Arrangement related question Question- $5$ boys and $5$ girls to be arranged in a row such that boys and girls are alternate and a particular boy and a particular girl are never together. Try- taking the total number of cases to be $5!×5!×2$ then making groups such as $b2g2$ $b3g3$ $b4g4$ $b5g5$ and subtracting the case of permutations of b1g1 from original . Is this the right and an answer would be very much appreciated.

Your $2 \cdot 5!^2$ is correct for the number of ways to seat them ignoring the restriction of the two who won't sit together. Now subtract the number of ways with them sitting together. The first one you come to in the row can be either the boy or girl and can be in nine places.