4 professors and 7 students are to be seated on 11 chairs. the probability that every professor is between 2 students

Question

There are 4 professors and 7 students to be sited on 11 chairs. What is a probability that every professor is between 2 students?

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My thoughts: There is $11!$ outcomes. Good ones are: first we arrange the professors on $4!$ ways, then we put 7 chairs so that on left and right side is at least one chair and between every two professor is one chair. That we can do on ${6\choose 2}$ ways by solving the equation $$a+b+c+d+e = 7$$ where $a,b,c,d,e$ are natural numbers.

Now we arrange the students on empty chairs, and that we can do on $7!$ ways. So the probability is $$P = {4!\cdot 15\cdot 7!\over 11!}$$

Is that correct?

Answer 1

Looks good. I get the same result by reasoning as follows:

• $7$ students in a row give $6$ "slots" where to place the professors in between.
• There are $\binom 6 4$ ways to choose the slots.
• The students and professors give $7!\cdot 4!$ arrangements for each way of choosing the "slots" $$\frac{\binom 6 4 \cdot 4! \cdot 7!}{11!}$$