<s>[INST] ways in which $A,B$ refuse to be the member of same team 
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 member of same team, is 

what i try
Method $(1)$
Ways in which $A,B$ not a member = Total -ways in which both $A,B$ included
$$=\binom{5}{2}\cdot \binom{7}{3}-\binom{5}{2}\cdot \binom{5}{1}=300$$
Method $(2)$
Ways in which $A,B$ not a member $$=\binom{5}{2}\cdot \binom{5}{3}=100$$
above we have excluded $2$ boys because they are not a member
But answer given is $300$
please explain me How i am wrong in $(2)$ Method
 [/INST] As John Omielan has pointed out, your second answer is incorrect since you omitted those cases in which exactly one of the boys $A$ or $B$ is a member of the team.  Let's correct your count.
$A$ is a member, but $B$ is not:  Since $A$ is on the team and $B$ is not, we must choose two of the five girls and two of the five boys other than $A$ or $B$, which can be done in
$$\binom{5}{2}\binom{5}{2}$$
ways.
$B$ is a member, but $A$ is not:  By symmetry, this can also be done in
$$\binom{5}{2}\binom{5}{2}$$
ways.
Neither $A$ nor $B$ is a member:  This is what you actually calculated.  We must choose two of the five girls and three of the other five boys, which can be done in
$$\binom{5}{2}\binom{5}{3}$$
ways.
Total:  Since the three cases are mutually exclusive and exhaustive, the number of teams with two girls and three boys which can be formed given that boys $A$ and $B$ refuse to be a member of the same team is
$$\binom{5}{2}\binom{5}{2} + \binom{5}{2}\binom{5}{2} + \binom{5}{2}\binom{5}{3} = 300$$
which agrees with the answer you obtained by subtracting the number of teams that include both $A$ and $B$ from the total number of teams that could be formed without restriction.
</s>