When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.

Correct Option: B

When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways.
∴ Total number of ways = 4! = 24

Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).

Correct Option: A

Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).
∴ Total number of ways = 3! X 2! = 12

Total number of words = 5! = 120
combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways.

Correct Option: B

Total number of words = 5! = 120
combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways
So, when vowels are together, then number of words = 3! x 3! = 36
there4; Required number of ways, when vowels being never together =120 - 36 = 84

We have If ^{2n + 1}P_{n - 1} / ^{2n - 1}P_n = 3/5
⇒ 5 x ^{2n + 1}P_{n - 1} = 3 x ^{2n - 1}P_n
⇒ {5 x (2n + 1)!} / {(n + 2 )!} = 3 x {(2n - 1)!} / {(n - 1)!}

Correct Option: A

We have If ^{2n + 1}P_{n - 1} / ^{2n - 1}P_n = 3/5
⇒ 5 x ^{2n + 1}P_{n - 1} = 3 x ^{2n - 1}P_n
⇒ {5 x (2n + 1)!} / {(n + 2 )!} = 3 x {(2n - 1)!} / {(n - 1)!}
⇒ 10(2n + 1) = 3(n + 2) (n + 1)
⇒ 3n² - 11n - 4 = 0
⇒(3n + 1 ) (n - 4) = 0
∴ n = 4

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60

Correct Option: B

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60