Bunuel wrote:
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
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.