Arranging 5 pairs of brothers in ten tables

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Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.



What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?



$$\ frac 5 choose 3 cdot 3cdot 2cdot 4 choose 12 choose 110 choose 2,2,2,2,2 / 5! $$



What I thought was picking $ 3 $ pairs out of the $ 5 $ then seating each of them at a different table. Therefore, $ 5 choose 3 cdot 3 cdot 2 cdot 1$ and then I'm left with $ 4 $ people which is two pairs and picking one of them $ 4 choose 1 cdot 2 choose 1 $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.



The sample space I thought would be $ 10 choose 2,2,2,2,2 / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups.



I really don't know what I am doing wrong here.






probability combinatorics







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    up vote
    2
    down vote

    favorite












    Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.



    What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?



    $$\ frac 5 choose 3 cdot 3cdot 2cdot 4 choose 12 choose 110 choose 2,2,2,2,2 / 5! $$



    What I thought was picking $ 3 $ pairs out of the $ 5 $ then seating each of them at a different table. Therefore, $ 5 choose 3 cdot 3 cdot 2 cdot 1$ and then I'm left with $ 4 $ people which is two pairs and picking one of them $ 4 choose 1 cdot 2 choose 1 $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.



    The sample space I thought would be $ 10 choose 2,2,2,2,2 / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups.



    I really don't know what I am doing wrong here.






    probability combinatorics







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.



      What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?



      $$\ frac 5 choose 3 cdot 3cdot 2cdot 4 choose 12 choose 110 choose 2,2,2,2,2 / 5! $$



      What I thought was picking $ 3 $ pairs out of the $ 5 $ then seating each of them at a different table. Therefore, $ 5 choose 3 cdot 3 cdot 2 cdot 1$ and then I'm left with $ 4 $ people which is two pairs and picking one of them $ 4 choose 1 cdot 2 choose 1 $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.



      The sample space I thought would be $ 10 choose 2,2,2,2,2 / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups.



      I really don't know what I am doing wrong here.






      probability combinatorics







      share|cite|improve this question















      Suppose I have $10$ people composed of $5$ pairs of brothers. They are seated randomly around $5$ tables. Each table has only two places.



      What is the possibility that exactly $3$ pairs of brothers will be sitting at the same table?



      $$\ frac 5 choose 3 cdot 3cdot 2cdot 4 choose 12 choose 110 choose 2,2,2,2,2 / 5! $$



      What I thought was picking $ 3 $ pairs out of the $ 5 $ then seating each of them at a different table. Therefore, $ 5 choose 3 cdot 3 cdot 2 cdot 1$ and then I'm left with $ 4 $ people which is two pairs and picking one of them $ 4 choose 1 cdot 2 choose 1 $, because after picking one of the people, I have only two options left for the second one and none for the pairing the last two.



      The sample space I thought would be $ 10 choose 2,2,2,2,2 / 5! $ because it's the number of ways to divide $10$ people into $5$ groups of $2$ with no difference between the groups.



      I really don't know what I am doing wrong here.






      probability combinatorics




      probability combinatorics






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      share|cite|improve this question













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      edited 1 hour ago









      N. F. Taussig

      41.8k93253




      41.8k93253










      asked 1 hour ago









      bm1125

      49116




      49116




















          2 Answers
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          up vote
          3
          down vote













          Let's take as our sample space the number of ways of placing two people at each table, which is
          $$binom102, 2, 2, 2, 2 = binom102binom82binom62binom42binom22$$



          For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
          $$binom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2$$
          Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
          $$fracdbinom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2dbinom102dbinom82dbinom62dbinom42dbinom22$$






          share|cite|improve this answer



























            up vote
            3
            down vote













            First calculate the number of possible arrangements. We place 10 men in 10 places, therefore
            $$|Omega| = 10!$$
            Now we pick 3 pairs out of 5 ($binom53$), pick 3 tables out of 5 ($binom53$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)



            Therefore we have:
            $$|A|=binom53cdot binom53cdot 3!cdot 2!^3cdot 4cdot 2cdot 2!=binom53^2cdot 4!cdot 2^5$$



            Of course:
            $$P(A)=frac$$






            share|cite|improve this answer






















            • superb reasoning!
              – Satish Ramanathan
              47 mins ago










            Your Answer




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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            Let's take as our sample space the number of ways of placing two people at each table, which is
            $$binom102, 2, 2, 2, 2 = binom102binom82binom62binom42binom22$$



            For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
            $$binom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2$$
            Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
            $$fracdbinom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2dbinom102dbinom82dbinom62dbinom42dbinom22$$






            share|cite|improve this answer
























              up vote
              3
              down vote













              Let's take as our sample space the number of ways of placing two people at each table, which is
              $$binom102, 2, 2, 2, 2 = binom102binom82binom62binom42binom22$$



              For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
              $$binom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2$$
              Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
              $$fracdbinom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2dbinom102dbinom82dbinom62dbinom42dbinom22$$






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                Let's take as our sample space the number of ways of placing two people at each table, which is
                $$binom102, 2, 2, 2, 2 = binom102binom82binom62binom42binom22$$



                For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
                $$binom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2$$
                Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
                $$fracdbinom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2dbinom102dbinom82dbinom62dbinom42dbinom22$$






                share|cite|improve this answer












                Let's take as our sample space the number of ways of placing two people at each table, which is
                $$binom102, 2, 2, 2, 2 = binom102binom82binom62binom42binom22$$



                For the favorable cases, choose which three of the five tables will receive a pair of brothers. Choose a pair of brothers to sit at each of those tables. That leaves four people and two tables. Choose which of those two tables the youngest of those four people sits at. One of the other three people will be his brother. To ensure that there are exactly three pairs of brothers sitting together, we must choose one of the other two people to sit with the youngest person remaining. The final two people must sit together at the remaining table. Hence, the number of favorable cases is
                $$binom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2$$
                Thus, the probability that exactly three pairs of brothers will be seated together at the five tables if they are seated randomly is
                $$fracdbinom53 cdot 5 cdot 4 cdot 3 cdot 2 cdot 2dbinom102dbinom82dbinom62dbinom42dbinom22$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                N. F. Taussig

                41.8k93253




                41.8k93253




















                    up vote
                    3
                    down vote













                    First calculate the number of possible arrangements. We place 10 men in 10 places, therefore
                    $$|Omega| = 10!$$
                    Now we pick 3 pairs out of 5 ($binom53$), pick 3 tables out of 5 ($binom53$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)



                    Therefore we have:
                    $$|A|=binom53cdot binom53cdot 3!cdot 2!^3cdot 4cdot 2cdot 2!=binom53^2cdot 4!cdot 2^5$$



                    Of course:
                    $$P(A)=frac$$






                    share|cite|improve this answer






















                    • superb reasoning!
                      – Satish Ramanathan
                      47 mins ago














                    up vote
                    3
                    down vote













                    First calculate the number of possible arrangements. We place 10 men in 10 places, therefore
                    $$|Omega| = 10!$$
                    Now we pick 3 pairs out of 5 ($binom53$), pick 3 tables out of 5 ($binom53$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)



                    Therefore we have:
                    $$|A|=binom53cdot binom53cdot 3!cdot 2!^3cdot 4cdot 2cdot 2!=binom53^2cdot 4!cdot 2^5$$



                    Of course:
                    $$P(A)=frac$$






                    share|cite|improve this answer






















                    • superb reasoning!
                      – Satish Ramanathan
                      47 mins ago












                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    First calculate the number of possible arrangements. We place 10 men in 10 places, therefore
                    $$|Omega| = 10!$$
                    Now we pick 3 pairs out of 5 ($binom53$), pick 3 tables out of 5 ($binom53$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)



                    Therefore we have:
                    $$|A|=binom53cdot binom53cdot 3!cdot 2!^3cdot 4cdot 2cdot 2!=binom53^2cdot 4!cdot 2^5$$



                    Of course:
                    $$P(A)=frac$$






                    share|cite|improve this answer














                    First calculate the number of possible arrangements. We place 10 men in 10 places, therefore
                    $$|Omega| = 10!$$
                    Now we pick 3 pairs out of 5 ($binom53$), pick 3 tables out of 5 ($binom53$), place the pairs in these tables ($3!$), arrange each of these pairs in 2 places ($2!^3$), place first of the remaining men (they can be arranged in a query in one set way, for example alphabetically) in one of the remaining places ($4$), select one of the remaining men, which is not his brother, and sat him in opposite place ($2$), and at last - place the remaining men in 2 places ($2!$)



                    Therefore we have:
                    $$|A|=binom53cdot binom53cdot 3!cdot 2!^3cdot 4cdot 2cdot 2!=binom53^2cdot 4!cdot 2^5$$



                    Of course:
                    $$P(A)=frac$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 45 mins ago

























                    answered 54 mins ago









                    Jaroslaw Matlak

                    4,187830




                    4,187830











                    • superb reasoning!
                      – Satish Ramanathan
                      47 mins ago
















                    • superb reasoning!
                      – Satish Ramanathan
                      47 mins ago















                    superb reasoning!
                    – Satish Ramanathan
                    47 mins ago




                    superb reasoning!
                    – Satish Ramanathan
                    47 mins ago

















                     

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