Prove $sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$ for $ngeq 2$












1












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How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







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  • 1




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    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17
















1












$begingroup$


How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17














1












1








1





$begingroup$


How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







share|cite|improve this question











$endgroup$




How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation




discrete-mathematics proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 10:21









greedoid

41.3k1150102




41.3k1150102










asked Dec 6 '18 at 10:15









TimgascdTimgascd

303




303








  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17














  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17








1




1




$begingroup$
I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
$endgroup$
– Ronald
Dec 6 '18 at 10:17




$begingroup$
I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
$endgroup$
– Ronald
Dec 6 '18 at 10:17










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint:



For $r(r-1)ne0$



$$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



Now in
$$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



put $a=b=1, m=n-2$



Some observations :




  1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


  2. The proposition trivially holds true for $n=0,1$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
    $endgroup$
    – Timgascd
    Dec 6 '18 at 10:41










  • $begingroup$
    @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 10:55












  • $begingroup$
    Thanks a lot. I got it
    $endgroup$
    – Timgascd
    Dec 6 '18 at 11:04



















1












$begingroup$

Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



$$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



$$n(n-1)2^{n-2}$$ and this is the answer to your question.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04
















      0












      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04














      0












      0








      0





      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$



      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$








      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 6 '18 at 10:20









      lab bhattacharjeelab bhattacharjee

      225k15157275




      225k15157275












      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04


















      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04
















      $begingroup$
      Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
      $endgroup$
      – Timgascd
      Dec 6 '18 at 10:41




      $begingroup$
      Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
      $endgroup$
      – Timgascd
      Dec 6 '18 at 10:41












      $begingroup$
      @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
      $endgroup$
      – lab bhattacharjee
      Dec 6 '18 at 10:55






      $begingroup$
      @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
      $endgroup$
      – lab bhattacharjee
      Dec 6 '18 at 10:55














      $begingroup$
      Thanks a lot. I got it
      $endgroup$
      – Timgascd
      Dec 6 '18 at 11:04




      $begingroup$
      Thanks a lot. I got it
      $endgroup$
      – Timgascd
      Dec 6 '18 at 11:04











      1












      $begingroup$

      Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



      Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



      $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



      On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



      $$n(n-1)2^{n-2}$$ and this is the answer to your question.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



        Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



        $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



        On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



        $$n(n-1)2^{n-2}$$ and this is the answer to your question.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



          Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



          $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



          On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



          $$n(n-1)2^{n-2}$$ and this is the answer to your question.






          share|cite|improve this answer









          $endgroup$



          Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



          Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



          $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



          On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



          $$n(n-1)2^{n-2}$$ and this is the answer to your question.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 10:20









          greedoidgreedoid

          41.3k1150102




          41.3k1150102























              0












              $begingroup$

              Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 10:20









                  lhflhf

                  165k10171395




                  165k10171395






























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