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 $20 \cdot 144 = 2880$.
Simple "gap" method
$- C - C - C - C -$
Position blocks of 2 and 1 vowel in two gaps and permute in $_5P_2\cdot 3!$ = 120 ways
Permute consonants in their places to give 120*4! = 2880
In your solution you fix the position of the vowels as if, for example, they're always the first and second letters of the word. So you have to multiply 720 for 6 and then subtract two times the words in which all vowels are together
