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)$\,2\times {{\left( 6! \right)}^{2}}$ and (ii) $12!\,-7!6!$
(b) (i) ${{\left( 6! \right)}^{2}}$and (ii) $12!\,+7!6!$
(c) (i) ${{\left( 6! \right)}^{2}}$and (ii) $12!\,-7!6!$
(d) (i) $2\times {{\left( 6! \right)}^{2}}$ and (ii) $12!\,+7!6!$

Answer 1

Hint: Here, use the general form of permutations $^{n}{{P}_{r}}$ or when n = r, $n!={{\,}^{n}}{{P}_{n}}$ and find the solution for the first sub-question. For the next one, subtract the number of ways of sitting together from the total number of ways to find the number of ways where girls are never together with the help of multiplication rule.

Complete step-by-step answer:
In this question, we need to the find the arrangement of 6 boys and 6 girls that can be seated in a row, according to the conditions given,
Let us consider that there are 12 seats which will be occupied by 6 boys and 6 girls,

01

02

03

04

05

06

07

08

09

10

11

12

(i) All the girls sit together and all the boys sit together
Let us suppose that the 6 boys are considered as one unit and 6 girls are considered as another unit. This means there are 2 units. When, these 6 boys sit together and 6 girls sit together, hence we can find the number of permutations by using $^{n}{{P}_{r}}$, (notation for the number of permutations $r$ from $n$ distinct values)
Number of ways $=2\times \left( 6! \right)\times \left( 6! \right)$............................ ($\because n!={{\,}^{n}}{{P}_{n}}$)
Here, 2 indicates the units, one being 6 boys and the other being 6 girls and $6!={{\,}^{6}}{{P}_{6}}$ is the number of ways both the girls and the boys can arrange themselves.
Therefore, the number of ways 6 boys can sit together and 6 girls can sit together is $2\times {{\left( 6! \right)}^{2}}$
(ii) All the girls are never together
We can find the required value by subtracting the number of ways where the girls can sit together from the total number of ways of occupying the seats.
That elaborates to, Total number of ways = Girls sitting together + Girls never sitting together
Therefore, Girls never sitting together = Total number of ways – Girls sitting together
Here, we know the total number of ways the girls $12!$
Now, when girls sit together it will be by considering them as one unit, so we need to arrange the 6 boys and one unit of girls together = 6+1 = 7, hence they have 7! ways and the girls can rearrange themselves by 6! ways by the help of multiplication rule in permutations.
Therefore, the number of ways when girls are sitting together is $7!.6!$
So, now the number of ways of girls never sitting together = $12! – 7!6!$
Hence, the answers are (i) $2\times {{\left( 6! \right)}^{2}}$ and (ii) $12! – 7!6!$

Note: In this question, for the second sub-question, you could directly find out the number of ways girls are never sitting together but it would be confusing and requires time, but the method we used is easy to understand and saves a lot of time.