Question

How many different words can be formed of the letters of MALENKOV so that no two vowels are together?

Answer 1

Answer:

36000

Step-by-step explanation:

MALENKOV has 3 vowels (A, E, O) and 5 consonants (M, L, N,K,V)

total no of ways to form words=8!

=40320

no of ways in which two vowels are together

3

c × 6! ×2!=3×720×2=4320

2

no of ways to form words in which no two vowels are together =40320-4320

=36000

Answer 2

Here, from the total possible no. of words, we will subtract the no. of words in which atleast two vowels are together.

Total no. of words is 8! = 40320.

In the word MALENKOV there are 3 vowels, A, E and O. Let me consider any two vowels as a single object, so that the 8 letter word will shrink to a 7 letter one.

One such arrangement is given below.

AEMLNKOV

Here AE is considered as a single object. O can replace any of them. So this single object can be formed in 3P2 = 6 ways.

Generally the letters can be arranged in 7! = 5040 ways (Remember, AE is considered as a single object).

So the required no. of arrangements is 5040 × 6 = 30240.

Now we will subtract this from the whole, so we get 40320 - 30240 = 10080.

Hence 10080 is the answer.