|A) 2(6!)||B) 6! x 7|
|C) 6! x ⁷P₆||D) None|
We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! x ⁷P₆