The number of ways of arranging the letters o | vedclass

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(C) The word $BANANA$ contains $6$ letters: $A, A, A, B, N, N$.Total arrangements of $BANANA = \frac{6!}{3!2!1!} = \frac{720}{6 	imes 2} = 60$.To fin

Vedclass · Accepted Answer

$40$

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(C) The word $BANANA$ contains $6$ letters: $A, A, A, B, N, N$.Total arrangements of $BANANA = \frac{6!}{3!2!1!} = \frac{720}{6 	imes 2} = 60$.To find the number of arrangements where the two $N$s do not come together,we subtract the arrangements where the two $N$s are together from the total arrangements.Treat the two $N$s as a single unit $(NN)$. Now we have $5$ units: $A, A, A, B, (NN)$.Arrangements with $N$s together $= \frac{5!}{3!1!1!} = \frac{120}{6} = 20$.Number of ways where $N$s do not come together $= 60 - 20 = 40$.

The number of ways of arranging the letters of the word $BANANA$ so that the two $N$s do not come together is:

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