If all the letters of the word 'GANGARAM' are | vedclass

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(A) The word $'GANGARAM'$ contains $8$ letters: $G, G, A, N, G, A, R, A, M$. The vowels are ${A, A, A}$ and consonants are ${G, G, N, R, M}$.Step $1$:

Vedclass · Accepted Answer

$1320$

Vedclass · Answer

(A) The word $'GANGARAM'$ contains $8$ letters: $G, G, A, N, G, A, R, A, M$. The vowels are ${A, A, A}$ and consonants are ${G, G, N, R, M}$.Step $1$: Arrange the consonants ${G, G, N, R, M}$ such that no two $'G'$ are together.Total arrangements of ${G, G, N, R, M}$ is $\frac{5!}{2!} = 60$.Arrangements where both $'G'$ are together is $4! = 24$.So,arrangements where no two $'G'$ are together is $60 - 24 = 36$.Step $2$: In these $36$ arrangements,there are $6$ gaps created (including ends) to place the vowels.We need exactly two vowels together. Let the vowels be $A_1, A_2, A_3$. We select $2$ vowels to be together as a block $(AA)$ and the third $A$ must be separate.Number of ways to select $2$ vowels out of $3$ is $^3C_2 = 3$.Treat $(AA)$ as one unit and $A$ as another unit. We have $2$ units to place in $6$ gaps.Number of ways to place these $2$ units in $6$ gaps is $^6P_2 = 30$.Total ways = $36 	imes 3 	imes 30 = 3240$.Wait,the standard approach for this specific constraint is: Total arrangements with exactly two vowels together (ignoring $G$ constraint) minus arrangements where two vowels are together $AND$ two $G$ are together.Calculation: $\frac{5!}{2!} 	imes ^6C_2 	imes 2! - 4! 	imes ^5C_2 	imes 2! = 60 	imes 15 	imes 2 - 24 	imes 10 	imes 2 = 1800 - 480 = 1320$.

If all the letters of the word $'GANGARAM'$ are arranged,then the number of words in which exactly two vowels are together but no two $'G'$ occur together is-

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