Given that $binom{16}r =binom{16}{2r + 1}$, what is the value of $r$?












-1












$begingroup$



Solve
$$displaystylebinom{16}r =binom{16}{2r + 1}.$$




From what I understand:



$$displaystylefrac{16!}{r!(16-r)!}=frac{16!}{(2r+1)!(16-(2r+1))!}$$



$$implies r!(16-r)! = (2r+1)!(16-(2r+1))!$$



But I am stuck at this point.
If there is any straightforward way to solving this problem, please describe it.










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$endgroup$








  • 1




    $begingroup$
    algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:24






  • 2




    $begingroup$
    Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
    $endgroup$
    – farruhota
    Dec 15 '18 at 9:32






  • 1




    $begingroup$
    Don't forget that any negative integer $r$ is a trivial solution.
    $endgroup$
    – user10354138
    Dec 16 '18 at 8:46










  • $begingroup$
    @user10354138, and any integer greater than 16.
    $endgroup$
    – Peter Taylor
    Dec 17 '18 at 14:50
















-1












$begingroup$



Solve
$$displaystylebinom{16}r =binom{16}{2r + 1}.$$




From what I understand:



$$displaystylefrac{16!}{r!(16-r)!}=frac{16!}{(2r+1)!(16-(2r+1))!}$$



$$implies r!(16-r)! = (2r+1)!(16-(2r+1))!$$



But I am stuck at this point.
If there is any straightforward way to solving this problem, please describe it.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:24






  • 2




    $begingroup$
    Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
    $endgroup$
    – farruhota
    Dec 15 '18 at 9:32






  • 1




    $begingroup$
    Don't forget that any negative integer $r$ is a trivial solution.
    $endgroup$
    – user10354138
    Dec 16 '18 at 8:46










  • $begingroup$
    @user10354138, and any integer greater than 16.
    $endgroup$
    – Peter Taylor
    Dec 17 '18 at 14:50














-1












-1








-1


0



$begingroup$



Solve
$$displaystylebinom{16}r =binom{16}{2r + 1}.$$




From what I understand:



$$displaystylefrac{16!}{r!(16-r)!}=frac{16!}{(2r+1)!(16-(2r+1))!}$$



$$implies r!(16-r)! = (2r+1)!(16-(2r+1))!$$



But I am stuck at this point.
If there is any straightforward way to solving this problem, please describe it.










share|cite|improve this question











$endgroup$





Solve
$$displaystylebinom{16}r =binom{16}{2r + 1}.$$




From what I understand:



$$displaystylefrac{16!}{r!(16-r)!}=frac{16!}{(2r+1)!(16-(2r+1))!}$$



$$implies r!(16-r)! = (2r+1)!(16-(2r+1))!$$



But I am stuck at this point.
If there is any straightforward way to solving this problem, please describe it.







combinatorics binomial-coefficients






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 14:07









Brahadeesh

6,47942363




6,47942363










asked Dec 15 '18 at 9:24









Rajdeep BiswasRajdeep Biswas

284




284








  • 1




    $begingroup$
    algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:24






  • 2




    $begingroup$
    Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
    $endgroup$
    – farruhota
    Dec 15 '18 at 9:32






  • 1




    $begingroup$
    Don't forget that any negative integer $r$ is a trivial solution.
    $endgroup$
    – user10354138
    Dec 16 '18 at 8:46










  • $begingroup$
    @user10354138, and any integer greater than 16.
    $endgroup$
    – Peter Taylor
    Dec 17 '18 at 14:50














  • 1




    $begingroup$
    algebra.com/algebra/homework/Permutations/…
    $endgroup$
    – lab bhattacharjee
    Dec 15 '18 at 9:24






  • 2




    $begingroup$
    Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
    $endgroup$
    – farruhota
    Dec 15 '18 at 9:32






  • 1




    $begingroup$
    Don't forget that any negative integer $r$ is a trivial solution.
    $endgroup$
    – user10354138
    Dec 16 '18 at 8:46










  • $begingroup$
    @user10354138, and any integer greater than 16.
    $endgroup$
    – Peter Taylor
    Dec 17 '18 at 14:50








1




1




$begingroup$
algebra.com/algebra/homework/Permutations/…
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:24




$begingroup$
algebra.com/algebra/homework/Permutations/…
$endgroup$
– lab bhattacharjee
Dec 15 '18 at 9:24




2




2




$begingroup$
Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
$endgroup$
– farruhota
Dec 15 '18 at 9:32




$begingroup$
Use ${nchoose m}={nchoose n-m}$ and $rne 2r+1$ to find $r=16-(2r+1)Rightarrow r=5$.
$endgroup$
– farruhota
Dec 15 '18 at 9:32




1




1




$begingroup$
Don't forget that any negative integer $r$ is a trivial solution.
$endgroup$
– user10354138
Dec 16 '18 at 8:46




$begingroup$
Don't forget that any negative integer $r$ is a trivial solution.
$endgroup$
– user10354138
Dec 16 '18 at 8:46












$begingroup$
@user10354138, and any integer greater than 16.
$endgroup$
– Peter Taylor
Dec 17 '18 at 14:50




$begingroup$
@user10354138, and any integer greater than 16.
$endgroup$
– Peter Taylor
Dec 17 '18 at 14:50










2 Answers
2






active

oldest

votes


















2












$begingroup$

This is a great question! If all you have learnt is the definition of
$binom nm$ as
$$
binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
$$

then this problem seems quite imposing. How is one to check for which values of $r$
the equation
$$
r!(16-r)! = (2r+1)! (16-(2r+1))!
$$

holds?!



However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.



Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
Indeed, just plug into the formula $(dagger)$
and check that LHS = RHS.



However, you cannot immediately use this to solve
$$r = 2r+1$$
and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.



As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
$$
binom nr = binom nl.
$$

Then, using the definition we get
$$
begin{align}
&&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
iff&& l!(n-l)! &= r!(n-r)!\
iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
end{align}
$$



Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
$$
a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
$$



And, et voila! We can now conclude that $(*)$ holds if and only if
$$
r+1 = n-l+1 iff r = n-l
$$

as we wanted to show.





With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      This is a great question! If all you have learnt is the definition of
      $binom nm$ as
      $$
      binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
      $$

      then this problem seems quite imposing. How is one to check for which values of $r$
      the equation
      $$
      r!(16-r)! = (2r+1)! (16-(2r+1))!
      $$

      holds?!



      However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.



      Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
      Indeed, just plug into the formula $(dagger)$
      and check that LHS = RHS.



      However, you cannot immediately use this to solve
      $$r = 2r+1$$
      and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.



      As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
      $$
      binom nr = binom nl.
      $$

      Then, using the definition we get
      $$
      begin{align}
      &&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
      iff&& l!(n-l)! &= r!(n-r)!\
      iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
      iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
      end{align}
      $$



      Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
      $$
      a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
      $$



      And, et voila! We can now conclude that $(*)$ holds if and only if
      $$
      r+1 = n-l+1 iff r = n-l
      $$

      as we wanted to show.





      With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        This is a great question! If all you have learnt is the definition of
        $binom nm$ as
        $$
        binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
        $$

        then this problem seems quite imposing. How is one to check for which values of $r$
        the equation
        $$
        r!(16-r)! = (2r+1)! (16-(2r+1))!
        $$

        holds?!



        However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.



        Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
        Indeed, just plug into the formula $(dagger)$
        and check that LHS = RHS.



        However, you cannot immediately use this to solve
        $$r = 2r+1$$
        and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.



        As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
        $$
        binom nr = binom nl.
        $$

        Then, using the definition we get
        $$
        begin{align}
        &&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
        iff&& l!(n-l)! &= r!(n-r)!\
        iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
        iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
        end{align}
        $$



        Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
        $$
        a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
        $$



        And, et voila! We can now conclude that $(*)$ holds if and only if
        $$
        r+1 = n-l+1 iff r = n-l
        $$

        as we wanted to show.





        With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          This is a great question! If all you have learnt is the definition of
          $binom nm$ as
          $$
          binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
          $$

          then this problem seems quite imposing. How is one to check for which values of $r$
          the equation
          $$
          r!(16-r)! = (2r+1)! (16-(2r+1))!
          $$

          holds?!



          However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.



          Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
          Indeed, just plug into the formula $(dagger)$
          and check that LHS = RHS.



          However, you cannot immediately use this to solve
          $$r = 2r+1$$
          and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.



          As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
          $$
          binom nr = binom nl.
          $$

          Then, using the definition we get
          $$
          begin{align}
          &&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
          iff&& l!(n-l)! &= r!(n-r)!\
          iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
          iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
          end{align}
          $$



          Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
          $$
          a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
          $$



          And, et voila! We can now conclude that $(*)$ holds if and only if
          $$
          r+1 = n-l+1 iff r = n-l
          $$

          as we wanted to show.





          With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)






          share|cite|improve this answer









          $endgroup$



          This is a great question! If all you have learnt is the definition of
          $binom nm$ as
          $$
          binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
          $$

          then this problem seems quite imposing. How is one to check for which values of $r$
          the equation
          $$
          r!(16-r)! = (2r+1)! (16-(2r+1))!
          $$

          holds?!



          However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.



          Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
          Indeed, just plug into the formula $(dagger)$
          and check that LHS = RHS.



          However, you cannot immediately use this to solve
          $$r = 2r+1$$
          and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.



          As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
          $$
          binom nr = binom nl.
          $$

          Then, using the definition we get
          $$
          begin{align}
          &&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
          iff&& l!(n-l)! &= r!(n-r)!\
          iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
          iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
          end{align}
          $$



          Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
          $$
          a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
          $$



          And, et voila! We can now conclude that $(*)$ holds if and only if
          $$
          r+1 = n-l+1 iff r = n-l
          $$

          as we wanted to show.





          With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 12:53









          BrahadeeshBrahadeesh

          6,47942363




          6,47942363























              2












              $begingroup$

              Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$






                  share|cite|improve this answer









                  $endgroup$



                  Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 14:09









                  AmbretteOrriseyAmbretteOrrisey

                  54210




                  54210






























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