# In how many ways can $5$ students and $3$ teacher sit around a table so that no two teachers are together?

In how many ways can $5$ students and $3$ teacher sit around a table so that no two teachers are together?

My attempt:

$5$ student can sit $(5-1)!$ in round table.

A teacher can sit between two student, there are such $5$ places, which order of sitting place of teacher is matter, so total number of ways $= (5-1)! \times ^5P_3 = 24\times60= 1440$ ways.

Can you explain in formal/alternative way, please?

• Your approach strikes me as optimal. – N. F. Taussig May 29 '16 at 10:23

I would do the problem exactly the way you did it.

We can line up the five students in $5!$ ways, leaving spaces between them and at the ends of the row in which to insert the teachers. There are six such spaces, four between successive students and two at the ends of the row. We can insert the three teachers in $P(6, 3) = 6 \cdot 5 \cdot 4$ ways. This gives us $$5! \cdot 6 \cdot 5 \cdot 4$$ linear arrangements of students and teachers in which no two teachers are consecutive.

However, since we wish to arrange the students and teachers around a circular table so that no two teachers sit in consecutive seats, we must exclude those linear arrangements in which teachers are at both ends of the row. There are three ways to select the teacher at the left end of the row, two ways to select the teacher at the right end of the row, and four ways to place the remaining teacher in one of the four spaces between successive students. Hence, there are $$3 \cdot 2 \cdot 4 \cdot 5!$$ linear arrangements in which teachers are at both ends of the row.

Hence, there are $$6 \cdot 5 \cdot 4 \cdot 5! - 3 \cdot 2 \cdot 4 \cdot 5! = (120 - 24)5! = 96 \cdot 5!$$ linear arrangements of teachers and students so that no two teachers are consecutive and teachers are not at both ends of the row.

These linear arrangements correspond to the permissible ways we can seat the students and teachers around the table. To account for rotational invariance, we divide the number of linear arrangements by $8$, which yields $$\frac{96 \cdot 5!}{8} = 12 \cdot 5! = 1440$$ permissible seating arrangements around a circular table, as you found.

• Thanks for nice explanation. – 1 0 May 29 '16 at 10:52

Students $S_1,\dots,S_5$, teachers $T_1,T_2,T_3$. Suppose we put $S_1$ into a particular seat. Then there are $4!=24$ ways of choosing the anticlockwise order for the remaining students. There are two 5 possible positions for the teachers. We can choose which of them to leave unoccupied in ${5\choose2}=10$ ways, then place the teachers in the others in $3!=6$ ways. So we have reached 1440 ways.

But now for each of these ways we could rotate everyone round the table (keeping the order the same), so a grand total of $8\cdot1440=11520$ ways.

• In circular permutation problems, arrangements obtained by a rotation are considered equivalent. Hence, you do not need to multiply by $8$. – N. F. Taussig May 29 '16 at 10:25
• Someone voted your answer but I think answer should be $1440$. Am I correct? – 1 0 May 29 '16 at 10:55
• Who knows, the question is ambiguous. That is why I gave both answers. Table questions often regard arrangements which can be rotated into each other as identical, but the question does not say whether this is the case or not. – almagest May 29 '16 at 11:40
• Exactly, seems quite vague.. If the above is a problem of "circular permutation", then the answer is 1440. Otherwise, the answer is 11520. – user321268 May 29 '16 at 12:42