In how many ways can $4$ boys and $3$ girls be arranged in a row so that no two girls are together?

Question

Arrange 4 boys and 3 girls in a row so that no two girls are together.

I solved using this concept: If I have a total of $7$ seats, I can arrange boys in $7C4$ ways and girls in $7C3$ ways, then multiplying both by $3!$ and $4!$ as well which gives the result as: $$7C4 × 7C3 × 4! × 3! $$ But the answer actually is $5C3 × 4! × 3!$. Kindly help me where i am wrong.

Answer 1

You can use a simple method for these kinds of questions:
First arrange boys in four places $4! ways$

Now we have 5 gaps for girls in between boys. Now choose $5C3 * 3!$

So, $5C3 * 4! * 3!$ is the correct answer.

Edit:
5 gaps because _B_B_B_B_
Since girls can't be adjacent we can only place girls in the above gaps (denoted by "_"). There are 5 gaps possible.

Answer 2

A nice way to visualize things is to imagine nine numbered chairs in a row. Have the boys sit, in any order, in the $4$ even-numbered chairs. Then have Alice sit in one of the $5$ odd-numbered chairs, Barbara sit in one of the $4$ remaining odd-numbered chairs, and, finally, Carla sit in one of the $3$ remaining odd-numbered chairs. This ensures the none of the girls sit together, and can be done in a total of

$$4!\cdot5\cdot4\cdot3=1440$$

different ways. Now have them all stand up, step away from the chairs, and close ranks.