The letters of the word PROMISE are arranged so that no two of the vow

Question

The letters of the word PROMISE are arranged so that no two of the vowels should come together. Find total number of arrangements.

A. 7
B. 49 
C. 1.440 
D. 1,898
E. 4,320

soapbolt · Accepted Answer

Really Good Question

i learned it the hard way , hope my solution helps

__C__C__C__C__ 
Here is the arrangement of 4 Consonant such that between any two Consonant at most one Vowel can come but since the Vowels can also appear before the leftmost Consonant or after the rightmost Consonant so it gives us 5 places available for Vowels(represented by “___”)

But since we require only 3 places due to 3 vowels to be arranged so we will select the 3 places out of 5 in 5C3 ways = 10 ways

In the given arrangement the Consonants can be arranged in 4! Ways at the selected 4 places and,

All the vowels can also be arranged among themselves in 3! Ways

so Total ways to arrange the letters as per desired condition = 10*3!*4! = 1440

Answer: Option C

rahul6019 · Answer

This can also be solve as an overlapping set problem - 
7!(total arrangements without restrictions) - 3.2.6!(arrangements when any 2 of three are together) + 5!.6 (removing one duplicate overlapping of all three together from previous as this will happen twice) = 1440.

Ans C.

Archit3110 · Answer

Combine vowels together so we are left with 4 places for Consonants and 1 for vowels out of 7 we will have now 4+1 = 5 ways and since 3 vowels are given then the combination to arrange this array is 5c3
Now since no 2 vowels are to be together , so no. of ways to arrange C in 4 places 4 ! and vowels 3! = total arrangements hence possible = 5c3*4!*3! = 1440 option C

wizardofoddz · Answer

Let '__' denote the possible locations of the vowels in the resultant word.

Since no two vowels can be together, there must be at least one consonant between them.

__C1__C2__C3__C4__

Number of ways in which consonants can rearrange among themselves = 4! = 24 ways

Now, we have 5 different spots that can accommodate vowels.
Number of vowels in PROMISE is 3.

We have 5 spots; we need 3.

Does order of appearance of vowels matter? Of course! Since we have to count possible arrangements here, order does matter.

Therefore number of ways of choosing 3 spots from a pool of 5 (for our vowels) is 5P3 = 5!/2! = 60.

Therefore total number of ways in which the acceptable arrangements can be achieved = 24*60 = 1440 ways

navderm · Answer

I have two solutions I wanted to share.

1. Using straight calculation

We know the set of consonants = C = {P, R, M, S} and set of vowels = V = {O, I, E}.

Since no vowels should come together, all the consonants are placed in between them, or they form the start and end of the word.

Hence the format for the acceptable answer is : _ C _ C _ C _ C _ 
Now, there are 4 consonants and we are using all the consonants. The way 4 consonants can be arranged when using all of them is 4!

Talking about the vowels, we have 3 vowels and 5 place for them. So we have to pick 3 positions where we can put those 3 vowels. 
No. of ways we can pick 3 places from 5 available is 5C3.
Once we have picked 3 places, we need to arrange three vowels in the 3 places picked. We can do that in 3! ways.

Hence total no. of ways is : 4 ! * 5C3 * 3 ! = 1440

2. Using negation

no. of ways so that no two of the vowels are together = total no of ways - no. of ways so all 3 vowels are together - no. of ways so 2 of the vowels are together
 
total no of ways we can arrange C + V = 7 !

total no. of ways we can find when 3 vowels are together : 
Consider 3 vowels to be 1 new letter σ. Now we have 4 letters of consonants and σ making it 5 letters. 
We can arrange 5 letters in 5 ! ways. 
But σ has 3 letters inside and each arrangement between those 3 letters gives 1 new solution. 
So total solution : 5! * 3!

total no. of ways we can find when 2 vowels are together : 
We choose 2 vowels to form a new letter ζ. That can be done in 3C2 ways. 
Now we can choose 1 place from the 5 possible place for ζ by 5C1 . 
Within ζ, there can be 2! arrangements
Then there is 1 vowel remaining and 4 places remaining which we can arrange in 4 different ways.

So 7! - 5!*3! - (3C2 * 5C1 * 2! * 4) = 1440

I know 2nd option becomes cumbersome but was a great exercise in thinking through different way to do the problem.

Kinshook · Answer

Given: The letters of the word PROMISE are arranged so that no two of the vowels should come together.
Asked: Find total number of arrangements.

Consonents : - 
P - 1
R - 1
M - 1
S - 1

Vowels: - 
O - 1
I - 1
E - 1

Total arrangements = 7! = 5040
Total arragements in which 2 vowels are together = 3C2*2C1*6! - 6*5!= 6*6!  - 6*5! = 6*5*5! = 3600
Total arragements in which no two of the vowels should come together = 5040 - 3600 = 1440

IMO C

samarpan.g28 · Answer

Since no two vowels should be together, this means, all the vowels are separated from each other by the consonants. There are 4 consonants P,R,M,S and 3 vowels O,I,E.
_C_C_C_C_. In the blanks, we can fill with vowels.
Consonants can be arranged in 4!=24 ways.
Vowels can be filled in 5C3=10 ways and arranged among themselves in 3!=6 ways so a total of 10*6=60 ways.
Therefore, PROMISE can be arranged in 24*60=1440 ways. Option (C) is correct.

The letters of the word PROMISE are arranged so that no two of the vow

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The letters of the word PROMISE are arranged so that no two of the vow

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