How to prove a combinatorial identity with a combinatorial argument

Question

I am trying to prove the following identity using a a combinatorial argument: $$\dbinom{n+r+1}{r}=\sum_{k=0}^{r}\dbinom{n+k}{k}$$

Answer 1

${n+k \choose k}={n+k \choose n}$= Coeff. of $x^n ~in~ (1+x)^{n+k}$ So $$\sum_{k=0}^{r} {n+k \choose n}= Coeffi.~ of~x^n ~ in ~(1+x)^n\sum_{k=0}^r (1+x)^k$$ $$=Coeffi. ~ of ~ x^n ~ in (1+x)^n\frac{(1+x)^{r+1}-1}{1+x-1}$$ $$=Coeffi. ~ of ~ x^{n+1}~in ~[(1+x)^{n+r+1}-(1+x)^n]={n+r+1 \choose n+1}={n+r+1 \choose r}.$$