Management Aptitude Test Sep 2013 Paper

(a) There are ten places, where the boys can be seated. Now, ${\mathit{a}}_2$ cannot come at 9th and 10th positions otherwise condition is not fulfilled.
∴ Total number of ways the boys can be arranged in 8 places regarding ${\mathit{a}}_2$ = 8!
Similarly for ${\mathit{a}}_5$, he cannot be seated in first place and at the 10th place, then possible ways of arrangement = 8!
Now, ${\mathit{a}}_7$ cannot be seated on 1st and 2nd places, then possible number of ways = 8!
∴ Total number of w s = 8! + 8! + 8!
= 3 x 8 !
$\frac{3\times 3\times 8!}{3}$ = $\frac{9\times 8!}{3}=\frac{9!}{3}$