In how many ways all the letters of the word WEAPONS can be arranged

Solution

Given:
In this question, we are given

• A specific word is given: WEAPONS, of which all the letters can be arranged to form different words

To find:
We need to determine

• How many such arrangements are possible, where no two vowels will be together

Approach and Working:
In the given word “WEAPONS”, there are 4 consonants: W, P, N, S and 3 vowels: E, A, O.

• When we say that no two vowels are together, it means there will be at least one consonant between any two vowels.

As there are 4 consonants, let us first put them, as shown in the following figure:



Now, for the 3 vowels, there will be actually 5 places where any of those vowels can be placed.

• Out of the 5 possible places, 3 will be chosen for the vowels in \(^5C_3\) ways.
• Also, the vowels can rearrange themselves in 3! ways and the consonants can rearrange themselves in 4! ways.
• Therefore, the total number of possible arrangements = \(^5C_3 * 3! * 4!\)

Hence, the correct answer is option E.

Answer: E

Hey Archit3110,
In your calculation, you missed those cases where two vowels can be together. You only removed the cases where all 3 vowels are together.

There are 7 letters in the word WEAPONS with 3 vowels and 4 consonant 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∗3!∗4! so the correct answer is E

And what is the best way to do it with that approach? (removing csaes from the total)

EgmatQuantExpert
yeah i made an error.. misread question.. edited thanks..

When we insert 1 vowel in the middle of each consonant and 1 vowel each at the corner, in total, it would make 9 spaces, however WEAPONS has only 7 spaces.

Below you find another approach, which I used to solve this question:

WEAPONS = 3 Vowels & 4 Not Vowels
No repeating letter => Total arrangement possibilities = 7!

What is the number of arrangement in which all three Vowels are together?
5!*3! = 6!
(3! refers to the arrangements between the Vowels as AEO, OEA...)

What is the number of arrangement in which two Vowels are together?
3*(6!*2!-6!) + 6! = 4*6!

6!*2! = arrangements in which two Vowels are together; However, we have already considered the arrangements with three Vowels => need to deduct by 6!)
This whole package is multiplied by 3, because we have 3 pairs of two vowels: AO, AE, OE. Finally we add 6! to compensate extra reductions.
For example for pair EA, EOA is additionally deducted. This is the case for 1/3*3*(6!*2!-6!)=6! arrangements


Final Calculation:
7! - 4*6! - 6! = 7*6!-5*6! = 2*6! = 5C3*3!*4!

E.

I see a flaw in the question.
Can't we assume that consonants and vowels ould also be arranged in the following ways given that the only restriction is that no 2 vowels may appear together?:

VCVCVCC
CCVCVCV
VCCVCVC
VCVCCCV

If this were the case, the arrangement CVCVCVC wouldn't be the only one. After all, the question is not giving any hints on restricting consonants from being together. So, wouldn't there be many more ways in which the letters could be arranged?
I would love to clear this out. Thanks

Sr CaptainOfMySoul, I use the same logic with you but I couldn't get the right answer because I miss this part: "Finally we add 6! to compensate extra reductions.
For example for pair EA, EOA is additionally deducted. This is the case for 1/3*3*(6!*2!-6!)=6! arrangements".
I still don't understand why (+ 6!) in 3*(6!*2!-6!) + 6!. I think you add 6! arrangements of 3 vowels in 2 times meanwhile we should add only 1.
My solution is: 7! - 3*(6!*2!-6!) - 6!
Please help explain and correct me. Thanks.

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