Question

If we permute 5 letters of the word 'lemon' in 5! ways. in how many words vowels do not come together?

Answer 1

Given word is "lemon". which has total 5 letters.

By properties of factorial we know that total possible 5-letter words that we can make using letters of "lemon" is 5! in which no letter is repeated.

There are two vovels "e" and "o" in given word "lemon"

say e and o are together then that can be assumed as 1 letter

so now we are arranging 4 letter to make words which will be 4!

Difference of both cases will produce possible number of 5 letter words that doesn't have vovels together

so the answer is given by 5!-4!=120-24=96

Hence final answer is 96.

Answer 2

Answer:

72

Step-by-step explanation:

the only addition to the explanation above is E and O itself can permute. So 2! is also added with 4!

Hence 5! - 4!.2! = 120-48 = 72