Question

How many seven lettered words can be formed with the letters of the word ${}^{\prime }MISTAK{E}^{\prime }$ such that no two vowels come together?

• A. $7!$
• B. $4!\times 3!$
• C. $4!{\times }^5{P}_3$
• D. $5!{\times }^4{P}_2$

Answer 1

The correct option is C $4!{\times }^5{P}_3$

First we arrange four consonants in $4!$ ways
We now have 5 places for 3 vowels, it can be done in ${}^5{P}_3$ ways
So total number of ways $=4!{\times }^5{P}_3$