The number of arrangements of the letters of | vedclass

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Question

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$. Here,$A$ appears $3$ times and $N$ appears $2$ times. Total number of arrangements = $

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The word $BANANA$ contains $6$ letters: $B, A, N, A, N, A$. Here,$A$ appears $3$ times and $N$ appears $2$ times. Total number of arrangements = $\frac{6!}{3! 	imes 2!} = \frac{720}{6 	imes 2} = 60$. Number of arrangements where two $N$s are together: Treat $(NN)$ as one unit. Now we have $5$ units: $B, A, A, A, (NN)$. Number of arrangements = $\frac{5!}{3! 	imes 1!} = \frac{120}{6} = 20$. Number of arrangements where two $N$s do not appear adjacently = (Total arrangements) - (Arrangements where $N$s are 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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