Question

Given,$12$ boys and $2$ girls are to be seated in a row such that there are at least $3$ boys between the $2$ girls. The number of ways this can be done is $m.12!$ where $m = ?$
A.${2.^{12}}{C_6}$
B.$20$
C.$^{11}{P_2}$
D.$^{11}{C_2}$

Answer 1

Hint: We need to analyze the given information first so that we can able to solve the problem. Here we are given some arrangements. And, we are asked to find the value of$m$. Also, it is given that the number of ways in which $12$ boys and $2$ girls are to be seated in a row such that there are at least $3$ boys between the $2$ girls is $m.12!$. We need to find the number of arrangements.

Complete step by step solution:
Here, we are given that there are $12$ boys and $2$ girls and they are to be seated in a row.
Total seats will be $14$ because there are $12$ boys and $2$ girls, adding them we get $14$.
Without applying any condition, we need to find the total number of arrangements.
Hence, the total number of arrangements will be$14!$
Now, we shall find the seating arrangements of two girls such that the girls are to be seated together.
Hence, \[1\_2,2\_3,3\_4,4\_5,5\_6,6\_7,7\_8,8\_9,9\_10,10\_11,11\_12,12\_13,13\_14\] are the required seating arrangements of two girls. That is one girl will be seated in$13$ ways and another girl will be seated in $13$ ways.
Therefore, the arrangements of two girls seated together
$ = 2 \times 13$
$ = 26$
Now, we need to place a boy between the two girls.
Hence, the seating will be$123,234,...$ and there will be$12$ways.
Therefore, the arrangements of one boy between two girls
$ = 2 \times 12$
$ = 24$
Now, we need to place two boys between the two girls.
Hence, the seating arrangement will be $1234,2345,...$ and there will be $11$ ways.
Therefore, the arrangements of two boys between two girls
$ = 2 \times 11$
$ = 22$
Now, we need to find the arrangement such that there are at least $3$ boys between the $2$ girls.
So, we shall follow the below steps.
The required arrangements = $14! - \left( {26 + 24 + 22} \right)12!$ (Here$12!$ is applied since the arrangement of boys varied)
$ = 14 \times 13 \times 12! - 72 \times 12!$
$ = \left( {14 \times 13 - 72} \right)12!$
$ = \left( {182 - 72} \right)12!$
$ = 110 \times 12!$
We are given $m.12!$.
Since we have found it $110 \times 12!$, we need to equate it with the given $m.12!$.
Thus $m.12!$$ = 110 \times 12!$
Hence, $m = 110$
$ = \dfrac{{11 \times 10 \times 9!}}{{9!}}$
$ = \dfrac{{11!}}{{\left( {11 - 2} \right)!}}$
\[{ = ^{11}}{P_2}\]
Hence, we get$m{ = ^{11}}{P_2}$.

Thus the option C is the correct answer.

Note:
First, we have found the number of arrangements of two girls together, arrangements of a boy and two boys between the girls. Then we subtract it from the total arrangements. Then we compare the resultant answer with the given$m.12!$. Here, $m = 110$is also correct. We need to proceed further because $m = 110$is not there in the options.