Question

Find the number of ways in which $6$ boys and $6$ girls can be seated in a row, so that
i) all the girls sit together and all the boys sit together
ii) all the girls are never together.

• A. (i) $2\times (6!{)}^2$ and (ii) $12!-7!6!$
• B. (i) $(6!{)}^2$ and (ii) $12!+7!6!$
• C. (i) $(6!{)}^2$ and (ii) $12!-7!6!$
• D. (i) $2\times (6!{)}^2$ and (ii) $12!+7!6!$

Answer 1

The correct option is A (i) $2\times (6!{)}^2$ and (ii) $12!-7!6!$

i) Considering boys and girls as two units, the number of permutations is

$2!\times 6!\times 6!=2\times (6!{)}^2$

ii) The total arrangements where all girls are not together is as follows : Total arrangement without any restriction - arrangement when all girls are together $=\left(12\right)!-7!6!$