Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
2  
It's not really a set theory question. –  tomasz Aug 8 '12 at 18:39
1  
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 –  hamohl Aug 8 '12 at 18:42
    
@tomasz what category would you suggest? –  hamohl Aug 8 '12 at 18:46
3  
For an entirely misogynistic hint to arrive at the general solution, begin your argument by treating all of the girls together as a single "person". Later separate them as individuals. –  Arthur Fischer Aug 8 '12 at 18:52
show 3 more comments

1 Answer

up vote 0 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.