[Math] In how many ways can 7 boys and 3 girls be arranged in a row so that the end positions are taken by the boys

Question

In how many ways can 7 different boys and 3 different girls be arranged in a row so that the end positions are taken by the boys and no 2 girls are sat next to each other.

I think I have an idea on how to solve this but my answer differs from the one in the solutions.

First of all choose the boys for the end positions, $7*6$ but since there are 2 end positions we get $7*6*2$. The rest of the boys can be seated in $5!$ ways.

Now consider the configuration $$BXBXBXBXBXBXB$$

Where $X$ is a potential position for a girl to sit. There are 6 such positions so for the girls its $\tbinom 63$

Putting it all together gives $$\tbinom 63 *5! *5 *4 *2$$

But the answer in the solutions is $3!*\tbinom 63$

Answer 1

You need to count ways to :

• Line up the seven boys in a row.
• Line up the three girls in a row.
• Select three of the six spaces between the boys to merge the lines.

The full answer is:

$$7!\,3!\,\binom{6}{3}$$