Prove by induction that $sumlimits_{k=m}^{ n}{nchoose k}{kchoose m}={nchoose m}2^{n-m}$.












1















Prove by induction that $displaystylesumlimits_{k=m}^{ n}{nchoose k}{kchoose m}={nchoose m}2^{n-m}$.




I can't figure out what is the base case. Could someone show the steps?






combinatorics discrete-mathematics summation induction binomial-coefficients







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  • 2




    The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
    – Daniel Fischer♦
    Jun 20 '15 at 22:58










  • If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
    – Martin Sleziak
    Nov 16 '16 at 4:49
















1















Prove by induction that $displaystylesumlimits_{k=m}^{ n}{nchoose k}{kchoose m}={nchoose m}2^{n-m}$.




I can't figure out what is the base case. Could someone show the steps?






combinatorics discrete-mathematics summation induction binomial-coefficients







share|cite|improve this question




















  • 2




    The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
    – Daniel Fischer♦
    Jun 20 '15 at 22:58










  • If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
    – Martin Sleziak
    Nov 16 '16 at 4:49














1












1








1


1






Prove by induction that $displaystylesumlimits_{k=m}^{ n}{nchoose k}{kchoose m}={nchoose m}2^{n-m}$.




I can't figure out what is the base case. Could someone show the steps?






combinatorics discrete-mathematics summation induction binomial-coefficients







share|cite|improve this question
















Prove by induction that $displaystylesumlimits_{k=m}^{ n}{nchoose k}{kchoose m}={nchoose m}2^{n-m}$.




I can't figure out what is the base case. Could someone show the steps?






combinatorics discrete-mathematics summation induction binomial-coefficients




combinatorics discrete-mathematics summation induction binomial-coefficients






share|cite|improve this question















share|cite|improve this question













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edited Dec 10 '18 at 9:51









Batominovski

33.8k33292




33.8k33292










asked Jun 20 '15 at 22:44









user300045

1,349418




1,349418








  • 2




    The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
    – Daniel Fischer♦
    Jun 20 '15 at 22:58










  • If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
    – Martin Sleziak
    Nov 16 '16 at 4:49














  • 2




    The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
    – Daniel Fischer♦
    Jun 20 '15 at 22:58










  • If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
    – Martin Sleziak
    Nov 16 '16 at 4:49








2




2




The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
– Daniel Fischer♦
Jun 20 '15 at 22:58




The obvious base case is $n = m$, and then induction on $n$. But a little rewriting of the terms of the sum makes the proof much [I think, maybe the induction is easier than I expect] easier.
– Daniel Fischer♦
Jun 20 '15 at 22:58












If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
– Martin Sleziak
Nov 16 '16 at 4:49




If you do not insist on induction, there are several posts with the same sum: math.stackexchange.com/questions/203198/… math.stackexchange.com/questions/380555/… math.stackexchange.com/questions/1640706/… I found the above questions using Approach0.
– Martin Sleziak
Nov 16 '16 at 4:49










2 Answers
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We may write the required equality as
$$sum_{k=0}^n,binom{n}{k},binom{k}{m}=binom{n}{m},2^{n-m}$$
for all integers $n,mgeq 0$, using the convention that $displaystylebinom{p}{q}=0$ for integers $p,qgeq 0$ such that $p<q$. We shall prove by induction on $m$ the generalized statement:
$$sum_{k=0}^n,binom{n}{k},binom{k}{m},x^{k-m}=binom{n}{m},(1+x)^{n-m},,$$
where $x:=1$ leads to the required equality. The basis step is $m=0$, where we have the well known Binomial Theorem $$sum_{k=0}^n,binom{n}{k},binom{k}{0},x^{k-0}=sum_{k=0}^n,binom{n}{k},x^k=(1+x)^n=binom{n}{0},(1+x)^{n-0},.$$



Now, suppose the identity is true when $m=s$ for some integer $sgeq 0$ (and for all integer $ngeq 0$). Therefore, we have $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},x^{k-s}=binom{n}{s},(1+x)^{n-s},.$$
for all integer $ngeq 0$. Taking the derivative with respect to $x$, we obtain
$$sum_{k=0}^{n},binom{n}{k},binom{k}{s},(k-s),x^{k-s-1}=binom{n}{s},(n-s),(1+x)^{n-s-1},.$$
Dividing both sides by $s+1$ to get
$$sum_{k=0}^n,binom{n}{k},binom{k}{s},frac{k-s}{s+1},x^{k-(s+1)}=binom{n}{s},frac{n-s}{s+1},(1+x)^{n-(s+1)},.$$
Since $displaystylebinom{p}{s},frac{p-s}{s+1}=binom{p}{s+1}$ for every integer $pgeq 0$, we get the desired equality
$$sum_{k=0}^n,binom{n}{k},binom{k}{s+1},x^{k-(s+1)}=binom{n}{s+1},(1+x)^{n-(s+1)},,$$
establishing that the required equality holds for $m=s+1$. By induction, we are done.




Alternatively, consider a task of choosing a committee from $n$ people, where $m$ committee members are given the chief status. If the committee has $k$ members, then there are $displaystylebinom{n}{k}$ ways to choose the committee and $displaystylebinom{k}{m}$ ways to choose the chiefs. Hence, the number of ways to perform this task is precisely $displaystylesumlimits_{k=m}^n,binom{n}{k},binom{k}{m}$. On the other hand, we may chose $m$ chiefs first, whereby there are are $displaystylebinom{n}{m}$ ways to do so. The non-chief committee members are then selected from $n-m$ remaining people, which can be done in $2^{n-m}$ ways. Hence, we can perform this task in $displaystylebinom{n}{m},2^{n-m}$ ways. The proof is now complete.






share|cite|improve this answer























  • +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
    – David Holden
    Jun 21 '15 at 0:38



















4














you can get by without induction if you observe:
$$
binom{n}{k} binom{k}{m} = binom{n}{m} binom{n-m}{k-m}
$$
then
$$
sum_{k=m}^n binom{n}{k} binom{k}{m} = binom{n}{m}sum_{j=0}^{n-m}binom{n-m}{j} = binom{n}{m}2^{n-m}
$$






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    We may write the required equality as
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m}=binom{n}{m},2^{n-m}$$
    for all integers $n,mgeq 0$, using the convention that $displaystylebinom{p}{q}=0$ for integers $p,qgeq 0$ such that $p<q$. We shall prove by induction on $m$ the generalized statement:
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m},x^{k-m}=binom{n}{m},(1+x)^{n-m},,$$
    where $x:=1$ leads to the required equality. The basis step is $m=0$, where we have the well known Binomial Theorem $$sum_{k=0}^n,binom{n}{k},binom{k}{0},x^{k-0}=sum_{k=0}^n,binom{n}{k},x^k=(1+x)^n=binom{n}{0},(1+x)^{n-0},.$$



    Now, suppose the identity is true when $m=s$ for some integer $sgeq 0$ (and for all integer $ngeq 0$). Therefore, we have $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},x^{k-s}=binom{n}{s},(1+x)^{n-s},.$$
    for all integer $ngeq 0$. Taking the derivative with respect to $x$, we obtain
    $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},(k-s),x^{k-s-1}=binom{n}{s},(n-s),(1+x)^{n-s-1},.$$
    Dividing both sides by $s+1$ to get
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s},frac{k-s}{s+1},x^{k-(s+1)}=binom{n}{s},frac{n-s}{s+1},(1+x)^{n-(s+1)},.$$
    Since $displaystylebinom{p}{s},frac{p-s}{s+1}=binom{p}{s+1}$ for every integer $pgeq 0$, we get the desired equality
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s+1},x^{k-(s+1)}=binom{n}{s+1},(1+x)^{n-(s+1)},,$$
    establishing that the required equality holds for $m=s+1$. By induction, we are done.




    Alternatively, consider a task of choosing a committee from $n$ people, where $m$ committee members are given the chief status. If the committee has $k$ members, then there are $displaystylebinom{n}{k}$ ways to choose the committee and $displaystylebinom{k}{m}$ ways to choose the chiefs. Hence, the number of ways to perform this task is precisely $displaystylesumlimits_{k=m}^n,binom{n}{k},binom{k}{m}$. On the other hand, we may chose $m$ chiefs first, whereby there are are $displaystylebinom{n}{m}$ ways to do so. The non-chief committee members are then selected from $n-m$ remaining people, which can be done in $2^{n-m}$ ways. Hence, we can perform this task in $displaystylebinom{n}{m},2^{n-m}$ ways. The proof is now complete.






    share|cite|improve this answer























    • +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
      – David Holden
      Jun 21 '15 at 0:38
















    5














    We may write the required equality as
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m}=binom{n}{m},2^{n-m}$$
    for all integers $n,mgeq 0$, using the convention that $displaystylebinom{p}{q}=0$ for integers $p,qgeq 0$ such that $p<q$. We shall prove by induction on $m$ the generalized statement:
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m},x^{k-m}=binom{n}{m},(1+x)^{n-m},,$$
    where $x:=1$ leads to the required equality. The basis step is $m=0$, where we have the well known Binomial Theorem $$sum_{k=0}^n,binom{n}{k},binom{k}{0},x^{k-0}=sum_{k=0}^n,binom{n}{k},x^k=(1+x)^n=binom{n}{0},(1+x)^{n-0},.$$



    Now, suppose the identity is true when $m=s$ for some integer $sgeq 0$ (and for all integer $ngeq 0$). Therefore, we have $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},x^{k-s}=binom{n}{s},(1+x)^{n-s},.$$
    for all integer $ngeq 0$. Taking the derivative with respect to $x$, we obtain
    $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},(k-s),x^{k-s-1}=binom{n}{s},(n-s),(1+x)^{n-s-1},.$$
    Dividing both sides by $s+1$ to get
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s},frac{k-s}{s+1},x^{k-(s+1)}=binom{n}{s},frac{n-s}{s+1},(1+x)^{n-(s+1)},.$$
    Since $displaystylebinom{p}{s},frac{p-s}{s+1}=binom{p}{s+1}$ for every integer $pgeq 0$, we get the desired equality
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s+1},x^{k-(s+1)}=binom{n}{s+1},(1+x)^{n-(s+1)},,$$
    establishing that the required equality holds for $m=s+1$. By induction, we are done.




    Alternatively, consider a task of choosing a committee from $n$ people, where $m$ committee members are given the chief status. If the committee has $k$ members, then there are $displaystylebinom{n}{k}$ ways to choose the committee and $displaystylebinom{k}{m}$ ways to choose the chiefs. Hence, the number of ways to perform this task is precisely $displaystylesumlimits_{k=m}^n,binom{n}{k},binom{k}{m}$. On the other hand, we may chose $m$ chiefs first, whereby there are are $displaystylebinom{n}{m}$ ways to do so. The non-chief committee members are then selected from $n-m$ remaining people, which can be done in $2^{n-m}$ ways. Hence, we can perform this task in $displaystylebinom{n}{m},2^{n-m}$ ways. The proof is now complete.






    share|cite|improve this answer























    • +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
      – David Holden
      Jun 21 '15 at 0:38














    5












    5








    5






    We may write the required equality as
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m}=binom{n}{m},2^{n-m}$$
    for all integers $n,mgeq 0$, using the convention that $displaystylebinom{p}{q}=0$ for integers $p,qgeq 0$ such that $p<q$. We shall prove by induction on $m$ the generalized statement:
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m},x^{k-m}=binom{n}{m},(1+x)^{n-m},,$$
    where $x:=1$ leads to the required equality. The basis step is $m=0$, where we have the well known Binomial Theorem $$sum_{k=0}^n,binom{n}{k},binom{k}{0},x^{k-0}=sum_{k=0}^n,binom{n}{k},x^k=(1+x)^n=binom{n}{0},(1+x)^{n-0},.$$



    Now, suppose the identity is true when $m=s$ for some integer $sgeq 0$ (and for all integer $ngeq 0$). Therefore, we have $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},x^{k-s}=binom{n}{s},(1+x)^{n-s},.$$
    for all integer $ngeq 0$. Taking the derivative with respect to $x$, we obtain
    $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},(k-s),x^{k-s-1}=binom{n}{s},(n-s),(1+x)^{n-s-1},.$$
    Dividing both sides by $s+1$ to get
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s},frac{k-s}{s+1},x^{k-(s+1)}=binom{n}{s},frac{n-s}{s+1},(1+x)^{n-(s+1)},.$$
    Since $displaystylebinom{p}{s},frac{p-s}{s+1}=binom{p}{s+1}$ for every integer $pgeq 0$, we get the desired equality
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s+1},x^{k-(s+1)}=binom{n}{s+1},(1+x)^{n-(s+1)},,$$
    establishing that the required equality holds for $m=s+1$. By induction, we are done.




    Alternatively, consider a task of choosing a committee from $n$ people, where $m$ committee members are given the chief status. If the committee has $k$ members, then there are $displaystylebinom{n}{k}$ ways to choose the committee and $displaystylebinom{k}{m}$ ways to choose the chiefs. Hence, the number of ways to perform this task is precisely $displaystylesumlimits_{k=m}^n,binom{n}{k},binom{k}{m}$. On the other hand, we may chose $m$ chiefs first, whereby there are are $displaystylebinom{n}{m}$ ways to do so. The non-chief committee members are then selected from $n-m$ remaining people, which can be done in $2^{n-m}$ ways. Hence, we can perform this task in $displaystylebinom{n}{m},2^{n-m}$ ways. The proof is now complete.






    share|cite|improve this answer














    We may write the required equality as
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m}=binom{n}{m},2^{n-m}$$
    for all integers $n,mgeq 0$, using the convention that $displaystylebinom{p}{q}=0$ for integers $p,qgeq 0$ such that $p<q$. We shall prove by induction on $m$ the generalized statement:
    $$sum_{k=0}^n,binom{n}{k},binom{k}{m},x^{k-m}=binom{n}{m},(1+x)^{n-m},,$$
    where $x:=1$ leads to the required equality. The basis step is $m=0$, where we have the well known Binomial Theorem $$sum_{k=0}^n,binom{n}{k},binom{k}{0},x^{k-0}=sum_{k=0}^n,binom{n}{k},x^k=(1+x)^n=binom{n}{0},(1+x)^{n-0},.$$



    Now, suppose the identity is true when $m=s$ for some integer $sgeq 0$ (and for all integer $ngeq 0$). Therefore, we have $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},x^{k-s}=binom{n}{s},(1+x)^{n-s},.$$
    for all integer $ngeq 0$. Taking the derivative with respect to $x$, we obtain
    $$sum_{k=0}^{n},binom{n}{k},binom{k}{s},(k-s),x^{k-s-1}=binom{n}{s},(n-s),(1+x)^{n-s-1},.$$
    Dividing both sides by $s+1$ to get
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s},frac{k-s}{s+1},x^{k-(s+1)}=binom{n}{s},frac{n-s}{s+1},(1+x)^{n-(s+1)},.$$
    Since $displaystylebinom{p}{s},frac{p-s}{s+1}=binom{p}{s+1}$ for every integer $pgeq 0$, we get the desired equality
    $$sum_{k=0}^n,binom{n}{k},binom{k}{s+1},x^{k-(s+1)}=binom{n}{s+1},(1+x)^{n-(s+1)},,$$
    establishing that the required equality holds for $m=s+1$. By induction, we are done.




    Alternatively, consider a task of choosing a committee from $n$ people, where $m$ committee members are given the chief status. If the committee has $k$ members, then there are $displaystylebinom{n}{k}$ ways to choose the committee and $displaystylebinom{k}{m}$ ways to choose the chiefs. Hence, the number of ways to perform this task is precisely $displaystylesumlimits_{k=m}^n,binom{n}{k},binom{k}{m}$. On the other hand, we may chose $m$ chiefs first, whereby there are are $displaystylebinom{n}{m}$ ways to do so. The non-chief committee members are then selected from $n-m$ remaining people, which can be done in $2^{n-m}$ ways. Hence, we can perform this task in $displaystylebinom{n}{m},2^{n-m}$ ways. The proof is now complete.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 10 '18 at 10:46

























    answered Jun 21 '15 at 0:24









    Batominovski

    33.8k33292




    33.8k33292












    • +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
      – David Holden
      Jun 21 '15 at 0:38


















    • +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
      – David Holden
      Jun 21 '15 at 0:38
















    +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
    – David Holden
    Jun 21 '15 at 0:38




    +1 for a nice combinatorial solution! when such a representation can be found, it is the most aesthetically satisfying demonstration
    – David Holden
    Jun 21 '15 at 0:38











    4














    you can get by without induction if you observe:
    $$
    binom{n}{k} binom{k}{m} = binom{n}{m} binom{n-m}{k-m}
    $$
    then
    $$
    sum_{k=m}^n binom{n}{k} binom{k}{m} = binom{n}{m}sum_{j=0}^{n-m}binom{n-m}{j} = binom{n}{m}2^{n-m}
    $$






    share|cite|improve this answer


























      4














      you can get by without induction if you observe:
      $$
      binom{n}{k} binom{k}{m} = binom{n}{m} binom{n-m}{k-m}
      $$
      then
      $$
      sum_{k=m}^n binom{n}{k} binom{k}{m} = binom{n}{m}sum_{j=0}^{n-m}binom{n-m}{j} = binom{n}{m}2^{n-m}
      $$






      share|cite|improve this answer
























        4












        4








        4






        you can get by without induction if you observe:
        $$
        binom{n}{k} binom{k}{m} = binom{n}{m} binom{n-m}{k-m}
        $$
        then
        $$
        sum_{k=m}^n binom{n}{k} binom{k}{m} = binom{n}{m}sum_{j=0}^{n-m}binom{n-m}{j} = binom{n}{m}2^{n-m}
        $$






        share|cite|improve this answer












        you can get by without induction if you observe:
        $$
        binom{n}{k} binom{k}{m} = binom{n}{m} binom{n-m}{k-m}
        $$
        then
        $$
        sum_{k=m}^n binom{n}{k} binom{k}{m} = binom{n}{m}sum_{j=0}^{n-m}binom{n-m}{j} = binom{n}{m}2^{n-m}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 21 '15 at 0:08









        David Holden

        14.7k21224




        14.7k21224






























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