How to simplify this binomial coefficients?












0












$begingroup$


For $a, b, r, n in mathbb N$ :
I have to simplify



$$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}$$



I tried by using factorials but it seemed more complicated






binomial-coefficients







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    For $a, b, r, n in mathbb N$ :
    I have to simplify



    $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}$$



    I tried by using factorials but it seemed more complicated






    binomial-coefficients







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      For $a, b, r, n in mathbb N$ :
      I have to simplify



      $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}$$



      I tried by using factorials but it seemed more complicated






      binomial-coefficients







      share|cite|improve this question









      $endgroup$




      For $a, b, r, n in mathbb N$ :
      I have to simplify



      $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}$$



      I tried by using factorials but it seemed more complicated






      binomial-coefficients




      binomial-coefficients






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 24 '18 at 9:57









      LamethysteLamethyste

      424




      424






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          Hint:



          Use that, by definition, $dbinom nk$ is the coefficients of $x^k$ in the expansion of $(1+x)^n$. So here, consider the expansion of each side in the equality and compare the coefficients of $x^r$:
          $$(1+x)^a(1+x)^b=(1+x)^{a+b}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thank you I'm going to try
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:10










          • $begingroup$
            @Lamethyste: I've added a detail to the hint.
            $endgroup$
            – Bernard
            Dec 24 '18 at 10:14










          • $begingroup$
            $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:33








          • 1




            $begingroup$
            I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
            $endgroup$
            – Bernard
            Dec 24 '18 at 11:03












          • $begingroup$
            sorry I could not answer you, but i think that I found something better, thank you for your help
            $endgroup$
            – Lamethyste
            Dec 31 '18 at 18:10












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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint:



          Use that, by definition, $dbinom nk$ is the coefficients of $x^k$ in the expansion of $(1+x)^n$. So here, consider the expansion of each side in the equality and compare the coefficients of $x^r$:
          $$(1+x)^a(1+x)^b=(1+x)^{a+b}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thank you I'm going to try
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:10










          • $begingroup$
            @Lamethyste: I've added a detail to the hint.
            $endgroup$
            – Bernard
            Dec 24 '18 at 10:14










          • $begingroup$
            $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:33








          • 1




            $begingroup$
            I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
            $endgroup$
            – Bernard
            Dec 24 '18 at 11:03












          • $begingroup$
            sorry I could not answer you, but i think that I found something better, thank you for your help
            $endgroup$
            – Lamethyste
            Dec 31 '18 at 18:10
















          1












          $begingroup$

          Hint:



          Use that, by definition, $dbinom nk$ is the coefficients of $x^k$ in the expansion of $(1+x)^n$. So here, consider the expansion of each side in the equality and compare the coefficients of $x^r$:
          $$(1+x)^a(1+x)^b=(1+x)^{a+b}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thank you I'm going to try
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:10










          • $begingroup$
            @Lamethyste: I've added a detail to the hint.
            $endgroup$
            – Bernard
            Dec 24 '18 at 10:14










          • $begingroup$
            $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:33








          • 1




            $begingroup$
            I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
            $endgroup$
            – Bernard
            Dec 24 '18 at 11:03












          • $begingroup$
            sorry I could not answer you, but i think that I found something better, thank you for your help
            $endgroup$
            – Lamethyste
            Dec 31 '18 at 18:10














          1












          1








          1





          $begingroup$

          Hint:



          Use that, by definition, $dbinom nk$ is the coefficients of $x^k$ in the expansion of $(1+x)^n$. So here, consider the expansion of each side in the equality and compare the coefficients of $x^r$:
          $$(1+x)^a(1+x)^b=(1+x)^{a+b}.$$






          share|cite|improve this answer











          $endgroup$



          Hint:



          Use that, by definition, $dbinom nk$ is the coefficients of $x^k$ in the expansion of $(1+x)^n$. So here, consider the expansion of each side in the equality and compare the coefficients of $x^r$:
          $$(1+x)^a(1+x)^b=(1+x)^{a+b}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 10:13

























          answered Dec 24 '18 at 10:07









          BernardBernard

          125k743118




          125k743118












          • $begingroup$
            thank you I'm going to try
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:10










          • $begingroup$
            @Lamethyste: I've added a detail to the hint.
            $endgroup$
            – Bernard
            Dec 24 '18 at 10:14










          • $begingroup$
            $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:33








          • 1




            $begingroup$
            I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
            $endgroup$
            – Bernard
            Dec 24 '18 at 11:03












          • $begingroup$
            sorry I could not answer you, but i think that I found something better, thank you for your help
            $endgroup$
            – Lamethyste
            Dec 31 '18 at 18:10


















          • $begingroup$
            thank you I'm going to try
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:10










          • $begingroup$
            @Lamethyste: I've added a detail to the hint.
            $endgroup$
            – Bernard
            Dec 24 '18 at 10:14










          • $begingroup$
            $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
            $endgroup$
            – Lamethyste
            Dec 24 '18 at 10:33








          • 1




            $begingroup$
            I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
            $endgroup$
            – Bernard
            Dec 24 '18 at 11:03












          • $begingroup$
            sorry I could not answer you, but i think that I found something better, thank you for your help
            $endgroup$
            – Lamethyste
            Dec 31 '18 at 18:10
















          $begingroup$
          thank you I'm going to try
          $endgroup$
          – Lamethyste
          Dec 24 '18 at 10:10




          $begingroup$
          thank you I'm going to try
          $endgroup$
          – Lamethyste
          Dec 24 '18 at 10:10












          $begingroup$
          @Lamethyste: I've added a detail to the hint.
          $endgroup$
          – Bernard
          Dec 24 '18 at 10:14




          $begingroup$
          @Lamethyste: I've added a detail to the hint.
          $endgroup$
          – Bernard
          Dec 24 '18 at 10:14












          $begingroup$
          $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
          $endgroup$
          – Lamethyste
          Dec 24 '18 at 10:33






          $begingroup$
          $(1+X)^r=(1+X)^{a+b}$, so $$sum_{k = 0}^{r}binom {r}kX^k=sum_{k = 0}^{r}(sum_{i = 0}^{k}binom {a}ibinom {b}{k-i})X^k$$ So for k=r :$binom {r}rX^r$=$sum_{i = 0}^{r}binom {a}rbinom {b}{r-i}X^r$, knowing that $binom {r}r=1$ $$sum_{k = 0}^{r}binom {a}kbinom {b}{r-k}=1$$
          $endgroup$
          – Lamethyste
          Dec 24 '18 at 10:33






          1




          1




          $begingroup$
          I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
          $endgroup$
          – Bernard
          Dec 24 '18 at 11:03






          $begingroup$
          I didn't suggest to expand $(1+x)^r$ but to calculate the coefficient of $x^r$ in both sides (which ultimately use the expansion of $(1+x)^{a+b}$). Anyway, a sum of binomial coefficients can't be equal to $1$, unless there is only one term, equal to $1$.
          $endgroup$
          – Bernard
          Dec 24 '18 at 11:03














          $begingroup$
          sorry I could not answer you, but i think that I found something better, thank you for your help
          $endgroup$
          – Lamethyste
          Dec 31 '18 at 18:10




          $begingroup$
          sorry I could not answer you, but i think that I found something better, thank you for your help
          $endgroup$
          – Lamethyste
          Dec 31 '18 at 18:10


















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