# Is my solution to this permutation question correct?

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)

My solution

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..

• It's not really a set theory question. – tomasz Aug 8 '12 at 18:39
• Note that $n!(n+1)=(n+1)!$ – celtschk Aug 8 '12 at 18:41
• sorry my bad, the chapter was all about sets and permutations, just presumed set theory – hampusohlsson Aug 8 '12 at 18:42
• @tomasz what category would you suggest? – hampusohlsson Aug 8 '12 at 18:46
• Your answer is correct. One way to argue it for general $m$ and $n$ is that we can have $0$ or $1$ or $2$ or $\dots$ $n$ boys to the left of the clump of girls, so $n+1$ choices. Multiply $n+1$ by $m!n!$ to account for ordering. There is a cuter argument that gets you an answer of $(n+1)!m!$ directly. – André Nicolas Aug 8 '12 at 18:47

The basic idea and the answer are correct, but for a complete solution, you should show it in abstract context, so start with arbitrary (unspecified) $m,n$ and try to show that you get $(n+1)!m!$ as a result.