how many different words can be formed of the letters of the word malenkov so that no two vowels are together?

There are 8 letters in the word 'MALENKOV', including 3 vowels (A, E, O) and 5 consonant's (M, L, N, K, V)

The total number of words formed by using all the eight letters of the word 'MALENKOV' is

So, the total number of words formed so that no vowels are together = Total number of words – Number of words in which vowel are always together  ....... (1)

Now, considering three vowels as one letter so we have 6 letters which can be arranged in ways

The vowels (A, E, O) can be put together in 3! ways.

So, the required number of words in which vowels are always together = 6! × 3!

= 720 × 6

= 4320

Thus, the total number of words formed so that no vowels are together = 40320 – 4320

= 36000