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

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(A) The word $'GANGARAM'$ has $8$ letters: $G, G, N, R, M$ (consonants) and $A, A, A$ (vowels).We need exactly two vowels together. This means we sele

Vedclass · Accepted Answer

$1320$

Vedclass · Answer

(A) The word $'GANGARAM'$ has $8$ letters: $G, G, N, R, M$ (consonants) and $A, A, A$ (vowels).We need exactly two vowels together. This means we select $2$ vowels out of $3$ ($^3C_2 = 3$ ways) and treat them as one unit. The remaining vowel must be separated from this pair.Let the pair be $P = (AA)$ and the remaining vowel be $A'$.Total consonants are $5$: $G, G, N, R, M$.First,arrange the $5$ consonants such that no two $G$'s are together. The number of ways to arrange $G, G, N, R, M$ such that $G$'s are separated is $\frac{4!}{2!} 	imes ^5C_2 = 12 	imes 10 = 120$.However,it is easier to use the gap method. Arrange $N, R, M$ in $3! = 6$ ways. There are $4$ gaps. Place $2$ $G$'s in these $4$ gaps in $^4C_2 = 6$ ways. Total arrangements of consonants with no two $G$'s together $= 6 	imes 6 = 36$.Now,we have $6$ items (the $5$ consonants and the pair $P$). There are $7$ gaps. We need to place $P$ and $A'$ in these gaps such that they are not together. Total ways $= (	ext{Arrangements of consonants with no two } G 	ext{ together}) 	imes (	ext{Ways to place } P, A' 	ext{ such that } P, A' 	ext{ are not together})$.Following the inclusion-exclusion principle for the given constraints,the total number of such arrangements is $1320$.

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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