Why can't I split combinations/events?












4












$begingroup$


I was recently struggling with a problem that read as so:




A club has 30 members work in business and 30 members that are professors. In how many ways can a committee of 8 be selected that has at least 3 in business and at least 3 professors?




For my answer, I first took care of the requirements, then grouped the rest together to get $$binom{30}{3}binom{30}{3}binom{54}{2}$$
Which ended up being wildly wrong from the correct answer which was
$$binom{30}{5}binom{30}{3}*2 +binom{30}{4}binom{30}{4}$$
I once again tried splitting the last combination by case to get
$$binom{30}{3}binom{30}{3}(binom{27}{1}binom{27}{1}+binom{27}{2}+binom{27}{2})$$
But this just turned out to be the same as my previous answer. After doing some research, I came to the conclusion that my answers were larger than the correct one because I was splitting the event into many smaller sections, and in doing so, overcounting cases, which is why
$$binom{10}{3}neq binom{10}{2}binom{8}1$$
I can accept this as a general principle, as the numbers aren't equal. However, it doesn't really logically make sense to me. What am I actually overcounting by splitting up a combination such as $binom{10}{3}$ into $binom{10}{1}binom{9}{1}binom{8}{1}$? Why does splitting and adding events together have no effect?






combinatorics







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    4












    $begingroup$


    I was recently struggling with a problem that read as so:




    A club has 30 members work in business and 30 members that are professors. In how many ways can a committee of 8 be selected that has at least 3 in business and at least 3 professors?




    For my answer, I first took care of the requirements, then grouped the rest together to get $$binom{30}{3}binom{30}{3}binom{54}{2}$$
    Which ended up being wildly wrong from the correct answer which was
    $$binom{30}{5}binom{30}{3}*2 +binom{30}{4}binom{30}{4}$$
    I once again tried splitting the last combination by case to get
    $$binom{30}{3}binom{30}{3}(binom{27}{1}binom{27}{1}+binom{27}{2}+binom{27}{2})$$
    But this just turned out to be the same as my previous answer. After doing some research, I came to the conclusion that my answers were larger than the correct one because I was splitting the event into many smaller sections, and in doing so, overcounting cases, which is why
    $$binom{10}{3}neq binom{10}{2}binom{8}1$$
    I can accept this as a general principle, as the numbers aren't equal. However, it doesn't really logically make sense to me. What am I actually overcounting by splitting up a combination such as $binom{10}{3}$ into $binom{10}{1}binom{9}{1}binom{8}{1}$? Why does splitting and adding events together have no effect?






    combinatorics







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      0



      $begingroup$


      I was recently struggling with a problem that read as so:




      A club has 30 members work in business and 30 members that are professors. In how many ways can a committee of 8 be selected that has at least 3 in business and at least 3 professors?




      For my answer, I first took care of the requirements, then grouped the rest together to get $$binom{30}{3}binom{30}{3}binom{54}{2}$$
      Which ended up being wildly wrong from the correct answer which was
      $$binom{30}{5}binom{30}{3}*2 +binom{30}{4}binom{30}{4}$$
      I once again tried splitting the last combination by case to get
      $$binom{30}{3}binom{30}{3}(binom{27}{1}binom{27}{1}+binom{27}{2}+binom{27}{2})$$
      But this just turned out to be the same as my previous answer. After doing some research, I came to the conclusion that my answers were larger than the correct one because I was splitting the event into many smaller sections, and in doing so, overcounting cases, which is why
      $$binom{10}{3}neq binom{10}{2}binom{8}1$$
      I can accept this as a general principle, as the numbers aren't equal. However, it doesn't really logically make sense to me. What am I actually overcounting by splitting up a combination such as $binom{10}{3}$ into $binom{10}{1}binom{9}{1}binom{8}{1}$? Why does splitting and adding events together have no effect?






      combinatorics







      share|cite|improve this question









      $endgroup$




      I was recently struggling with a problem that read as so:




      A club has 30 members work in business and 30 members that are professors. In how many ways can a committee of 8 be selected that has at least 3 in business and at least 3 professors?




      For my answer, I first took care of the requirements, then grouped the rest together to get $$binom{30}{3}binom{30}{3}binom{54}{2}$$
      Which ended up being wildly wrong from the correct answer which was
      $$binom{30}{5}binom{30}{3}*2 +binom{30}{4}binom{30}{4}$$
      I once again tried splitting the last combination by case to get
      $$binom{30}{3}binom{30}{3}(binom{27}{1}binom{27}{1}+binom{27}{2}+binom{27}{2})$$
      But this just turned out to be the same as my previous answer. After doing some research, I came to the conclusion that my answers were larger than the correct one because I was splitting the event into many smaller sections, and in doing so, overcounting cases, which is why
      $$binom{10}{3}neq binom{10}{2}binom{8}1$$
      I can accept this as a general principle, as the numbers aren't equal. However, it doesn't really logically make sense to me. What am I actually overcounting by splitting up a combination such as $binom{10}{3}$ into $binom{10}{1}binom{9}{1}binom{8}{1}$? Why does splitting and adding events together have no effect?






      combinatorics




      combinatorics






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      asked Jan 20 at 8:03









      CharlesCharles

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          $begingroup$

          Let $b_1,dots, b_{30}$ be the business men and $p_1,dots, p_{30}$ the professors and consider the committee $b_1,b_2,b_3,b_4,p_1,p_2,p_3,p_4$. By using the formula $$binom{30}{3}binom{30}{3}binom{54}{2}$$
          you will overcount that committee more than one time: one if you first choose the subsets ${b_1,b_2,b_3}$ and ${p_1,p_2,p_3}$ and then $b_4$ and $p_4$ among the remaining $54$ members, AND another time if you first choose the subsets ${b_1,b_2,b_4}$ and ${p_1,p_2,p_4}$ and then $b_3$ and $p_3$ among the remaining $54$ members. Actually you will overcount it $4cdot 4=16$ times.



          Instead consider the admissible compositions of the committe: $8=3+5=4+4=5+3$ where the first number is the number of businessmen and the second one the number of professors in the committee. Then the number of such committees is
          $$binom{30}{3}binom{30}{5} +binom{30}{4}binom{30}{4}+binom{30}{5}binom{30}{3}.$$






          share|cite|improve this answer











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          • $begingroup$
            @drhab Thanks!!
            $endgroup$
            – Robert Z
            Jan 20 at 8:22











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          oldest

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          active

          oldest

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          3












          $begingroup$

          Let $b_1,dots, b_{30}$ be the business men and $p_1,dots, p_{30}$ the professors and consider the committee $b_1,b_2,b_3,b_4,p_1,p_2,p_3,p_4$. By using the formula $$binom{30}{3}binom{30}{3}binom{54}{2}$$
          you will overcount that committee more than one time: one if you first choose the subsets ${b_1,b_2,b_3}$ and ${p_1,p_2,p_3}$ and then $b_4$ and $p_4$ among the remaining $54$ members, AND another time if you first choose the subsets ${b_1,b_2,b_4}$ and ${p_1,p_2,p_4}$ and then $b_3$ and $p_3$ among the remaining $54$ members. Actually you will overcount it $4cdot 4=16$ times.



          Instead consider the admissible compositions of the committe: $8=3+5=4+4=5+3$ where the first number is the number of businessmen and the second one the number of professors in the committee. Then the number of such committees is
          $$binom{30}{3}binom{30}{5} +binom{30}{4}binom{30}{4}+binom{30}{5}binom{30}{3}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @drhab Thanks!!
            $endgroup$
            – Robert Z
            Jan 20 at 8:22
















          3












          $begingroup$

          Let $b_1,dots, b_{30}$ be the business men and $p_1,dots, p_{30}$ the professors and consider the committee $b_1,b_2,b_3,b_4,p_1,p_2,p_3,p_4$. By using the formula $$binom{30}{3}binom{30}{3}binom{54}{2}$$
          you will overcount that committee more than one time: one if you first choose the subsets ${b_1,b_2,b_3}$ and ${p_1,p_2,p_3}$ and then $b_4$ and $p_4$ among the remaining $54$ members, AND another time if you first choose the subsets ${b_1,b_2,b_4}$ and ${p_1,p_2,p_4}$ and then $b_3$ and $p_3$ among the remaining $54$ members. Actually you will overcount it $4cdot 4=16$ times.



          Instead consider the admissible compositions of the committe: $8=3+5=4+4=5+3$ where the first number is the number of businessmen and the second one the number of professors in the committee. Then the number of such committees is
          $$binom{30}{3}binom{30}{5} +binom{30}{4}binom{30}{4}+binom{30}{5}binom{30}{3}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @drhab Thanks!!
            $endgroup$
            – Robert Z
            Jan 20 at 8:22














          3












          3








          3





          $begingroup$

          Let $b_1,dots, b_{30}$ be the business men and $p_1,dots, p_{30}$ the professors and consider the committee $b_1,b_2,b_3,b_4,p_1,p_2,p_3,p_4$. By using the formula $$binom{30}{3}binom{30}{3}binom{54}{2}$$
          you will overcount that committee more than one time: one if you first choose the subsets ${b_1,b_2,b_3}$ and ${p_1,p_2,p_3}$ and then $b_4$ and $p_4$ among the remaining $54$ members, AND another time if you first choose the subsets ${b_1,b_2,b_4}$ and ${p_1,p_2,p_4}$ and then $b_3$ and $p_3$ among the remaining $54$ members. Actually you will overcount it $4cdot 4=16$ times.



          Instead consider the admissible compositions of the committe: $8=3+5=4+4=5+3$ where the first number is the number of businessmen and the second one the number of professors in the committee. Then the number of such committees is
          $$binom{30}{3}binom{30}{5} +binom{30}{4}binom{30}{4}+binom{30}{5}binom{30}{3}.$$






          share|cite|improve this answer











          $endgroup$



          Let $b_1,dots, b_{30}$ be the business men and $p_1,dots, p_{30}$ the professors and consider the committee $b_1,b_2,b_3,b_4,p_1,p_2,p_3,p_4$. By using the formula $$binom{30}{3}binom{30}{3}binom{54}{2}$$
          you will overcount that committee more than one time: one if you first choose the subsets ${b_1,b_2,b_3}$ and ${p_1,p_2,p_3}$ and then $b_4$ and $p_4$ among the remaining $54$ members, AND another time if you first choose the subsets ${b_1,b_2,b_4}$ and ${p_1,p_2,p_4}$ and then $b_3$ and $p_3$ among the remaining $54$ members. Actually you will overcount it $4cdot 4=16$ times.



          Instead consider the admissible compositions of the committe: $8=3+5=4+4=5+3$ where the first number is the number of businessmen and the second one the number of professors in the committee. Then the number of such committees is
          $$binom{30}{3}binom{30}{5} +binom{30}{4}binom{30}{4}+binom{30}{5}binom{30}{3}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 20 at 8:22

























          answered Jan 20 at 8:14









          Robert ZRobert Z

          98.3k1067139




          98.3k1067139












          • $begingroup$
            @drhab Thanks!!
            $endgroup$
            – Robert Z
            Jan 20 at 8:22


















          • $begingroup$
            @drhab Thanks!!
            $endgroup$
            – Robert Z
            Jan 20 at 8:22
















          $begingroup$
          @drhab Thanks!!
          $endgroup$
          – Robert Z
          Jan 20 at 8:22




          $begingroup$
          @drhab Thanks!!
          $endgroup$
          – Robert Z
          Jan 20 at 8:22


















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