4 professors and 7 students are to be seated on 11 chairs. What is 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" $$\frac{\binom 6 4 \cdot 4! \cdot 7!}{11!}$$

Answer 2

Update: this is incorrect. See the comments at the end.

Mine is a little different. Let's first assume that all students are identical, and all professors are identical, so we only arrange patterns. Two students must sit at the two ends (otherwise, the professors at the ends would not be between students). Now we have 5 students and 4 professors, to sit in the middle 9 seats. Since there is only one more student than professors, at most, two students can sit together (ignore the two students at the ends). In the case in which no two students sitting together, there is only one pattern: +101010101+, where + stands for the students sitting at the ends; in the case where two students are sitting together, we have 5 locations for them. So, we have 6 patterns in total. Now, with every pattern, we have 4!7! ways to arrange students and professors. So, the probability for the good is $\frac{6\cdot 4!\cdot 7!}{11!}$.

Okay, after some thought, mine is incorrect. We may have more than two students sitting together. The prior one is simpler and more elegant, and more importantly, correct!