Question

In a class of 10 students, there are 3 girls A, B, C. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive.

Answer 1

Hint: First find the possible ways to arrange the seven boys of the class in a row and then set the girls in between then and on the corner side so, the possible positions of the three girls is 8. Then find the possible arrangement of the girls, it gives the desired result.

Complete step by step solution:
It is given in the problem that there are 10 girls in the class out of which, 3 are girls A, B, and C.
We have to find the number of ways can they be arranged in a row such that no three girls are consecutive.
We know that the total number of students in the class is 10.
Out of which the number of girls is 3 then the number of boys must be 7.
There are different ways of arranging the boys so that no two girls sit together. First, find the number of ways to arrange seven boys in a row.
Number of ways to arrange seven boys in a row${ = ^7}{P_7} = 7! = 5040$
Now, we have to arrange 3 girls in between the boys so that no two girls sit together, then each girl sits in between the boys or on the corner side. Assume $\boxed B$ as a boy and $\boxed G$as a girl, then take a look at the possible position where the girls can sit.
\[\boxed G - \boxed B - \boxed G - \boxed B - \boxed G - \boxed B - \boxed G - \boxed B - \boxed G - \boxed B - \boxed G - \boxed B - \boxed G - \boxed B - \boxed G\]
There are 8 possible positions where the girls can sit in a row so that no two girls sit together.
Then the number of ways to arrange the girls in a row is given as:
$ = {}^8{p_{_3}}$
On fining the above permutation value,
$ = \dfrac{{8!}}{{(8 - 3)!}}$
On simplification of the above terms,
$ = \dfrac{{8!}}{{5!}} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5!}}$
$ = 8 \times 7 \times 6$
$ = 336$
So, there are 336 ways to arrange the girls in the row so that no two girls sit together.
Hence, the number of arranging 10 students in such a way that no two girls sit together is given as:
$ = 5040 \times 336 $
On simplification, we get
$ = 1693440$ways.

$\therefore$There are $1693440$ ways to arrange ten students so that no two girls sit together.

Note:
Ensure that the number of places in which the girl can be arranged is 8 and not 6 because the positions at both corner sides are also included. Therefore the number of arrangements is given as ${}^8{p_{_3}}$.