Problem (1.8) A Path to Combinatorics Solution Verification. 
Problem (1.8): There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other.

My Attempt: Total number of ways in which both boys are girls can be arranged on a circular table is $13!$ and the number of ways in which any two boys can be together is $2!*12!$. Therefore the answer is $13!-2*12!=11*12!$. Is this answer correct? I am asking this because I am confused about the exact meaning of the line "no boys are together." Does this mean that no two/three/four boys are together? 
Note: Meanwhile I have made another attempt on the problem. We can place the $10$ girls on the circular table in $9!$ ways. Then we are left with $10$ spots which need to be filled with $4$ boys. This can be done in $P^{10}_4$ ways. Thus, the number of ways is $9!P^{10}_4$.  
 A: First way:


*

*Seat the girls.

*Since no two boys can sit together the boys will be inserted separately between the girls. Pick 4 girls to insert the boys to the right hand side of. 

*Consider how many different orders the boys may be inserted in the places indicated.


Second way:
Forget for a moment that the table is circular. Pretend that it's a linear table. The only difference is that there will be a 14-fold symmetry in a circular table (in other words any rotation of a solution is considered identical to the original). So we may as well make our calculations for a linear table and then divide by 14.


*

*Seat the girls

*Since no two boys can sit together the boys will be inserted separately between the girls. Pick 4 from the 11 positions available (including the left of the leftmost girl and right of the rightmost girl) but exclude the cases where both of the leftmost and rightmost positions are picked.

*Consider how many different orders the boys may be inserted in the places picked.

*Divide by 14 to account for the circular symmetry.

A: The question asks in how many ways can you arrange them such that you never have two boys adjacent to each other. The problem with your method is that you don't account for how many possible ways the boys sit adjacent to each other. First of all, their are $4$ boys, so there are alone ${4 \choose 2}*2$ ways for two to be adjacent. However, you could have multiple adjacent at once, but then you're counting that multiple times. There are situations in which only two are adjacent, or two pairs are adjacent, or three are adjacent, or all four are adjacent- truthfully, this method is two complicated.
Try this instead- order the boys first. There are $3!$ ways to order the boys in a circle. Then order the girls- this time, since you've already seated the boys, you can pick one boy (arbitrarily) and decide which girl sits to his right. Essentially, ordering the girls is not a circular permutation, so there are $10!$ ways to order them.
Finally, we decide how the girls are to sit with respect to the boys. This is known as a composition problem- we have four empty regions between a pair of boys, and we want at least one girl in each so that the boys are adjacent. In other words, we want a composition of $10$ into $4$ terms, none of which are $0$. (Ordinarily, in a composition problem, we work with identical objects, but since we already ordered the girls, we can treat them as identical, and then impose the ordering onto the composition.) This is just ${9 \choose 3}$.
The wikipedia link explains this, but the idea is that if I have $n$ candy bars, and I want to spread them among $k$ children such that each child gets at least, that's the same as putting the candy bars in a row, and then putting $k-1$ walls, each between a pair of candy bars. You can easily see that if I have $k-1$ walls, then I have $k$ different parts. Since there are $n-1$ spaces that exist between candy bars, this is ${n-1 \choose k-1}$.
Multiplying my terms together, this is $3!10!{9\choose 3}$.
A: Consider the 14 seats around the table. We can choose 4 non consecutive seats in $\frac{14}{14-4}\binom{14-4}{4}$ ways. We can place the boys and girls in $10!4!$ ways but to account for the circular permutation, we need to divide this by 14. Hence the number of ways is 
$$\frac{14}{10} \binom{10}{4}\frac{10!4!}{14} = 10! \cdot 9 \cdot 8 \cdot 7$$
