Question: Suppose there are m girls and n boys in a class. What is the number of ways of arranging them in a line so that all the girls are together? (Biggs, Discrete Mathematics 2nd ed, Exercise 10.7.5)
Say $m=4$ and $n=3$
The number of ways they can be arranged are:
G - a girl, B - a boy
position 1: [G G G G][B B B] position 2: [B][G G G G][B B] position 3: [B B][G G G G][B] position 4: [B B B][G G G G]
So in this case the group of girls can be placed in $4$ different positions. On each position the group of girls can internally be arranged in $4!$ different ways, and the boys can be arranged in $3!$ different ways.
Hence the total number of arrangements are $4!*3!*4$
Converting the answer back to variables, I end up with: $m!*n!*(n + 1)$
The book provides no solution, so I would really like to know if I came up with the right answer..