Evaluate the sum $$\sum_{i=0}^{n-1}\binom{i+m-1}{m-1}$$
I know the steps I have to follow but don't know how to apply the process to this example.
First I have to find $$A(x)=\sum_{i=0}^\infty a_{i}x^i$$ Then I have to determine $$B(x)=\sum_{i=0}^\infty b_{i}x^i$$ I have to take the product of A(x) and B(x) and $$A(x)B(x)=\sum_{k=0}^k c_{k}x^k$$ where $$c_{k}=\sum_{i=0}^k a_{i}b_{k-i}$$
so I'm trying to find $$c_{k}$$
Use the variable $k$ so as not to confuse this with complex numbers and set the upper limit to $n$ which does not change the problem as $n$ only appears in said upper limit of the sum. This gives $$\sum_{k=0}^n {k+m-1\choose m-1} = \frac{1}{n! (m-1)!} \sum_{k=0}^n n! \frac{(k+m-1)!}{k!} \\= \frac{1}{n! (m-1)!} \sum_{k=0}^n {n\choose k} (k+m-1)! (n-k)!$$
Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Now in the present case we clearly have $$A(z) = \sum_{q\ge 0} (q+m-1)! \frac{z^q}{q!} = (m-1)! \sum_{q\ge 0} {q+m-1\choose q} z^q = (m-1)! \frac{1}{(1-z)^m}.$$ by Newton's binomial series. Furthermore we see that $$B(z) = \sum_{q\ge 0} q! \frac{z^q}{q!} = \sum_{q\ge 0} z^q = \frac{1}{1-z}.$$ It follows that the sum is given by $$n! [z^n] A(z) B(z) = n! [z^n] (m-1)! \frac{1}{(1-z)^{m+1}} = n! \; (m-1)! {n+m\choose m}.$$ The conclusion is that the sum we started with (i.e. before we extracted the two factors to the front) is given by $$\frac{n! \; (m-1)!}{n! \; (m-1)!} {n+m\choose m} = {n+m\choose m}.$$