here's a question I had in an exam today:
Four people are checking 230 exams. In how many ways can you split the papers between the four of them if you want each one to check at least 15?
So, after checking $4 \cdot 15$ papers we are left with 170 and so the result is $$\binom{170+(4-1)}{4-1}=\binom{173}{3}$$ And here's the twist: Following the last question, in how many ways can you split the papers between the four of them, only now the papers are different and its important to count who checks how many. Meaning - The number of ways to divide 230 elements to 4 different groups.
My initial attemp was $$\sum_{x_1,x_2,x_3,x_4=230}\binom{230}{x_1,x_2,x_3,x_4}=4^{230}$$ But that's obviously a mistake since I wasn't counting the first 60 papers correctly. I think it probably involves using inclusion-exclusion, but I'm not sure how.
Would love to hear some ideas. Thanks!