Closed form using Binomial Expansion












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$begingroup$


I would like to obtain a closed form for the following:



$${n choose 0}{n choose k} + {n choose 1}{n-1 choose k-1} + {n choose 2}{n-2 choose k-2} + ... + {n choose k}{n-k choose 0}$$



To me, this looked like it closely resembles the Binomial Theorem, which I know to be: $(a+b)^k=sum_{k=0}^nbinom nk a^k b^{n-k},$ but I couldn't get it to work.



The $sum{n choose k}$ bit falls nicely in the Binomial expansion, but I don't know how to manipulate $a$ and $b$ to represent $sum_{i=0}{n-i choose k-i}$. How should I go about this?






combinatorics binomial-theorem







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    1












    $begingroup$


    I would like to obtain a closed form for the following:



    $${n choose 0}{n choose k} + {n choose 1}{n-1 choose k-1} + {n choose 2}{n-2 choose k-2} + ... + {n choose k}{n-k choose 0}$$



    To me, this looked like it closely resembles the Binomial Theorem, which I know to be: $(a+b)^k=sum_{k=0}^nbinom nk a^k b^{n-k},$ but I couldn't get it to work.



    The $sum{n choose k}$ bit falls nicely in the Binomial expansion, but I don't know how to manipulate $a$ and $b$ to represent $sum_{i=0}{n-i choose k-i}$. How should I go about this?






    combinatorics binomial-theorem







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I would like to obtain a closed form for the following:



      $${n choose 0}{n choose k} + {n choose 1}{n-1 choose k-1} + {n choose 2}{n-2 choose k-2} + ... + {n choose k}{n-k choose 0}$$



      To me, this looked like it closely resembles the Binomial Theorem, which I know to be: $(a+b)^k=sum_{k=0}^nbinom nk a^k b^{n-k},$ but I couldn't get it to work.



      The $sum{n choose k}$ bit falls nicely in the Binomial expansion, but I don't know how to manipulate $a$ and $b$ to represent $sum_{i=0}{n-i choose k-i}$. How should I go about this?






      combinatorics binomial-theorem







      share|cite|improve this question









      $endgroup$




      I would like to obtain a closed form for the following:



      $${n choose 0}{n choose k} + {n choose 1}{n-1 choose k-1} + {n choose 2}{n-2 choose k-2} + ... + {n choose k}{n-k choose 0}$$



      To me, this looked like it closely resembles the Binomial Theorem, which I know to be: $(a+b)^k=sum_{k=0}^nbinom nk a^k b^{n-k},$ but I couldn't get it to work.



      The $sum{n choose k}$ bit falls nicely in the Binomial expansion, but I don't know how to manipulate $a$ and $b$ to represent $sum_{i=0}{n-i choose k-i}$. How should I go about this?






      combinatorics binomial-theorem




      combinatorics binomial-theorem






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      asked Jan 23 at 23:42









      Corp. and Ltd.Corp. and Ltd.

      16914




      16914






















          2 Answers
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          6












          $begingroup$

          Your sum can be rewritten in sigma notation as
          $$ sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j}.$$
          We observe that for $j = 0, dots, k$, we have
          $$ binom{n}{j} binom{n-j}{k-j} = frac{n!}{j!(n-j)!} frac{(n-j)!}{(n-k)!(k-j)!} =
          frac{n!}{k! (n-k)!} frac{k!}{j!(k-j)!} = binom{n}{k} binom{k}{j}.$$           Summing over $j$ gives
          $$
          sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j} = sum_{j = 0}^k binom{n}{k} binom{k}{j} = 2^k binom{n}{k}.$$






          share|cite|improve this answer











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            4












            $begingroup$

            Think about what each term counts. $binom{n}{i}$ counts the number of ways to choose $i$ things from $n$ things, and $binom{n-i}{k-i}$ counts the number of ways to choose $k-i$ things from $n-i$ things. So we could interpret this as the following: start with a group of $n$ objects, then pick $i$ of them to be assigned to "group A" and pick $k-i$ of the remaining $n-i$ objects to be assigned to "group B". The number of ways to do this is $binom{n}{i}binom{n-i}{k-i}$.



            If we sum these from $i=0$ to $i=k$, then we are counting all the ways that we can choose some number of objects and assign them to "group A" and some number of objects and assign them to "group B" such that the total size of "group A" and "group B" is always $k$. If we want a closed form for this quantity, we should calculate it in another way. So we could first pick the $k$ objects that will be assigned to "group A" or "group B" (there are $binom{n}{k}$ ways to do this), then for each object, assign it to "group A" or "group B". There are $2^k$ ways to make these assignments. So,
            $$sum_{i=0}^k binom{n}{i}binom{n-i}{k-i}=binom{n}{k}2^k$$






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              6












              $begingroup$

              Your sum can be rewritten in sigma notation as
              $$ sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j}.$$
              We observe that for $j = 0, dots, k$, we have
              $$ binom{n}{j} binom{n-j}{k-j} = frac{n!}{j!(n-j)!} frac{(n-j)!}{(n-k)!(k-j)!} =
              frac{n!}{k! (n-k)!} frac{k!}{j!(k-j)!} = binom{n}{k} binom{k}{j}.$$               Summing over $j$ gives
              $$
              sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j} = sum_{j = 0}^k binom{n}{k} binom{k}{j} = 2^k binom{n}{k}.$$






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                Your sum can be rewritten in sigma notation as
                $$ sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j}.$$
                We observe that for $j = 0, dots, k$, we have
                $$ binom{n}{j} binom{n-j}{k-j} = frac{n!}{j!(n-j)!} frac{(n-j)!}{(n-k)!(k-j)!} =
                frac{n!}{k! (n-k)!} frac{k!}{j!(k-j)!} = binom{n}{k} binom{k}{j}.$$                 Summing over $j$ gives
                $$
                sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j} = sum_{j = 0}^k binom{n}{k} binom{k}{j} = 2^k binom{n}{k}.$$






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Your sum can be rewritten in sigma notation as
                  $$ sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j}.$$
                  We observe that for $j = 0, dots, k$, we have
                  $$ binom{n}{j} binom{n-j}{k-j} = frac{n!}{j!(n-j)!} frac{(n-j)!}{(n-k)!(k-j)!} =
                  frac{n!}{k! (n-k)!} frac{k!}{j!(k-j)!} = binom{n}{k} binom{k}{j}.$$                   Summing over $j$ gives
                  $$
                  sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j} = sum_{j = 0}^k binom{n}{k} binom{k}{j} = 2^k binom{n}{k}.$$






                  share|cite|improve this answer











                  $endgroup$



                  Your sum can be rewritten in sigma notation as
                  $$ sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j}.$$
                  We observe that for $j = 0, dots, k$, we have
                  $$ binom{n}{j} binom{n-j}{k-j} = frac{n!}{j!(n-j)!} frac{(n-j)!}{(n-k)!(k-j)!} =
                  frac{n!}{k! (n-k)!} frac{k!}{j!(k-j)!} = binom{n}{k} binom{k}{j}.$$                   Summing over $j$ gives
                  $$
                  sum_{j = 0}^k binom{n}{j} binom{n-j}{k-j} = sum_{j = 0}^k binom{n}{k} binom{k}{j} = 2^k binom{n}{k}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 6 at 4:34

























                  answered Jan 23 at 23:58









                  Jordan GreenJordan Green

                  1,133410




                  1,133410























                      4












                      $begingroup$

                      Think about what each term counts. $binom{n}{i}$ counts the number of ways to choose $i$ things from $n$ things, and $binom{n-i}{k-i}$ counts the number of ways to choose $k-i$ things from $n-i$ things. So we could interpret this as the following: start with a group of $n$ objects, then pick $i$ of them to be assigned to "group A" and pick $k-i$ of the remaining $n-i$ objects to be assigned to "group B". The number of ways to do this is $binom{n}{i}binom{n-i}{k-i}$.



                      If we sum these from $i=0$ to $i=k$, then we are counting all the ways that we can choose some number of objects and assign them to "group A" and some number of objects and assign them to "group B" such that the total size of "group A" and "group B" is always $k$. If we want a closed form for this quantity, we should calculate it in another way. So we could first pick the $k$ objects that will be assigned to "group A" or "group B" (there are $binom{n}{k}$ ways to do this), then for each object, assign it to "group A" or "group B". There are $2^k$ ways to make these assignments. So,
                      $$sum_{i=0}^k binom{n}{i}binom{n-i}{k-i}=binom{n}{k}2^k$$






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        Think about what each term counts. $binom{n}{i}$ counts the number of ways to choose $i$ things from $n$ things, and $binom{n-i}{k-i}$ counts the number of ways to choose $k-i$ things from $n-i$ things. So we could interpret this as the following: start with a group of $n$ objects, then pick $i$ of them to be assigned to "group A" and pick $k-i$ of the remaining $n-i$ objects to be assigned to "group B". The number of ways to do this is $binom{n}{i}binom{n-i}{k-i}$.



                        If we sum these from $i=0$ to $i=k$, then we are counting all the ways that we can choose some number of objects and assign them to "group A" and some number of objects and assign them to "group B" such that the total size of "group A" and "group B" is always $k$. If we want a closed form for this quantity, we should calculate it in another way. So we could first pick the $k$ objects that will be assigned to "group A" or "group B" (there are $binom{n}{k}$ ways to do this), then for each object, assign it to "group A" or "group B". There are $2^k$ ways to make these assignments. So,
                        $$sum_{i=0}^k binom{n}{i}binom{n-i}{k-i}=binom{n}{k}2^k$$






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Think about what each term counts. $binom{n}{i}$ counts the number of ways to choose $i$ things from $n$ things, and $binom{n-i}{k-i}$ counts the number of ways to choose $k-i$ things from $n-i$ things. So we could interpret this as the following: start with a group of $n$ objects, then pick $i$ of them to be assigned to "group A" and pick $k-i$ of the remaining $n-i$ objects to be assigned to "group B". The number of ways to do this is $binom{n}{i}binom{n-i}{k-i}$.



                          If we sum these from $i=0$ to $i=k$, then we are counting all the ways that we can choose some number of objects and assign them to "group A" and some number of objects and assign them to "group B" such that the total size of "group A" and "group B" is always $k$. If we want a closed form for this quantity, we should calculate it in another way. So we could first pick the $k$ objects that will be assigned to "group A" or "group B" (there are $binom{n}{k}$ ways to do this), then for each object, assign it to "group A" or "group B". There are $2^k$ ways to make these assignments. So,
                          $$sum_{i=0}^k binom{n}{i}binom{n-i}{k-i}=binom{n}{k}2^k$$






                          share|cite|improve this answer









                          $endgroup$



                          Think about what each term counts. $binom{n}{i}$ counts the number of ways to choose $i$ things from $n$ things, and $binom{n-i}{k-i}$ counts the number of ways to choose $k-i$ things from $n-i$ things. So we could interpret this as the following: start with a group of $n$ objects, then pick $i$ of them to be assigned to "group A" and pick $k-i$ of the remaining $n-i$ objects to be assigned to "group B". The number of ways to do this is $binom{n}{i}binom{n-i}{k-i}$.



                          If we sum these from $i=0$ to $i=k$, then we are counting all the ways that we can choose some number of objects and assign them to "group A" and some number of objects and assign them to "group B" such that the total size of "group A" and "group B" is always $k$. If we want a closed form for this quantity, we should calculate it in another way. So we could first pick the $k$ objects that will be assigned to "group A" or "group B" (there are $binom{n}{k}$ ways to do this), then for each object, assign it to "group A" or "group B". There are $2^k$ ways to make these assignments. So,
                          $$sum_{i=0}^k binom{n}{i}binom{n-i}{k-i}=binom{n}{k}2^k$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 24 at 0:00









                          kccukccu

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