Find the number of arrangements of the letter | vedclass

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Question

(A) The word $BANANA$ contains $6$ letters: $3$ $A$'s,$2$ $N$'s,and $1$ $B$.The total number of arrangements is $\frac{6!}{3!2!} = \frac{720}{6 	imes

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The word $BANANA$ contains $6$ letters: $3$ $A$'s,$2$ $N$'s,and $1$ $B$.The total number of arrangements is $\frac{6!}{3!2!} = \frac{720}{6 	imes 2} = 60$.To find the arrangements where the two $N$'s appear together,we treat the two $N$'s as a single unit $(NN)$. Now we have $5$ units: $(NN), A, A, A, B$.The number of such arrangements is $\frac{5!}{3!1!} = \frac{120}{6} = 20$.The number of arrangements where the two $N$'s do not appear together is the total arrangements minus the arrangements where they are together:$60 - 20 = 40$.

Find the number of arrangements of the letters of the word $BANANA$ in which the two $N$'s do not appear together.

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