The number of four-lettered words that can be | vedclass

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(B) The word $MAYANK$ contains $6$ letters: $M, A, Y, A, N, K$. The distinct letters are $M, Y, N, K$ and two $A$'s.To form a $4$-letter word containi

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) The word $MAYANK$ contains $6$ letters: $M, A, Y, A, N, K$. The distinct letters are $M, Y, N, K$ and two $A$'s.To form a $4$-letter word containing both $A$'s,we need to select $2$ more letters from the remaining $4$ distinct letters $(M, Y, N, K)$.Number of ways to select $2$ letters from $4$ is $^4C_2 = \frac{4 	imes 3}{2} = 6$.Now,we have $4$ letters: $A, A$ and two other letters (say $X, Y$).The total number of arrangements of these $4$ letters is $\frac{4!}{2!} = \frac{24}{2} = 12$.Since there are $6$ ways to choose the $2$ letters,the total number of words containing both $A$'s is $6 	imes 12 = 72$.Now,we find the number of words where both $A$'s are together. Treat $(AA)$ as one unit. We have $3$ units: $(AA), X, Y$.The number of arrangements is $3! = 6$.Since there are $6$ ways to choose the $2$ letters,the number of words where $A$'s are together is $6 	imes 6 = 36$.The number of words where $A$'s are never together is $72 - 36 = 36$.

The number of four-lettered 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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