Formation of Teams in Permutation and Combination











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A class has $n$ students , we have to form a team of the students including at least two and also excluding at least two students. The number of ways of forming the team is



My Approach : To include at least two students the required ways is



C($n$, $2$) $+$ C($n$,$3$) $+$ C($n$,$4$)...........$+$C($n$,$n-2$)
But I am not understanding how to calculate the number of ways of excluding at least two students with this..............



Please help.....






combinatorics permutations combinations binomial-theorem







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    up vote
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    down vote

    favorite
    1












    A class has $n$ students , we have to form a team of the students including at least two and also excluding at least two students. The number of ways of forming the team is



    My Approach : To include at least two students the required ways is



    C($n$, $2$) $+$ C($n$,$3$) $+$ C($n$,$4$)...........$+$C($n$,$n-2$)
    But I am not understanding how to calculate the number of ways of excluding at least two students with this..............



    Please help.....






    combinatorics permutations combinations binomial-theorem







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1





      A class has $n$ students , we have to form a team of the students including at least two and also excluding at least two students. The number of ways of forming the team is



      My Approach : To include at least two students the required ways is



      C($n$, $2$) $+$ C($n$,$3$) $+$ C($n$,$4$)...........$+$C($n$,$n-2$)
      But I am not understanding how to calculate the number of ways of excluding at least two students with this..............



      Please help.....






      combinatorics permutations combinations binomial-theorem







      share|cite|improve this question















      A class has $n$ students , we have to form a team of the students including at least two and also excluding at least two students. The number of ways of forming the team is



      My Approach : To include at least two students the required ways is



      C($n$, $2$) $+$ C($n$,$3$) $+$ C($n$,$4$)...........$+$C($n$,$n-2$)
      But I am not understanding how to calculate the number of ways of excluding at least two students with this..............



      Please help.....






      combinatorics permutations combinations binomial-theorem




      combinatorics permutations combinations binomial-theorem






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      edited Jul 23 '16 at 8:53









      N. F. Taussig

      43.3k93354




      43.3k93354










      asked Jul 23 '16 at 8:09









      saladi

      13010




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          3 Answers
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          down vote



          accepted










          Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.



          The Binomial Theorem states that
          $$(x + y)^n = sum_{k = 0}^n binom{n}{k}x^{n - k}y^k$$
          We can find
          $$sum_{k = 0}^{n} binom{n}{k}$$
          by substituting $1$ for both $x$ and $y$, which yields
          $$2^n = (1 + 1)^n = sum_{k = 0}^n binom{n}{k}1^{n - k}1^k = sum_{k = 0}^n binom{n}{k}$$
          Comparing this expression with your answer
          $$binom{n}{2} + binom{n}{3} + cdots + binom{n}{n - 3} + binom{n}{n - 2} = sum_{k = 2}^{n - 2} binom{n}{k}$$
          we have
          $$sum_{k = 2}^{n - 2} binom{n}{k} = sum_{k = 0}^{n} binom{n}{k} - left[binom{n}{0} + binom{n}{1} + binom{n}{n - 1} + binom{n}{n}right]$$
          Since
          $$binom{n}{0} = binom{n}{n} = 1$$
          and
          $$binom{n}{1} = binom{n}{n - 1} = n$$
          we obtain
          $$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$






          share|cite|improve this answer




























            up vote
            2
            down vote













            There are $2^n$ possible teams altogether, and there are



            $binom{n}{1}+binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $binom{n}{0}+binom{n}{n}=2$ teams with $0$ or $n$ students;



            so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.






            share|cite|improve this answer




























              up vote
              0
              down vote













              As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.



              ⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)



              nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n



              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn



              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1



              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)



              Substituting (II) in (I),



              Number of ways = 2n – 2n – 2



              Hence, number of ways of forming a team: (1) 2n – 2n – 2



              Hope it helps!






              share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

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              3 Answers
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              active

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              up vote
              2
              down vote



              accepted










              Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.



              The Binomial Theorem states that
              $$(x + y)^n = sum_{k = 0}^n binom{n}{k}x^{n - k}y^k$$
              We can find
              $$sum_{k = 0}^{n} binom{n}{k}$$
              by substituting $1$ for both $x$ and $y$, which yields
              $$2^n = (1 + 1)^n = sum_{k = 0}^n binom{n}{k}1^{n - k}1^k = sum_{k = 0}^n binom{n}{k}$$
              Comparing this expression with your answer
              $$binom{n}{2} + binom{n}{3} + cdots + binom{n}{n - 3} + binom{n}{n - 2} = sum_{k = 2}^{n - 2} binom{n}{k}$$
              we have
              $$sum_{k = 2}^{n - 2} binom{n}{k} = sum_{k = 0}^{n} binom{n}{k} - left[binom{n}{0} + binom{n}{1} + binom{n}{n - 1} + binom{n}{n}right]$$
              Since
              $$binom{n}{0} = binom{n}{n} = 1$$
              and
              $$binom{n}{1} = binom{n}{n - 1} = n$$
              we obtain
              $$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted










                Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.



                The Binomial Theorem states that
                $$(x + y)^n = sum_{k = 0}^n binom{n}{k}x^{n - k}y^k$$
                We can find
                $$sum_{k = 0}^{n} binom{n}{k}$$
                by substituting $1$ for both $x$ and $y$, which yields
                $$2^n = (1 + 1)^n = sum_{k = 0}^n binom{n}{k}1^{n - k}1^k = sum_{k = 0}^n binom{n}{k}$$
                Comparing this expression with your answer
                $$binom{n}{2} + binom{n}{3} + cdots + binom{n}{n - 3} + binom{n}{n - 2} = sum_{k = 2}^{n - 2} binom{n}{k}$$
                we have
                $$sum_{k = 2}^{n - 2} binom{n}{k} = sum_{k = 0}^{n} binom{n}{k} - left[binom{n}{0} + binom{n}{1} + binom{n}{n - 1} + binom{n}{n}right]$$
                Since
                $$binom{n}{0} = binom{n}{n} = 1$$
                and
                $$binom{n}{1} = binom{n}{n - 1} = n$$
                we obtain
                $$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$






                share|cite|improve this answer























                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.



                  The Binomial Theorem states that
                  $$(x + y)^n = sum_{k = 0}^n binom{n}{k}x^{n - k}y^k$$
                  We can find
                  $$sum_{k = 0}^{n} binom{n}{k}$$
                  by substituting $1$ for both $x$ and $y$, which yields
                  $$2^n = (1 + 1)^n = sum_{k = 0}^n binom{n}{k}1^{n - k}1^k = sum_{k = 0}^n binom{n}{k}$$
                  Comparing this expression with your answer
                  $$binom{n}{2} + binom{n}{3} + cdots + binom{n}{n - 3} + binom{n}{n - 2} = sum_{k = 2}^{n - 2} binom{n}{k}$$
                  we have
                  $$sum_{k = 2}^{n - 2} binom{n}{k} = sum_{k = 0}^{n} binom{n}{k} - left[binom{n}{0} + binom{n}{1} + binom{n}{n - 1} + binom{n}{n}right]$$
                  Since
                  $$binom{n}{0} = binom{n}{n} = 1$$
                  and
                  $$binom{n}{1} = binom{n}{n - 1} = n$$
                  we obtain
                  $$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$






                  share|cite|improve this answer












                  Your interpretation that the number of students selected is at least $2$ and at most $n - 2$ is correct, as is your answer. We can use the Binomial Theorem to obtain a closed form.



                  The Binomial Theorem states that
                  $$(x + y)^n = sum_{k = 0}^n binom{n}{k}x^{n - k}y^k$$
                  We can find
                  $$sum_{k = 0}^{n} binom{n}{k}$$
                  by substituting $1$ for both $x$ and $y$, which yields
                  $$2^n = (1 + 1)^n = sum_{k = 0}^n binom{n}{k}1^{n - k}1^k = sum_{k = 0}^n binom{n}{k}$$
                  Comparing this expression with your answer
                  $$binom{n}{2} + binom{n}{3} + cdots + binom{n}{n - 3} + binom{n}{n - 2} = sum_{k = 2}^{n - 2} binom{n}{k}$$
                  we have
                  $$sum_{k = 2}^{n - 2} binom{n}{k} = sum_{k = 0}^{n} binom{n}{k} - left[binom{n}{0} + binom{n}{1} + binom{n}{n - 1} + binom{n}{n}right]$$
                  Since
                  $$binom{n}{0} = binom{n}{n} = 1$$
                  and
                  $$binom{n}{1} = binom{n}{n - 1} = n$$
                  we obtain
                  $$sum_{k = 2}^{n - 2} binom{n}{k} = 2^n - 2n - 2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 23 '16 at 8:53









                  N. F. Taussig

                  43.3k93354




                  43.3k93354






















                      up vote
                      2
                      down vote













                      There are $2^n$ possible teams altogether, and there are



                      $binom{n}{1}+binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $binom{n}{0}+binom{n}{n}=2$ teams with $0$ or $n$ students;



                      so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        There are $2^n$ possible teams altogether, and there are



                        $binom{n}{1}+binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $binom{n}{0}+binom{n}{n}=2$ teams with $0$ or $n$ students;



                        so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          There are $2^n$ possible teams altogether, and there are



                          $binom{n}{1}+binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $binom{n}{0}+binom{n}{n}=2$ teams with $0$ or $n$ students;



                          so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.






                          share|cite|improve this answer












                          There are $2^n$ possible teams altogether, and there are



                          $binom{n}{1}+binom{n}{n-1}=2n$ teams with $1$ or $n-1$ students and $binom{n}{0}+binom{n}{n}=2$ teams with $0$ or $n$ students;



                          so there are $2^n-2n-2$ teams including at least 2 students and excluding at least 2 students.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jul 23 '16 at 20:20









                          user84413

                          22.8k11848




                          22.8k11848






















                              up vote
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                              down vote













                              As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.



                              ⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)



                              nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)



                              Substituting (II) in (I),



                              Number of ways = 2n – 2n – 2



                              Hence, number of ways of forming a team: (1) 2n – 2n – 2



                              Hope it helps!






                              share|cite|improve this answer





















                              • Please use Latex. Learn latex from here.
                                – tarit goswami
                                Nov 22 at 19:32

















                              up vote
                              0
                              down vote













                              As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.



                              ⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)



                              nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)



                              Substituting (II) in (I),



                              Number of ways = 2n – 2n – 2



                              Hence, number of ways of forming a team: (1) 2n – 2n – 2



                              Hope it helps!






                              share|cite|improve this answer





















                              • Please use Latex. Learn latex from here.
                                – tarit goswami
                                Nov 22 at 19:32















                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.



                              ⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)



                              nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)



                              Substituting (II) in (I),



                              Number of ways = 2n – 2n – 2



                              Hence, number of ways of forming a team: (1) 2n – 2n – 2



                              Hope it helps!






                              share|cite|improve this answer












                              As team should consist of at least 2 and at most (n – 2) students out of n students, so required number of ways = 2 students or 3 students ........or (n – 3) students or (n – 2)students.



                              ⇒ number or ways = nC2 + nC3 + nC4 + ...... nCn – 3 + nCn – 2 ...... (I)



                              nC0 + nC1 + nC2 + nC3 + ........ + nCn – 2 + nCn – 1 + nCn = 2n



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – nC0 – nC1 – nCn–1 – nCn



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 1 – n – n – 1



                              ⇒ nC2 + nC3 + ......... + nCn – 2 = 2n – 2n – 2 ......... (II)



                              Substituting (II) in (I),



                              Number of ways = 2n – 2n – 2



                              Hence, number of ways of forming a team: (1) 2n – 2n – 2



                              Hope it helps!







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 22 at 19:27









                              Krishna

                              1




                              1












                              • Please use Latex. Learn latex from here.
                                – tarit goswami
                                Nov 22 at 19:32




















                              • Please use Latex. Learn latex from here.
                                – tarit goswami
                                Nov 22 at 19:32


















                              Please use Latex. Learn latex from here.
                              – tarit goswami
                              Nov 22 at 19:32






                              Please use Latex. Learn latex from here.
                              – tarit goswami
                              Nov 22 at 19:32




















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