There's a problem in my combinatorics class: there're 5 people who need to complete 4 tasks. They decided that each task will be completed by a team of 2 people. In how many ways can we assign a team to a task if it's not allowed that there will be a person which didn't do any task? The problem has to be solved using inclusion-exclusion principle (IEP).
I have trouble solving the problem but this is what I have so far: 1) we can choose the teams in $\binom{5}{2}$ ways which equals $10$. All of the tasks can theoretically by completed by only one team so we have $10^4$ possibilities. 2) using IEP let's subtract from $10^4$ the case where at least one team will not complete any tasks ($9^4$) etc. So we would get something like this: $$10^4-9^4+8^4-7^4+6^4-5^4+4^4-3^4+2^4-1^4=5995$$
Am I in the right direction?