Does there exist a similar identity to $binom{n}{k} = binom{n}{n-k}$?












1














I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







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  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12


















1














I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







share|cite|improve this question




















  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12
















1












1








1







I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients







share|cite|improve this question















I know that $$binom{n}{k} = binom{n}{n-k}$$



My question is does there exist a similar identity where you change the top of the choose function, o any similar to the identity above?



An example: Can you define $m$ in terms of $n$ and $k$ such that $$binom{n}{k}=binom{n+m}{k}$$






combinatorics soft-question binomial-coefficients




combinatorics soft-question binomial-coefficients






share|cite|improve this question















share|cite|improve this question













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edited Nov 24 at 3:41









Shaun

8,507113580




8,507113580










asked Nov 24 at 2:57









Shrey Joshi

19512




19512








  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12
















  • 2




    No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
    – David Peterson
    Nov 24 at 3:12










2




2




No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
– David Peterson
Nov 24 at 3:12






No, because, for instance, $$5={5choose 1} neq {5+mchoose 1}=5+m$$ for any $mneq 0$. This becomes obvious once you review the construction of pascal's triangle.
– David Peterson
Nov 24 at 3:12












2 Answers
2






active

oldest

votes


















2














It is standard to define
$$
binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
$$
as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






share|cite|improve this answer





























    0














    No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



    An algebraic proof of the same fact is the following-



    Multiply both LHS and RHS by $k!$, we get,
    $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
    Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



    We can see this directly from the Pascal's identity-
    $$binom nk+binom n{k-1} =binom {n+1}k$$
    $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



    Hope this is helpful.






    share|cite|improve this answer

















    • 1




      What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
      – Shrey Joshi
      Nov 24 at 3:37











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    It is standard to define
    $$
    binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
    $$
    as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






    share|cite|improve this answer


























      2














      It is standard to define
      $$
      binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
      $$
      as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






      share|cite|improve this answer
























        2












        2








        2






        It is standard to define
        $$
        binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
        $$
        as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)






        share|cite|improve this answer












        It is standard to define
        $$
        binom{-n}{k}=(-1)^kbinom{n+k-1}{k}
        $$
        as, e.g., here. (Whether or not this needs to be a separate definition depends on how you defined the binomial coefficient in the first place, but it's pretty much always going to be the right choice.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 4:09









        Micah

        29.6k1363105




        29.6k1363105























            0














            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer

















            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37
















            0














            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer

















            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37














            0












            0








            0






            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.






            share|cite|improve this answer












            No, there is no such identity and in fact, there cannot be such identity. Because LHS is the number of ways of choosing k objects from n objects and RHS is the number of ways of choosing k objects out of n+m given objects which is obviously higher because you can still choose the k objects out of the n initial number of objects and now also can choose some objects out of the remaining objects.



            An algebraic proof of the same fact is the following-



            Multiply both LHS and RHS by $k!$, we get,
            $$(n)(n-1)…(n-k+1) = (n+m)(n+m-1)…(n+m-k+1)$$
            Each term of the RHS is strictly greater than each term of the LHS, therefore equality cannot hold.



            We can see this directly from the Pascal's identity-
            $$binom nk+binom n{k-1} =binom {n+1}k$$
            $$Longrightarrow binom {n+1}k - binom nk = binom n{k-1}>0$$



            Hope this is helpful.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 3:13









            Martund

            1,351212




            1,351212








            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37














            • 1




              What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
              – Shrey Joshi
              Nov 24 at 3:37








            1




            1




            What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
            – Shrey Joshi
            Nov 24 at 3:37




            What about $binom{n}{k}=binom{n+a}{k+b}$? I am not asking specifically the identity mentioned as an example in the original question, but rather anything similar.
            – Shrey Joshi
            Nov 24 at 3:37


















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