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# A class has 3 teachers Mr. X, Ms. Y and Mrs. Z and 6 students A, B, C, D, E and F. Number of ways in which they can be seated in a line of 9 chairs, if between any 2 teachers there are exactly 2 students isA. $18\left( {6!} \right)$B. $12\left( {6!} \right)$C. $24\left( {6!} \right)$D. $6\left( {6!} \right)$

Last updated date: 02nd Aug 2024
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Hint: Here, we will find the possible seating arrangements using the given information. We will find the number of ways of sitting for each case. We will add the results of all cases to find the solution.

Complete step by step solution:
We want to seat the teachers and students in such a way that there are exactly two students between 2 teachers. We will represent a teacher with T and a student with S and we will make possible cases of seating.
TSSTSSTSS ……………………………$\left( 1 \right)$
STSSTSSTS ……………………………$\left( 2 \right)$
SSTSSTSST ……………………………$\left( 3 \right)$

We can see that there are three ways to seat the students and teachers such that there are exactly 2 students between teachers.
We will find the number of ways of seating for each case using the formula for the number of $n$ - combinations of $n$ number of things.
In the first case, the teachers can be seated only on the 1st, 4th and 7th chairs. We can seat them in 3! ways. The students can be seated on the rest of the 6 seats in 6! Ways.
So total number of ways of seating in the case $\left( 1 \right)$is:
Number of ways of seating $= 3! \cdot \left( {6!} \right)$
Computing the factorial, we get
$\Rightarrow$ Number of ways of seating $= 3 \times 2 \times 1 \times 6!$
$\Rightarrow$ Number of ways of seating $= 6\left( {6!} \right)$
In the second case, the teachers can be seated only on the 2nd, 5th and 8th chairs. We can seat them in 3! ways. The students can be seated on the rest of the 6 seats in 6! Ways.
So total number of ways of seating in the case $\left( 2 \right)$ is:
Number of ways of seating $= 3! \cdot \left( {6!} \right)$
Computing the factorial, we get
$\Rightarrow$ Number of ways of seating $= 3 \times 2 \times 1 \times 6!$
$\Rightarrow$ Number of ways of seating $= 6\left( {6!} \right)$
In the third case, the teachers can be seated only on the 3rd, 6th and 9th chairs. We can seat them in 3! ways. The students can be seated on the rest of the 6 seats in 6! Ways.
So total number of ways of seating in the case $\left( 3 \right)$ is:
Number of ways of seating $= 3! \cdot \left( {6!} \right)$
Computing the factorial, we get
$\Rightarrow$ Number of ways of seating $= 3 \times 2 \times 1 \times 6!$
$\Rightarrow$ Number of ways of seating $= 6\left( {6!} \right)$
We will add results of all three cases to find the total number of ways of seating:
Total number of ways $= 6\left( {3!} \right) + 6\left( {3!} \right) + 6\left( {3!} \right) = 18\left( {3!} \right)$

$\therefore$ Option A is the correct answer.

Note:
1) We know that the number of $n$ - combinations of $n$ number of things is $n!$; this can be derived using the formula for permutations:
2) Number of $r$ - combinations of $n$ number of things is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.
3) If we substitute $n$ for $r$ in the above formula, we get:
$\begin{array}{l} \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{0!}}\\ \Rightarrow {}^n{P_r} = n!\end{array}$
4) We can see that this is the same as the formula we have used.