In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at...












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$begingroup$




  • In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?




I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below



$$binom{6}{2}$$



This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that



$$binom{4}{1}binom{3}{1}$$



We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.



$$binom{6}{2}binom{4}{2}binom{2}{2}$$



Finally, I'm combining all these results and we have that



$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$



I've gone wrong somewhere. Could you assist me with that?






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  • 1




    $begingroup$
    Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 19:27
















1












$begingroup$




  • In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?




I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below



$$binom{6}{2}$$



This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that



$$binom{4}{1}binom{3}{1}$$



We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.



$$binom{6}{2}binom{4}{2}binom{2}{2}$$



Finally, I'm combining all these results and we have that



$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$



I've gone wrong somewhere. Could you assist me with that?






combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 19:27














1












1








1





$begingroup$




  • In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?




I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below



$$binom{6}{2}$$



This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that



$$binom{4}{1}binom{3}{1}$$



We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.



$$binom{6}{2}binom{4}{2}binom{2}{2}$$



Finally, I'm combining all these results and we have that



$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$



I've gone wrong somewhere. Could you assist me with that?






combinations







share|cite|improve this question









$endgroup$






  • In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?




I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below



$$binom{6}{2}$$



This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that



$$binom{4}{1}binom{3}{1}$$



We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least.



$$binom{6}{2}binom{4}{2}binom{2}{2}$$



Finally, I'm combining all these results and we have that



$$binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{1}binom{3}{1} = 90 + 180 = 270$$



I've gone wrong somewhere. Could you assist me with that?






combinations




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asked Nov 27 '18 at 19:18









EnzoEnzo

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1677








  • 1




    $begingroup$
    Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 19:27














  • 1




    $begingroup$
    Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 19:27








1




1




$begingroup$
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
$endgroup$
– Shubham Johri
Nov 27 '18 at 19:27




$begingroup$
Both Michael and Carl can't get 1 toy each, because then you'd only end up distributing 4 toys, not 6
$endgroup$
– Shubham Johri
Nov 27 '18 at 19:27










2 Answers
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$begingroup$

Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.



If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
$$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    John get 2, that is ${6choose 2}$.



    Now from the rest of 4 Carl gets:



    $1$ toy, that is ${4choose 1}$ way or



    $3$ toys, that is ${4choose 3}$ ways or



    $2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      1












      $begingroup$

      Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.



      If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
      $$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$






      share|cite|improve this answer









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        1












        $begingroup$

        Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.



        If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
        $$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.



          If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
          $$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$






          share|cite|improve this answer









          $endgroup$



          Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.



          If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys. Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is
          $$binom{6}{2}binom{4}{1}binom{3}{3} + binom{6}{2}binom{4}{2}binom{2}{2} + binom{6}{2}binom{4}{3}binom{1}{1}$$







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Nov 27 '18 at 19:28









          N. F. TaussigN. F. Taussig

          44k93356




          44k93356























              1












              $begingroup$

              John get 2, that is ${6choose 2}$.



              Now from the rest of 4 Carl gets:



              $1$ toy, that is ${4choose 1}$ way or



              $3$ toys, that is ${4choose 3}$ ways or



              $2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                John get 2, that is ${6choose 2}$.



                Now from the rest of 4 Carl gets:



                $1$ toy, that is ${4choose 1}$ way or



                $3$ toys, that is ${4choose 3}$ ways or



                $2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  John get 2, that is ${6choose 2}$.



                  Now from the rest of 4 Carl gets:



                  $1$ toy, that is ${4choose 1}$ way or



                  $3$ toys, that is ${4choose 3}$ ways or



                  $2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.






                  share|cite|improve this answer









                  $endgroup$



                  John get 2, that is ${6choose 2}$.



                  Now from the rest of 4 Carl gets:



                  $1$ toy, that is ${4choose 1}$ way or



                  $3$ toys, that is ${4choose 3}$ ways or



                  $2$ toys, that is ${4choose 2}$ ways. So the number of all distributions is $${6choose 2}Big( {4choose 1}+ {4choose 2} + {4choose 3}Big)$$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 19:27









                  greedoidgreedoid

                  40.2k114799




                  40.2k114799






























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