Question

The number of arrangements that can be formed out of 'LOGARITHM' so that no two vowels come together is

• A. $6!{}^7{P}_3$
• B. $6!7!$
• C. $6!3!$
• D. $7!3!$

Answer 1

The correct option is A $6!{}^7{P}_3$

for no 2 vowels to appear together, we can place vowels in spaces

between consonants such as $⊻C⊻C⊻C⊻C⊻C⊻C⊻$

Now, 6 constants can be placed in 6 ! ways (L, G, R, T, H, M)

7 spaces can be filled with 3 vowels (O, A, I) in ${}^7{P}_3$ ways

Total no of arrangements = $6{!}^7{P}_3$

$\therefore$ Option A is correct.