A man parks his car among n cars standing in a row, not at either of the ends, on his return he finds that exactly m of the n cars are still there, probability that both the cars parked on two sides of his car, have left is $\frac{(n-m)(n-m-1)}{(n-A)(n-B)}$ then $A+B$ is 

Number of ways in which remaining $\text{'}m-1\text{'}$ cars can take their places
      (excluding the car of man)  ${=}^{n-1}{C}_{m-1}$
       No.of ways in which remaining $(m-1)$ cars can take places keeping the two places on two sides of his car vacant  ${=}^{n-3}{C}_{m-1}$
Prob $=\frac{{ }^{n-3}{C}_{m-1}}{{ }^{n-1}{C}_{m-1}}=\frac{(n-m)(n-m-1)}{(n-1)(n-2)}$
$\begin{array}{l}\Rightarrow A=1\\ B=2\\ A+B=3\end{array}$