Number of combinations of teachers and students

Question

I need to find the number of ways of arranging 10 teachers and 50 students in a line, such that there are at least 3 students between each pair of teachers. All of the people are distinct.

My Attempt: There are $10!$ ways of arranging the teachers, and $\frac{50!}{23!}$ ways to pick $27$ students to put in between the teachers. I'm not sure, though, what to do with the rest of the student. They can go anywhere, so are there ${37 + 23 \choose 23}$ ways of placing them? I'm not sure whether this is the correct count.

Answer 1

Lets attack the 27 students first.

Lets make a picture... and reduce the length just a little bit.

ooooTxxxooTxxxTxxxooToooo

The T's are the teachers. There need to be at least 3 students between each pair of teachers (the x's, above) And the o's are these 23 students we have freedom to place.

Lets get rid of the x's.

ooooTooTTooToooo

When there are 10 T's and 23 o's $33\choose 10$ describes the number of ways to place o's and T's 50! ways to but all the students in a line 10! ways to put all the teachers in a line.

${33\choose 10}50!10!$

Answer 2

Your reasoning is sound up to the point where you are not sure what to do with the rest of thes students.

You have $23$ students remaining. They have to be placed either side of the teachers. So there are $23!$ ways of arranging the students, and $24$ ways to choose how many of them are to the left of the first teacher. So $24!$ altogether.

Now let's take some shortcuts to this problem. First, why did we get $24!$ above? It comes out naturally if we view the problem as arranging $24$ objects, $23$ of which are the additional students and the $24$th as the preselected sequence of $10$ teachers and $27$ students.

An even better shortcut is to note that there are $10!$ ways of arranging the teachers, and $50!$ ways of arranging the students. All we need to consider on top of that is where the first teacher occurs - which we did above and there are $24$ possible places. Overall this gives $50!10!24$ arrangements.

Answer 3

Pretend we put down placemats to organise the line-up, ensuring that at least 3 student placemats are inserted between each teacher.

So we must arrange $10$ red and $23$ blue placemats in any order, with $27$ green placemats inserted in relatively-fixed positions ($3$ each following all but the last red), then arrange $50$ students and $10$ teachers among their placemats.   How shall we count the ways...?

$$\begin{align}&\dfrac{37!}{10!\cdotp 23!}\cdotp 10!\cdotp 50!\\[1ex]=\\[1ex]&\dfrac{50!\cdot 37!}{23!}\end{align}$$