The number of arrangements of the letters of | vedclass

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(A) The word $MANAMA$ contains $6$ letters: $M$ appears $2$ times and $A$ appears $3$ times,and $N$ appears $1$ time.First,we arrange the letters othe

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$40$

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(A) The word $MANAMA$ contains $6$ letters: $M$ appears $2$ times and $A$ appears $3$ times,and $N$ appears $1$ time.First,we arrange the letters other than $M$,which are $A, A, A, N$.The number of ways to arrange these $4$ letters is $\frac{4!}{3!1!} = 4$.These $4$ letters create $5$ possible gaps (including the ends) where the $2$ $M$'s can be placed so that they are not adjacent.The number of ways to choose $2$ gaps out of $5$ is $\binom{5}{2} = 10$.Since the $2$ $M$'s are identical,the number of ways to place them is $\binom{5}{2} = 10$.Therefore,the total number of arrangements is $4 	imes 10 = 40$.

The number of arrangements of the letters of the word $MANAMA$ in which two $M$'s do not appear adjacent is

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