Q.There are 5 girls and 2 boys, then the probability that no two boys are sitting together for a photograph is A) $\frac{1}{21}$ B) $\frac{4}{7}$ C) $\frac{2}{7}$ D) $\frac{5}{7}$

Total arrangements of 7 people in a row: $7!=5040$. Arrange the 5 girls first ($5!$ ways), creating 6 gaps (before, between, and after them) into which the 2 boys can be placed so no two boys sit together: choose and order 2 of the 6 gaps, $P(6,2)=6\times5=30$ ways. Favourable arrangements $=5!\times30=120\times30=3600$. $P=3600/5040=5/7$.

Gap method for 'no two together': arrange the majority group first, then place the minority group into the gaps it creates.

Placing the boys into only 5 gaps (between girls) instead of 6 (also counting the two end positions), which undercounts the favourable arrangements.