In how many ways can a team of $4$ boys and $4$ girls be selected out of $6$ boys and $5$ girls if a particular boy and girl refuse to work together? 
In how many ways can a team of $4$ boys and $4$ girls be selected out of $6$ boys and $5$ girls if a particular boy and a particular girl refuse to work together.

My attempt:
Case 1: That particular boy is not considered. So, 4 boys to be selected out of 5 boys. 4 girls to be selected out of 5 girls. So, $\dbinom{5}{4} \cdot \dbinom{5}{4}=25$
Case 2: That particular girl is not considered. So, 4 boys to be selected out of 6 boys. And 4 girls to be selected out of 4 girls. So, $\dbinom{6}{4}\cdot\dbinom{4}{4}=15$
Case 3: That particular boy and girl are not considered. So, 4 boys out of 5 and 4 girls out of 4. So, $\dbinom{5}{4}\cdot\dbinom{4}{4}=5$
So, total cases: $45$.
Answer is given as $35$.
What's my mistake?
 A: You ought to subtract case 3 from cases 1 and 2, not add it.
Out of the 25 possibilities in case 1, some of them are also counted in case 2. Case 3 tells you exactly that this overlap is 5 large. So adding cases 1 and 2, you get 40 possibilities. But five cases appear twice each among these 40. Thus the number of actual possibilities is 35.
Alternatively, just count all possibilities without restrictions, then subtract the number of invalid choices, where both the boy and the girl were chosen.
A: You count the cases where both of them are not selected (case 3) also in case 1 and case 2. Your cases should be mutually exclusive:
Case 1: BOY is not chosen, GIRL is: ${5 \choose 4}{4 \choose 3}$
Case 2: BOY is chosen, GIRL isn't. ${5 \choose 3}{4 \choose 4}$
Case 3: BOY & GIRL are not chosen. ${5 \choose 4}{4 \choose 4}$
Total: $20+10+5=35$.
A: Total cases are 40 and refused cases five so 40 -5 = 35
Or      no of cases by all students = 75 and
No of cases with unplayed students=40 so answer
Is 35
