While there are some similar problems on this site, I can't seem to find one in which the index $i$ itself has a bearing on the upper bound of each element $x_i$ (assume that $x_i$ is a positive whole number). I would like to verify that I am going in the correct direction with this problem and get some advice on how to continue.
Here is what I have so far:
It's relatively trivial for me to see that
$P(\sum_{i=1}^5x_i=30)=\binom{5+30-1}{30}$.
I am pretty sure that
$P(\forall i, i\leq x_i)=P(\sum_{i=1}^5x_i=15)=\binom{5+15-1}{15}$
The problem I run into is figuring out $P(\forall i, i\leq x_i\leq3i)$. Here is the direction I was heading:
$P(\forall i, i\leq x_i\leq3i)=P(\forall i, i\leq x_i)-P(\forall i, i\leq x_i \cap\exists j, x_j>3j)=$
$=\binom{5+15-1}{15}-\sum_{n=0}^5(-1)^n\binom{5}{n}|P((\forall i, i\leq x_i) \cap (x_1>3\cdot1)\cap...\cap(x_n>3\cdot n))|$
Is this direction correct (regardless of how inefficient it is)? Is there a faster or simpler method?
The expression $\sum_{i=1}^5x_i=30$ with constraints $i\leq x_i \leq 3i\ $ can be better visualized if we write it as $$x_1 + x_2 +x_3 +x_4 +x_5 = 30$$ with the constraints $1\le x_1\le 3,$ $\ \ 2\le x_2\le 6,\ \ $$\ \ 3\le x_4\le 9,\ \ $$\ \ 4\le x_4\le 12,\ \ $ and $\ \ 5\le x_5\le 15.\ \ $
This is equivalent to solving the system $$y_1 +y_2+y_3+y_4+y_5 = 15 $$ with $\ \ 0\le y_1\le 2,\ \ $$\ \ 0\le y_2\le 4,\ \ $$\ \ 0\le y_3\le 6,\ \ $$\ \ 0\le y_4\le 8,\ \ $ and $\ \ 0\le y_5\le 10.$
This is now a standard problem that can be solved by using the Principle of Inclusion-Exclusion. Are you able to take it from here?