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
 A: In your Method ($1$), you calculate the number of teams with all combinations of boys and girls minus just the cases where $A$ and $B$ are both on a team at the same time (however, in your method description, you left out this important qualifier). This is where your $\binom{5}{1}$ comes from, since in those cases where $A$ and $B$ are both on one team at the same time, there is then just $1$ more boy left to choose from among the remaining group of $5$ boys.
However, in your Method ($2$), you have excluded both $A$ and $B$ from being a member on any team. This is more strict than just not having them both be on any one team at the same time, but still individually being on some different teams, which is why its result of $100$ is less than the correct result of $300$.
A: 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.
