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 $\frac{10!}{3!3!4!}$ 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 $7\times4!\times6!$.