There are 4 mathematicians and 7 computer scientists. We need to create a team of experts that has 5 members. At least 2 members should be mathematicians. What is the number of ways we can choose? The solution is 301. Can someone explain why?
4 mathematicians (Ms for short) and 7 computer scientists (CSs for short) can be members. There are $11!/5!(11−5)!=462$ combinations for assigning membership. We need to subtract those combinations with 0 or 1 M. For 0 Ms it's easy: there is only one combination with 0 Ms i.e. the one with 5 CSs. As for those with 1 M: since there are four Ms there are at least 4 different combinations. In other 4 places (one place is taken by M) we need to distribute 7 CSs. $7!/(7−4)!=7!/3!=840$ permutations there are for that. $840$ permutations of CSs for each distinct M is $3360$. There are 4 places for CSs, so to bring $3360$ from permutations to combinations we need to divide by $4!$. $3360/24=140$. Then I concluded that there are $462−1−140=321$ combinations that contain at least 2 Ms. As you can see, I am very close ($321−301$ is only $20$). So... can you detect where am I wrong?