From the letters of the word BANANA find the number of arrangements such that
i) all A’S come together
ii) no two A’S come together

Given word BANANA contains 6 letters in which 3 A’s, 2 N’s are alike and one B
(i) All the A’s are together:-
Treat the 3 A’s as one unit then we have 3+1 = 4 units in which 2 N’s are alike and rest are different

$\therefore$They can be arranged in$\frac{4!}{2!}$ways The 3 A’s among them selves can be arranged in $\frac{3!}{3!}=1$

$\therefore$ Req. no. of arrangements ways$=\frac{4!}{2!}\times 1=12$

(ii) No. two A’s come together:-
First we arrange the remaining 3 letters in which 2 N’s and one is different in $\frac{3!}{2!}$ways $-$B $-$ N $-$ N $-$Then we can find 4 gaps among them the 3 A’s can be arranged in these 4 gaps in $\frac{{ }^4{P}_3}{3!}$ways
$\therefore$Req. no. of arrangements$=\frac{3!}{2!}\times \frac{{ }^4{P}_3}{3!}=3\times 4=12$