5 boys 3 girls in row a such that not all girls together. Two different approaches give different solutions.

Question

Question: In how many ways $5$ boys and $3$ girls can be seated in a row so that all $3$ girls do not sit together?

Attempt 1:
Find all the possible ways (which is $8!$), and subtract the total number of ways that the 3 girls can sit together.

G = Girl, B = Boy

Number of ways in GGG $ =3!$

Consider (GGG) as one unit. Then,

Number of ways in (GGG)BBBBB $= 6!$

Therefore, total number of ways that the 3 girls can sit together $= 3! \times 6! = 4320$

Answer $= 8! - 4320 = 40320 - 4320 = 36000$

Attempt 2:
Place boys in by keeping gaps for girls such that a gap is smaller for all 3 girls.

_ _ B _ _ B _ _ B _ _ B _ _ B _ _

Each gap has space for two girls. Therefore, all $3$ girls cannot sit together. And now, we can assign the $3$ girl to $3$ spaces from the above $12$ spaces in the gaps.

Total number of ways that girls can choose spaces $= 12 \times 11 \times 10 = 1320$

Total number of ways that boys can sit $= 5! = 120 $

Answer $= 1320 \times 120 = 158400$

Why we get two different answers to two approaches? I guess that the Attempt 1 is correct; If so why attempt 2 is wrong?

Note: By referring to following question I could get more understanding how to deal with the question: In how many ways can 5 boys and 3 girls be seated in a row such that no two girls are together?. But I could not get what is wrong with my attempt 2 by referring that. So please answer mainly focusing on how the Attempt 2 become fail?

Answer 1

In the second attempt, you say that there are $12\times 11\times 10$ ways for 3 girls to choose a space. However, there are two ways to place one girl between two boys: B G _ B and B _ G B (same goes for the edge cases on the left of the leftmost boy _ _ B and on the right of the rightmost boy B _ _). The calculation $12\times 11\times 10$ counts them separately, while for the context of this exercise, such placements should be considered the same: it's one girl between two boys (or, one girl next to a boy).

Answer 2

In attempt 2, you are overcounting. For example, when two girls are in the first two positions, only $5$ (not $10$) seats remain for the last girl.

You should consider two cases: 1) 1 double and 1 single; 2) all single:

$$P(3,2){6\choose 1}{5\choose 1}+P(6,3)=180+120=300$$ Hence: $$300*5!=300*120=36000.$$