Who could show me how many terms are in the expansion of $$(t_1+t_2+...+t_n)^p$$ I totally confused myself and I cannot figure it out:-(
I was inspired by GDumphart. Thanks so much! It should be the number of options that $k_1+k_2+...+k_n = p$. The answer should be (n+p-1) choose (p).
$$(t_{1}+t_{2}+\cdots+t_{n})^{p}=\sum_{k_{1}+\cdots+k_{n}=p}\frac{p!}{k_{1}!k_{2}!\ldots k_{n}!}t_{1}^{k_{1}}t_{2}^{k_{2}}\ldots t_{n}^{k_{n}}.$$ So there are $\binom{n+p-1}{p}$ terms = number of ways you can put $p$ identical balls into $n$ many cells.
Clarification: In how many ways can we write $k_{1}+k_{2}+\cdots+k_{n}=p$?. Where each $k_{i}$ is a non negative integer. It is exactly the number of possible ways to put $p$ identical balls into $n$ cells. Now treat $n$ cells as $n-1$ dividers. Then it is the number of permutations of $p$ identical balls and $n-1$ many identical dividers. Then the number of permutations is
$$\frac{(p+n-1)!}{p!(n-1)!}=\binom{p+n-1}{p}=\binom{p+n-1}{n-1}.$$