Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

We have a total of 5 girls and 7 boys in the class. Among the boys, two particular boys are named A and B, and they do not want to be in the same team. We want teams that always contain exactly 2 girls and 3 boys.

We first count the number of teams without any restriction and then subtract the “bad” teams that contain both A and B.

The combination formula is stated as: $$ ^nC_r \;=\; \frac{n!}{r!\,(n-r)!} $$ which gives the number of ways to choose $$r$$ objects out of $$n$$ objects.

Unrestricted count: For girls, we need 2 out of 5. $$ \text{Ways for girls} \;=\; ^5C_2 $$ Now, by direct substitution, $$ ^5C_2 \;=\; \frac{5!}{2!\,3!} \;=\; \frac{120}{2 \times 6} \;=\; 10. $$

For boys, we need 3 out of 7. $$ \text{Ways for boys} \;=\; ^7C_3 $$ Again using the formula, $$ ^7C_3 \;=\; \frac{7!}{3!\,4!} \;=\; \frac{5040}{6 \times 24} \;=\; 35. $$

The Product Rule says the total number of independent choices equals the product of the individual numbers of ways. So, the total number of unrestricted teams is $$ 10 \times 35 \;=\; 350. $$

Now we remove the “bad” teams in which both special boys A and B appear together.

Fixing A and B in the team leaves room for only one more boy. The remaining boys are 7 − 2 = 5 in number.

The number of ways to choose that one additional boy is $$ ^5C_1 = 5. $$

The girls are still chosen in the unrestricted way, namely $$ ^5C_2 = 10. $$

So, the number of “bad” teams containing both A and B is $$ 10 \times 5 \;=\; 50. $$

Finally, by the Subtraction Principle, the required number of valid teams is $$ 350 \;-\; 50 \;=\; 300. $$

Hence, the correct answer is Option A.