I have a problem when trying to understand the way the number of combinations can be found in this problem:
Three boys and three girls are to sit on a bench for a photograph. Find the number of ways this can be done if the three girls must sit apart.
Since the question before that asked to find the number of ways this can be done if the three girls must sit together, I thought that this value, which turns out to be $144 = 4!\times 3!$, should be somehow used in this question too.
So my approach is to, instead of finding the number of ways the girls can sit apart, I take the total number of was the three boys and three girls can sit, $6!$, and subtract the number of ways the girls should not sit.
The problem I have is to find the number of ways they can cannot sit or the number of ways they can sit so that they sit apart.
Some of my workings:
$$\{ab_1b_2b_3g_3\} \qquad a= \{g_1g_2\} \qquad \implies 6!-5!\times2!=480$$
The answer is:
Finding more than one position that the girls can sit (M1)
Counting exactly four positions (A1)
number of ways =
$4\times 3!\times 3! =144$
Best Answer
Well take that there are $4!$ combinations when $ag_3$ are together(take $n=\{ag_3\}$ so $\{nb_1b_2b_3\}$ has 4! permutations,so the number of exactly two girls together is $5!-4!=96$.Now you can take 2 girls in 3 ways $\{g_1g_2\},\{g_1g_3\},\{g_2g_3\}$ now the number of combinations so that there are exactly two girls together is $3*(5!*2!-96)$ and there are $4!*3!=144$ combinations with exactly 3 girls sitting together now $6!-3*(5!*2!-96)-144=144$ is the number of combinations when girls are setting apart(we substracted the number of all girls sitting together and exactly two girls sitting together)