The number of four-letter words that can be f | vedclass

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(B) The word '$MAYANK$' consists of $6$ letters: $M, A, Y, A, N, K$. The distinct letters are ${M, Y, N, K}$ and the repeated letter is ${A, A}$.To fo

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$36$

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(B) The word '$MAYANK$' consists of $6$ letters: $M, A, Y, A, N, K$. The distinct letters are ${M, Y, N, K}$ and the repeated letter is ${A, A}$.To form a $4$-letter word containing both '$A$'s,we need to select $2$ more letters from the remaining $4$ distinct letters ${M, Y, N, K}$.The number of ways to select $2$ letters from $4$ is $^4C_2 = 6$.Now,we have $4$ letters in total (e.g.,$A, A, X, Y$). The number of ways to arrange these $4$ letters is $\frac{4!}{2!} = 12$.Total words containing both '$A$'s = $6 	imes 12 = 72$.Now,we calculate the number of words where both '$A$'s are together. Treat $(AA)$ as one unit. We have $3$ units to arrange: $(AA), X, Y$. The number of ways to arrange these is $3! = 6$.Since we have $6$ pairs of letters ${X, Y}$,the number of words where both '$A$'s are together is $6 	imes 6 = 36$.The number of words where both '$A$'s are $NOT$ together = (Total words with both '$A$'s) - (Words with '$A$'s together) = $72 - 36 = 36$.

The number of four-letter words that can be formed from the letters of the word '$MAYANK$' such that both '$A$'s are included but never together is equal to:

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