In how many ways all the letters of the word WEAPONS can be arranged such that in any of the arrangements, no two vowels

Arranging all the letters of the word WEAPONS such that no two vowels will be together


There are 7 letters in the word WEAPONS with 3 vowels -> E, A, and O
All possible arrangement of WEAPONS = 8!
There are 4 consonants present -> WPNS


The vowels can be arranged among themselves in 3! ways
The consonants can also be arranged among themselves in 4! ways
_W_P_N_S_


There are 5 possible positions for the 3 vowels to occupy beside the consonants such that no two vowels will be together, and this means they can be arranged in 5C3*3!*4!

=>5C3*3!*4!
Answer = E

Vincen wrote: ↑

Thu Sep 03, 2020 5:41 am

In how many ways all the letters of the word WEAPONS can be arranged such that in any of the arrangements, no two vowels will be together?

A. \(^5C_3\cdot 3!\)

B. \(3!\cdot 4!\)

C. \(^5C_3\cdot 4!\)

D. \(7!-5!\cdot 3!\)

E. \(^5C_3\cdot 3!\cdot 4!\)

Answer: E

Source: e-GMAT

There are \(7\) letters in the word WEAPONS with \(3\) vowels and \(4\) consonants of the \(5\) possible places, \(3\) will be chosen for the vowels in \(5C35C3\) ways. The vowels can rearrange themselves in \(3!\) ways and the consonants can rearrange themselves in \(4!\) ways.

Hence, number of arrangements of letters such that all vowels are all together is \(5C3\cdot 3! \cdot 4!\), so the correct answer is E