Number of ways to sit 6 girls and 6 boys together with no two girls together.

Question

As the title of the question explains:

What I thought on the very first instant was that we will make them sit alternate hence the answer will be 2 * 6! * 6!

But then in some book I saw the explanation that there are 7 places where a Girl can sit out of the given 12 seats. Hence the answer is 7C6 * 6! * 6!.

But I don't get how are there 7 places for girls to sit.

I understand that howcome my original assumption of alternate placement was wrong. Because you can do

G B B G B G B G B G B G  

And so on. But the authors explanation of 7 seats for girls is still not clear to me.

Cheers.

Answer 1

Let the boys arrange themselves in $6!$ ways. Girls can sit in between or besides them such that maximum one girl is between any $2$ boys.

$$\text{1 B 2 B 3 B 4 B 5 B 6 B 7}$$

Clearly, there are $1-7$ places for $6$ girls. $7$ places. I bet you can take from here

Answer 2

Another way to look at this problem is to say that either the boys and girls are alternating (2 ways), or there is a pair of two boys somewhere in between two of out of the 6 girls (5 ways to choose where the pair is, since the pair cannot be on the end). So there are 7 possible cases before you choose the ordering of the boys and girls.