I have encountered this problem on a past paper:
In how many ways can 675 identical envelopes be divided, in packages of 25, among four student groups so that each group gets at least 150, but no more than 200, of the envelopes?
I so far have that x1+x2+x3+x4=(?) such that 150 <= xi <= 200 for 1 <= i <= 4
What is the correct value for (?)? I originally thought it would be 675/25=27 as there are 675 identical envelopes into packages of 25. But now i am thinking it should be C(675+25-1,24) since there are C(n+k-1,k-1) ways to put the envelopes into the packages in the first place.
I was wondering if, seeing as 27 is a much nicer number to deal with, you could use the inclusion-exclusion principle to find x1+x2+x3+x4=27 such that 150/25 <= xi <= 200/25 and then multiply this answer by C(675+25-1,24)?
Can I have some guidance please? I am also open to use of other methods!
Thank you in advance!
Let $x_i$ be the number of packages Group $i$ gets. Then we want $x_1+\cdots+x_4=27$, with $6\le x_i\le 8$. This was in your post. We need to count the integer solutions. That's all, no later multiplication by anything, since envelopes, and packages of $25$, are identical.
To solve the problem, imagine first giving each student group $6$ packages. This can be done in only $1$ way. That means we have $3$ packages left. We want to distribute these packages so that no one gets all $3$. If you want to use equations, we are solving $y_1+\cdots+y_4=3$ with condition $0\le y_i\le 2$.