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Combinatorics problem, right solution?



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manaragraph theory /combinatorics committee existenceDifferent Perspectives of Multinomial Theorem & PartitionsCombination with at least two people.Is the assumption of the question violates itselfCounting word problemWhat is the minimum number of people required?Combinatorics: How should we count lists?combinatorics select committee different jobsIntuitive understanding of Pascal's IdentityWhy aren't these two solutions equivalent? Combinatorics problem












3












$begingroup$


We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.



For the remaining two places, I could have $2$ more people of a single profession. This is $binom{5}{2}+binom{6}{2}+binom{3}{2}$ possibilities.



I could also have two people of different professions; a doctor and a laywer, $binom{5}{1}binom{3}{1}$; a doctor and an engineer, $binom{6}{1}binom{3}{1}$; or an engineer and a laywer $binom{6}{1}binom{5}{1}$.



This adds up to $binom{5}{2}+binom{6}{2}+binom{3}{2}+binom{5}{1}binom{3}{1}+binom{6}{1}binom{3}{1}+binom{6}{1}binom{5}{1}=91$ possible committees.



I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.



Thanks in advance!










share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected it, I apologize. The result is the same though, it was just a typo.
    $endgroup$
    – Lafinur
    2 hours ago






  • 1




    $begingroup$
    That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
    $endgroup$
    – lulu
    2 hours ago










  • $begingroup$
    When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
    $endgroup$
    – Austin Mohr
    40 mins ago


















3












$begingroup$


We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.



For the remaining two places, I could have $2$ more people of a single profession. This is $binom{5}{2}+binom{6}{2}+binom{3}{2}$ possibilities.



I could also have two people of different professions; a doctor and a laywer, $binom{5}{1}binom{3}{1}$; a doctor and an engineer, $binom{6}{1}binom{3}{1}$; or an engineer and a laywer $binom{6}{1}binom{5}{1}$.



This adds up to $binom{5}{2}+binom{6}{2}+binom{3}{2}+binom{5}{1}binom{3}{1}+binom{6}{1}binom{3}{1}+binom{6}{1}binom{5}{1}=91$ possible committees.



I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.



Thanks in advance!










share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected it, I apologize. The result is the same though, it was just a typo.
    $endgroup$
    – Lafinur
    2 hours ago






  • 1




    $begingroup$
    That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
    $endgroup$
    – lulu
    2 hours ago










  • $begingroup$
    When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
    $endgroup$
    – Austin Mohr
    40 mins ago
















3












3








3





$begingroup$


We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.



For the remaining two places, I could have $2$ more people of a single profession. This is $binom{5}{2}+binom{6}{2}+binom{3}{2}$ possibilities.



I could also have two people of different professions; a doctor and a laywer, $binom{5}{1}binom{3}{1}$; a doctor and an engineer, $binom{6}{1}binom{3}{1}$; or an engineer and a laywer $binom{6}{1}binom{5}{1}$.



This adds up to $binom{5}{2}+binom{6}{2}+binom{3}{2}+binom{5}{1}binom{3}{1}+binom{6}{1}binom{3}{1}+binom{6}{1}binom{5}{1}=91$ possible committees.



I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.



Thanks in advance!










share|cite|improve this question











$endgroup$




We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.



For the remaining two places, I could have $2$ more people of a single profession. This is $binom{5}{2}+binom{6}{2}+binom{3}{2}$ possibilities.



I could also have two people of different professions; a doctor and a laywer, $binom{5}{1}binom{3}{1}$; a doctor and an engineer, $binom{6}{1}binom{3}{1}$; or an engineer and a laywer $binom{6}{1}binom{5}{1}$.



This adds up to $binom{5}{2}+binom{6}{2}+binom{3}{2}+binom{5}{1}binom{3}{1}+binom{6}{1}binom{3}{1}+binom{6}{1}binom{5}{1}=91$ possible committees.



I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.



Thanks in advance!







combinatorics discrete-mathematics order-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







Lafinur

















asked 2 hours ago









LafinurLafinur

18211




18211












  • $begingroup$
    I corrected it, I apologize. The result is the same though, it was just a typo.
    $endgroup$
    – Lafinur
    2 hours ago






  • 1




    $begingroup$
    That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
    $endgroup$
    – lulu
    2 hours ago










  • $begingroup$
    When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
    $endgroup$
    – Austin Mohr
    40 mins ago




















  • $begingroup$
    I corrected it, I apologize. The result is the same though, it was just a typo.
    $endgroup$
    – Lafinur
    2 hours ago






  • 1




    $begingroup$
    That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
    $endgroup$
    – lulu
    2 hours ago










  • $begingroup$
    When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
    $endgroup$
    – Austin Mohr
    40 mins ago


















$begingroup$
I corrected it, I apologize. The result is the same though, it was just a typo.
$endgroup$
– Lafinur
2 hours ago




$begingroup$
I corrected it, I apologize. The result is the same though, it was just a typo.
$endgroup$
– Lafinur
2 hours ago




1




1




$begingroup$
That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
$endgroup$
– lulu
2 hours ago




$begingroup$
That number seems far too small. If you ignore the restriction there are $binom {17}5=6188$ possible combinations.
$endgroup$
– lulu
2 hours ago












$begingroup$
When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
$endgroup$
– Austin Mohr
40 mins ago






$begingroup$
When you say, for example, "pick a lawyer and then later possibly pick more lawyers", you're guaranteed to introduce an overcount. You might pick LawyerA first and then LawyerB or you might pick LawyerB first and then LawyerA. These are precisely the same subset of lawyers, but your counting scheme considers them to be different.
$endgroup$
– Austin Mohr
40 mins ago












4 Answers
4






active

oldest

votes


















7












$begingroup$

I think Inclusion Exclusion is an easier approach.



If we ignore the restriction, there are $binom {17}5$ ways to choose the group.



We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.



We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$



Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.



    The result is
    $$
    binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
    $$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Let $A={(l,e,d)in{1,2,3,4,5,6}times{1,2,3,4,5,6,7}times{1,2,3,4}mid l+e+d=5}$



      Then to be found is $$sum_{(l,e,d)in A}binom{6}{l}binom{7}{e}binom{4}{d}$$



      Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:




      • $5=3+1+1$

      • $5=2+2+1$

      • $5=2+1+2$

      • $5=1+3+1$

      • $5=1+2+2$

      • $5=1+1+3$


      This provides you a view on set $A$ and shows that the summation has $6$ terms.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        @lulu Thank you lulu! Repaired.
        $endgroup$
        – drhab
        2 hours ago



















      1












      $begingroup$

      There are 6 types of possible committees:
      begin{array} {|r|r|r|}
      hline
      3&1&1 \
      hline
      1&3&1 \
      hline
      1&1&3 \
      hline
      2&2&1 \
      hline
      2&1&2 \
      hline
      1&2&2 \
      hline
      end{array}



      For each type of committee, it is necessary to calculate the different groups of people that compose it:



      begin{array} {|c|c|c|c|}
      hline
      3&1&1& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
      hline
      1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
      hline
      1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
      hline
      2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
      hline
      2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
      hline
      1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
      hline
      end{array}



      Adding 6 gives a total of different committees:



      $$ 560+840+168+1260+630+756 = 4214 $$






      share|cite|improve this answer









      $endgroup$














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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        I think Inclusion Exclusion is an easier approach.



        If we ignore the restriction, there are $binom {17}5$ ways to choose the group.



        We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.



        We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$



        Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          I think Inclusion Exclusion is an easier approach.



          If we ignore the restriction, there are $binom {17}5$ ways to choose the group.



          We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.



          We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$



          Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            I think Inclusion Exclusion is an easier approach.



            If we ignore the restriction, there are $binom {17}5$ ways to choose the group.



            We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.



            We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$



            Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$






            share|cite|improve this answer









            $endgroup$



            I think Inclusion Exclusion is an easier approach.



            If we ignore the restriction, there are $binom {17}5$ ways to choose the group.



            We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.



            We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$



            Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            lulululu

            44.1k25182




            44.1k25182























                3












                $begingroup$

                I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.



                The result is
                $$
                binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
                $$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.



                  The result is
                  $$
                  binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.



                    The result is
                    $$
                    binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.



                    The result is
                    $$
                    binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Mike EarnestMike Earnest

                    28.6k22255




                    28.6k22255























                        1












                        $begingroup$

                        Let $A={(l,e,d)in{1,2,3,4,5,6}times{1,2,3,4,5,6,7}times{1,2,3,4}mid l+e+d=5}$



                        Then to be found is $$sum_{(l,e,d)in A}binom{6}{l}binom{7}{e}binom{4}{d}$$



                        Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:




                        • $5=3+1+1$

                        • $5=2+2+1$

                        • $5=2+1+2$

                        • $5=1+3+1$

                        • $5=1+2+2$

                        • $5=1+1+3$


                        This provides you a view on set $A$ and shows that the summation has $6$ terms.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          @lulu Thank you lulu! Repaired.
                          $endgroup$
                          – drhab
                          2 hours ago
















                        1












                        $begingroup$

                        Let $A={(l,e,d)in{1,2,3,4,5,6}times{1,2,3,4,5,6,7}times{1,2,3,4}mid l+e+d=5}$



                        Then to be found is $$sum_{(l,e,d)in A}binom{6}{l}binom{7}{e}binom{4}{d}$$



                        Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:




                        • $5=3+1+1$

                        • $5=2+2+1$

                        • $5=2+1+2$

                        • $5=1+3+1$

                        • $5=1+2+2$

                        • $5=1+1+3$


                        This provides you a view on set $A$ and shows that the summation has $6$ terms.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          @lulu Thank you lulu! Repaired.
                          $endgroup$
                          – drhab
                          2 hours ago














                        1












                        1








                        1





                        $begingroup$

                        Let $A={(l,e,d)in{1,2,3,4,5,6}times{1,2,3,4,5,6,7}times{1,2,3,4}mid l+e+d=5}$



                        Then to be found is $$sum_{(l,e,d)in A}binom{6}{l}binom{7}{e}binom{4}{d}$$



                        Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:




                        • $5=3+1+1$

                        • $5=2+2+1$

                        • $5=2+1+2$

                        • $5=1+3+1$

                        • $5=1+2+2$

                        • $5=1+1+3$


                        This provides you a view on set $A$ and shows that the summation has $6$ terms.






                        share|cite|improve this answer











                        $endgroup$



                        Let $A={(l,e,d)in{1,2,3,4,5,6}times{1,2,3,4,5,6,7}times{1,2,3,4}mid l+e+d=5}$



                        Then to be found is $$sum_{(l,e,d)in A}binom{6}{l}binom{7}{e}binom{4}{d}$$



                        Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:




                        • $5=3+1+1$

                        • $5=2+2+1$

                        • $5=2+1+2$

                        • $5=1+3+1$

                        • $5=1+2+2$

                        • $5=1+1+3$


                        This provides you a view on set $A$ and shows that the summation has $6$ terms.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 2 hours ago

























                        answered 2 hours ago









                        drhabdrhab

                        105k545136




                        105k545136












                        • $begingroup$
                          @lulu Thank you lulu! Repaired.
                          $endgroup$
                          – drhab
                          2 hours ago


















                        • $begingroup$
                          @lulu Thank you lulu! Repaired.
                          $endgroup$
                          – drhab
                          2 hours ago
















                        $begingroup$
                        @lulu Thank you lulu! Repaired.
                        $endgroup$
                        – drhab
                        2 hours ago




                        $begingroup$
                        @lulu Thank you lulu! Repaired.
                        $endgroup$
                        – drhab
                        2 hours ago











                        1












                        $begingroup$

                        There are 6 types of possible committees:
                        begin{array} {|r|r|r|}
                        hline
                        3&1&1 \
                        hline
                        1&3&1 \
                        hline
                        1&1&3 \
                        hline
                        2&2&1 \
                        hline
                        2&1&2 \
                        hline
                        1&2&2 \
                        hline
                        end{array}



                        For each type of committee, it is necessary to calculate the different groups of people that compose it:



                        begin{array} {|c|c|c|c|}
                        hline
                        3&1&1& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
                        hline
                        1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
                        hline
                        1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
                        hline
                        2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
                        hline
                        2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
                        hline
                        1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
                        hline
                        end{array}



                        Adding 6 gives a total of different committees:



                        $$ 560+840+168+1260+630+756 = 4214 $$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          There are 6 types of possible committees:
                          begin{array} {|r|r|r|}
                          hline
                          3&1&1 \
                          hline
                          1&3&1 \
                          hline
                          1&1&3 \
                          hline
                          2&2&1 \
                          hline
                          2&1&2 \
                          hline
                          1&2&2 \
                          hline
                          end{array}



                          For each type of committee, it is necessary to calculate the different groups of people that compose it:



                          begin{array} {|c|c|c|c|}
                          hline
                          3&1&1& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
                          hline
                          1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
                          hline
                          1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
                          hline
                          2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
                          hline
                          2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
                          hline
                          1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
                          hline
                          end{array}



                          Adding 6 gives a total of different committees:



                          $$ 560+840+168+1260+630+756 = 4214 $$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            There are 6 types of possible committees:
                            begin{array} {|r|r|r|}
                            hline
                            3&1&1 \
                            hline
                            1&3&1 \
                            hline
                            1&1&3 \
                            hline
                            2&2&1 \
                            hline
                            2&1&2 \
                            hline
                            1&2&2 \
                            hline
                            end{array}



                            For each type of committee, it is necessary to calculate the different groups of people that compose it:



                            begin{array} {|c|c|c|c|}
                            hline
                            3&1&1& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
                            hline
                            1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
                            hline
                            1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
                            hline
                            2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
                            hline
                            2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
                            hline
                            1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
                            hline
                            end{array}



                            Adding 6 gives a total of different committees:



                            $$ 560+840+168+1260+630+756 = 4214 $$






                            share|cite|improve this answer









                            $endgroup$



                            There are 6 types of possible committees:
                            begin{array} {|r|r|r|}
                            hline
                            3&1&1 \
                            hline
                            1&3&1 \
                            hline
                            1&1&3 \
                            hline
                            2&2&1 \
                            hline
                            2&1&2 \
                            hline
                            1&2&2 \
                            hline
                            end{array}



                            For each type of committee, it is necessary to calculate the different groups of people that compose it:



                            begin{array} {|c|c|c|c|}
                            hline
                            3&1&1& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
                            hline
                            1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
                            hline
                            1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
                            hline
                            2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
                            hline
                            2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
                            hline
                            1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
                            hline
                            end{array}



                            Adding 6 gives a total of different committees:



                            $$ 560+840+168+1260+630+756 = 4214 $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Angel MorenoAngel Moreno

                            45225




                            45225






























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