Find the number of ways of arranging 6 boys and 6 girls in a row. In how many of these arrangements.
i) All girls sit together
ii) No two girls sit together
iii) Boys and girls sit alternately

Sol: No of boys = 6
No. of girls = 6
Total no. of persons =12, and then can be arranged in a rows in 12! ways

(i) All girls sit together

Sol: Treat the 6 girls as one with they we have 6 boys + 1 unit of girls = 7 units They can be arranged in 7! ways. Now the 6
girls among themselves can be permuted in 6! ways
$\therefore$ Req. no. of arrangements = 7! × 6! 

(ii) No two girls sit together

Sol: First we arrange 6 boys in a row in 6! ways and 7 gaps are formed among them as shown in the given below
$-$ B$-$ B $-$B $-$B $-$B $-$B$-$
Now the 6 girls can be arranged in these 7 gaps in ⁷P₆ ways
$\therefore$ Req. no. of arrangements $=6!{\times }^7{P}_6$

(iii) Boys and girls sit alternately

Let us take 12 places then row may begin with either a boy or a girl That is 2 ways
B     G    B    G    B      G      B     G     B      G      B      G
G     B    G    B     G     B      G     B     G      B     G      B
If it begins with a boy, then all odd places will be occupied by boys and the even places occupied by the girls. The 6 boys can be arranged in the 6 odd places in 6! ways and the 6 girls can be arranged in the 6 even places in 6! ways
Req. no. of arrangements = 2 × 6! × 6!