Find the number of arrangements of letters
of word “ASSOCIATIONS” such that
(i) all S’s come together
(ii) no 2 A’s come together

The given word “ASSOCIATIONS” has 12
letters in which there are 2 A’s, 3 S’s, 2 O’s, 2 I’s
are alike and rest are different
$\therefore$ They can be arranged in  $\frac{12!}{2!.3!.2!.2!}$ ways
(i) All S’s come together:-
Treat 3 S’s as one unit  then we have 9 + 1 unit
= 10 units in which there are 2 A’s, 2 O’s, 2I’s
are alike and rest are different they can be
arranged in $\frac{10!}{2!.2!.2!}$ ways. The 3 S’s among
them selves can be arranged in $\frac{3!}{3!}=1$ ways
 $\therefore$Req. no. of arrangements = $\frac{10!}{2!.2!.2!}\times 1$

(ii) 2 A’s do not come together:-
First arrange the remaining 10 letters in which
there are 3 S’s, 2 O’s, 2 I’s are alike and rest are
different in $\frac{10!}{3!.2!.2!}$ ways
Then we have 11 gaps between them the 2 A’s
can be arranged in these 11 gaps in $\frac{{ }^{11}{P}_2}{2!}$ways
$\therefore$ Req. no. of arrangements = $\frac{10!}{3!.2!.2!}\times \frac{{ }^{11}{P}_2}{2!}$