commitees containing only one from the twain (so don't subtract these).






                share|cite|improve this answer













                My thought process so far is to find total combinations subtract the constraint scenarios.




                Yes, but the constraint(or forbidden) scenario is to "select a commitee containing both A and B".   Count that and subtract it from the total.



                You are allowed to select commitees containing only one from the twain (so don't subtract these).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 8:47









                Graham Kemp

                84.2k43378




                84.2k43378






















                    up vote
                    1
                    down vote













                    You may construct the number of "forbidden" committees as follows:





                    • $A$ and $5$ other males and $B$ and 3 other females: $color{blue}{binom{13}{5}cdotbinom{11}{3}}$


                    Subtracting from the total number of committees you get the "allowed" committees:
                    $$binom{14}{6}cdotbinom{12}{4} - color{blue}{binom{13}{5}cdotbinom{11}{3}}$$






                    share|cite|improve this answer





















                    • ahhh i see, thank you for your help
                      – Georges Shimmon
                      Nov 16 at 9:26















                    up vote
                    1
                    down vote













                    You may construct the number of "forbidden" committees as follows:





                    • $A$ and $5$ other males and $B$ and 3 other females: $color{blue}{binom{13}{5}cdotbinom{11}{3}}$


                    Subtracting from the total number of committees you get the "allowed" committees:
                    $$binom{14}{6}cdotbinom{12}{4} - color{blue}{binom{13}{5}cdotbinom{11}{3}}$$






                    share|cite|improve this answer





















                    • ahhh i see, thank you for your help
                      – Georges Shimmon
                      Nov 16 at 9:26













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    You may construct the number of "forbidden" committees as follows:





                    • $A$ and $5$ other males and $B$ and 3 other females: $color{blue}{binom{13}{5}cdotbinom{11}{3}}$


                    Subtracting from the total number of committees you get the "allowed" committees:
                    $$binom{14}{6}cdotbinom{12}{4} - color{blue}{binom{13}{5}cdotbinom{11}{3}}$$






                    share|cite|improve this answer












                    You may construct the number of "forbidden" committees as follows:





                    • $A$ and $5$ other males and $B$ and 3 other females: $color{blue}{binom{13}{5}cdotbinom{11}{3}}$


                    Subtracting from the total number of committees you get the "allowed" committees:
                    $$binom{14}{6}cdotbinom{12}{4} - color{blue}{binom{13}{5}cdotbinom{11}{3}}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 16 at 8:50









                    trancelocation

                    8,1191519




                    8,1191519












                    • ahhh i see, thank you for your help
                      – Georges Shimmon
                      Nov 16 at 9:26


















                    • ahhh i see, thank you for your help
                      – Georges Shimmon
                      Nov 16 at 9:26
















                    ahhh i see, thank you for your help
                    – Georges Shimmon
                    Nov 16 at 9:26




                    ahhh i see, thank you for your help
                    – Georges Shimmon
                    Nov 16 at 9:26


















                     

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