# [Math] In how many words the letter of word RAINBOW be arranged so that only 2 vowels always remain together

permutations

My Approach:

RAINBOW has 4 Consonants and 3 vowels.

Out of 3Vowels 2 vowel are selected and arranged in 3P2 ways

and the rest letters are arranged in 5! ways(1vowel and 2 consonants)

The Required arrangement is: 3P2*5!=720

But the Ans is 2880

Second Approach:

Out of 7 possible ways subract the one having either all vowel together or no vowel are together.

7!-(5!*3!+4!*5P3!)=2880 // i got the answer through this approach

What i have done wrong in ist approach and why the approach was wrong?

Let $V$ be a vowel and $C$ be a consonant.
In any word (for example: V V C C C V C) there are $4!$ possibilities to rotate $C$'s, and $3!$ possibilities to rotate $V$'s. This yields $3! \cdot 4! = 144$ possibilities to rotate $C$'s and $V$'s in a word.
When two $V$'s are in the first and second place, there are $4$ ways to put the third $V$ (same goes when two $V$'s are at the end). When two $V$'s are in the middle, there are $3$ ways to put the third $V$. You have 4 situations when two $V$'s are in the middle. So the number of permitted settings is $4 + 3 + 3 + 3 + 3 + 4 = 20$.
Now the total number of possiblities is $20 \cdot 144 = 2880$.