Question

A class has three teachers Mr. X, Ms. Y and Mrs. Z and six students A,B,C,D,E,F . Numbers of ways in which they can be seated in a line of 9 chairs, if between any two teachers there are exactly two students is,
A) $18(6!)$
B) $12(6!)$
C) $24(6!)$
D) $6(6!)$

Answer 1

Hint: This question will be done by using fundamental counting principle and is defined by the way we figure out the number of outcomes of a probability problem. In this you need to multiply the events together to get the total number of outcomes. The formula is given by: -
If you have an event ‘a’ and another event ‘b’ then all the different outcomes for the events is \[a \times b\]
So in this question we will make different arrangements in which there are two students in between two teachers.

Complete step-by-step answer:
For simplification let denote the teachers as ‘T’ and students as ‘S’
Number of chairs= 9
Students= A,B,C,D,E,F
Now we will arrange according to the condition given in question.
TSS,TSS,TSS= $3! \times 6!$
SST,SST,SST= $3! \times 6!$
STS,STS,STS=$3! \times 6!$
Now total number of ways = Sum of above three ways
$ \Rightarrow 3! \times 6! + 3! \times 6! + 3! \times 6!$
$ \Rightarrow 3 \times 3! \times 6!$
Now we will expand the factorial value.
$ \Rightarrow 3 \times 3 \times 2 \times 6!$
$ \Rightarrow 18(6!)$
Hence total number of arrangements that can be made if between any two teachers there are exactly two students is $18(6!)$

Therefore correct option is (A)

Note: Students may find difficulty in expanding the factorial so here is below mentioning one example to show how to expand it:-
Let we have to expand $6!$ then we will start multiplying number and then one less number up to 1 like $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$