What is the sum of number of ways of choosing $n$ elements from $(n+r)$ elements where $r$ is fixed and $n$ varies from $1$ to $m$ ? Can this be reduced to a formula ?
$$ \sum ^m _{n=1} \binom{n + r}n $$
Answer
Yes your formula is corrct and:
$$\sum_{n=1}^m \dbinom{n+r}n=\sum_{n=1}^m\dbinom{n+r}{r}$$
and you can prove by induction that this $\dbinom{m+r+1}{r+1}-1$
No comments:
Post a Comment