There are ten girls and four boys in Mr. Fat's Combinatorics class, Find the number of. . . 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? (Source : Titu Andreescu Combinatorics Textbook).
Here is my solution to this problem, however the number I got as an answer was extremely large and I felt I might have made a mistake so I need MSE to help verify and (probably) point out the error(s) in my solutions.
Solution 1. Let us use the Subtraction Principle in this case. There are 10 + 4 persons to be sitted around a circle, there are at most 13! ways to carry out this task.  Now.  Let us find all such sitting arrangements in which it is compulsory for two boys to sit next to each other, in this case, we can regard the two boys as an entity,  say boys $B_1$ $B_2$ as 
$X$ and boys $B_3$ $B_4$ as $Y$. Then in this case, the problem is this reduced to finding the number of possible cyclic permutations of the set P = {X, Y,  1, . . 10}, where 1, 2, 3, ... 10 denotes the girls.  The number of such cyclic permutations is at most 11!. But since we can permutate 2 boys from 4 boys in at most $_4P_2$/2 = 6 ways.  The number of such cyclic permutations is at most 11!$\cdot 6$ ways. 
Hence,  by the Subtraction Principle, there are at most 13! - 6(11!) = 5987520000 ways to arrange 10 girls and 14 boys in a circle such that no two boys are sitted together. 
Solution 2. In this case I used a direct approach by considering possible cases.
Case 1. A boy sits between every two girls. In this case we can select 7 out of 10 girl occupy the first 7 sits around the circle. The number of ways to permute 7 girls from 10 is at most $10P_7$/7 = 240 ways. 
Since 6! Was is the number of ways to sit them around a circle,  these 7 girls can occupy this sit in at most 240(6!) ways. Now the remaining 4 boys can sit anywhere provided that no two persons occupy the same seat, their sitting arrangements can be done in at most 6! ways. Hence for this case, we have a total of 240(6!)(6!) ways possible sitting arrangements. 
Case 2. A boy sits between every two girls, in this case,  we treat the two girls as a single entity like before, say X and Y.  And we can permute two girls from 10 in at most $\10P_2$/2 =45ways. These 5 entities can sit around the 5 available sits in at most 4! Ways.  This leaves us with a total of 45(4!) ways to execute the task. Now. The remaining 4 boys can sit on the remaining four sits around the circle in at most 3! ways.  This leaves us with a total of 45(4!)(3!) ways.  Obviously these are the only two cases we consider, hence for this solution, the answer is thus
240(6!)(6!) + 45(3!)(4!) = 12448080. But this answer doesn't correspond with my initial answer.  What's my mistake? 
 A: In your first approach, you have not taken into account the possibility that there can be more than one pair of boys sitting in adjacent seats and have not arranged the boys within the pair.  In your second approach, you have to take into account that a girl may be seated next to one, two, or no boys.
For the sake of comparison, we will first examine a simple solution, then correct your first approach.
Method 1:  Suppose Amelia is one of the girls.  Since the only relative order of the seats matters in a circular permutation, we will use Amelia as a reference point.  Seat Amelia.  The other nine girls can be arranged in $9!$ ways as we move clockwise around the circle from Amelia.  To separate the boys, we will seat them in four of the ten spaces to the immediate left of each girl. Once we choose these four spaces, the boys can be arranged in those spaces in $4!$ ways.  Hence, there are 
$$9!\binom{10}{4}4!$$
admissible seating arrangements.
Method 2:  We exclude those seating arrangements in which a pair of boys sit in adjacent seats from the total number of seating arrangements.
Once again, we seat Amelia.  The other $13$ students can be seated in $13!$ ways as we move clockwise around the circle from Amelia.
From these, we exclude those seating arrangements in which there are one or more pairs of boys in adjacent seats.
A pair of boys in adjacent seats:  Seat Amelia.  Choose which two of the four boys will sit in adjacent seats.  We have $12$ objects to arrange, the block of two boys, the other two boys, and the other nine girls. The objects can be arranged in $12!$ ways as we move clockwise around the circle from Amelia.  The boys can be arranged within the block in $2!$ ways.  Hence, there are 
$$\binom{4}{2}12!2!$$
such seating arrangements.
However, we have counted seating arrangements in which there are two pairs of boys sitting in adjacent seats twice, once for each way of designating one pairs as the pair of boys sitting in adjacent seats.  Since we only want to subtract such cases once, we must add them back.
Two pairs of boys who sit in adjacent seats:  This can occur in two ways.
Two overlapping pairs of boys sitting in adjacent seats:  This means that three boys sit in consecutive seats.  Seat Amelia.  Choose which three of the four boys will sit together.  We have eleven objects to arrange, the block of three boys, the other boy, and the other nine girls.  The objects can be arranged in $11!$ ways as we move clockwise around the circle from Amelia. The three boys can be arranged within the block in $3!$ ways.  Hence, there are 
$$\binom{4}{3}11!3!$$
such seating arrangements.
Two separate pairs of boys sitting in adjacent seats:  Seat Amelia.  This case is tricky since the two blocks are the same size, so we have to avoid counting each case twice.  There are three ways to choose which boy will be paired with the boy who appears  first in an alphabetical list.  This choice also determines the other pair of boys.  We have $11$ objects to arrange, the two pairs of boys and the other nine girls.  The objects can be arranged in $11!$ ways as we move clockwise around the circle from Amelia.  Each pair of boys can be arranged within their blocks in $2!$ ways.  Hence, there are 
$$3 \cdot 11! \cdot 2!2!$$
such seating arrangements.
Three pairs of boys sitting in adjacent seats:  This can only occur if all four boys sit in consecutive seats.  Seat Amelia.  We have ten objects to arrange, the block of boys and the other nine girls.  The objects can be arranged in $10!$ ways as we move clockwise around the circle from Amelia.  The boys can be arranged within the block in $4!$ ways.  Hence, there are 
$$10!4!$$
such seating arrangements.
By the Inclusion-Exclusion Principle, the number of admissible seating arrangements is 
$$13! - \binom{4}{2}12!2! + \binom{4}{3}11!3! + 3 \cdot 11!2!2! - 10!4!$$
