The number of arrangements of the letters of | vedclass

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Question

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$. The frequencies are: $A: 3, N: 2, B: 1$.Total number of arrangements = $\frac{6!}{3! \

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$. The frequencies are: $A: 3, N: 2, B: 1$.Total number of arrangements = $\frac{6!}{3! 	imes 2!} = \frac{720}{6 	imes 2} = 60$.Now,consider the case where the two $N$s are together. Treat $(NN)$ as a single unit. The letters are now ${B, A, A, A, (NN)}$.Number of arrangements with $N$s together = $\frac{5!}{3! 	imes 1!} = \frac{120}{6} = 20$.Number of arrangements where $N$s do not appear adjacently = (Total arrangements) - (Arrangements with $N$s together) = $60 - 20 = 40$.

The number of arrangements of the letters of the word $BANANA$ in which two $N$s do not appear adjacently is

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