Introductory Combinatorics Question Have a pretty standard combinatorics question that is causing me some confusion: _During a School's Open Day there are 3 first year, 3 second year and 4 third year students who have agreed to help. How many different ways can they arrive at the common room in the morning for their briefing_ ( **a** ) assuming they all arrive at random? ( **b** ) if the first person to enter the room is a first year student, and the last person to enter the room is a third year student? ( **c** ) if all of the third year students arrive in sequence, one after another? Unsure if ( **a** ) is 10!/(3!3!4!) or simply just 10! Also unsure of how to approach ( **b** ) and ( **c** ) Any help would be appreciated.

The students are obviously "distinct", so it's $10!$ for (a): all permutations yield a different arrival order.

For (b), there are $3$ choices for the first, $4$ for the last, and $8!$ for the remaining, so $3\times4\times8!$.

(c) is slightly more tricky: first, there are $4!$ possibilities for the third year students. And obviously $6!$ for the others (just pretend you remove the sequence of 3rd year students from the arrival order). Now, for each of the $6!$ possibilities, the sequence of 3rd year students can be anywhere among the others, and there are $7$ possibilities (before every one else, after the first, ... or after everyone else). So $7\times4!\times6!$.