Sum of certain binomial coefficients

Question

$$\sum_{k=0}^{m} \frac{(q+k)!}{k!q!}$$

I do not know how to even start this problem. Any general tips on these types of problems will also be welcomed.

Answer 1

Hint: Each term is just $\dbinom{q+k}{k} = \dbinom{q+k}q$. Then just hockey stick identity.

Answer 2

Partial sums along diagonals in Pascal's Triangle are simply entries in the next diagonal. I can see how someone might call that a hockey stick, the shape fits. One example is the cumulative number of gifts on day $n$ in The Twelve Days of Christmas song. A little care is needed as far as what $n$ needs to be in terms of your $m.$ See for yourself:

Answer 3

Taking into account Will Jagy's nice remark, you can establish that $$\sum_{k=0}^{m} \frac{(q+k)!}{k!q!}=\frac{ (m+q+1)!}{m! (q+1)! }$$