Permutation Question with two independent conditions 
In how many ways can $8$ boys $(B_1,B_2,...,B_8)$ and $5$ girls we arranged linearly such that $B_1$ and $B_2$ are NOT together and exactly four girls are together?

I can solve when either of the conditions is imposed, the former would give $13!-2!\cdot12!$ and the latter would be ${5 \choose 4} \cdot 4! \cdot 10! - 5! \cdot 9!$
However I'm not getting how both these conditions together are meant to be evaluated. I thought it won't follow from principle of inclusion, as these are two independent cases.

 Answer given is $5! \cdot 8! \cdot 58$

 A: Solving them seperately won't help here. You will have to consider block type arrangements, and then count the permutations. Consider the group of girls that need to be together as a grouped object $X$. Now, you have to permute 8 boys, 1 girl and this group, such that $B_1$ and $B_2$ are not adjacent, and neither is the girl and this group object. This can be written as 
$$\text {Number of ways that B1 and B2 are not adjacent and the girl and X are not adjacent} = \text{Total ways} - \text{Ways such that B1 and B2 are adjacent} - \text{Ways such that girl and group are adjacent} + \text{Ways such that boys are adjacent and girl is adjacent to group}$$
You should be able to calculate each of these fairly easily 
A: Without inclusion exclusion:


*

*Select the group of four girls and permute them in $\binom{5}{4}4!=5!$ ways.

*There are two option for the guys, either $B_1$ and $B_2$ are together or not. If the former is true, then you have to put one of the group of girls exactly in the middle. So there are $2$ ways to choose which of the two groups of girls goes in the middle and $8$ ways to put them in any slot of the boys. You have to multiply this by the number of ways boys are together: $7!\cdot 2!.$ So this gives $2\cdot 8 \cdot 7!\cdot 2=4\cdot 8!$

*Now the bows have to be non together in $8!-7!\cdot 2!$ and we can put the girls in any possition we like, there are $9$ slots so in $\binom{9}{2}\cdot 2$ ways by choosing the slots and permuting the group of girls. Putting this process together:
$$5!\cdot (4\cdot 8!+\binom{9}{2}\cdot 2\cdot (8!-7!2!))=5!\cdot 8!\cdot (4+\binom{9}{2}\cdot 2-9\cdot 2)=5!\cdot 8!\cdot (76-18)$$
