In how many words the letter of word RAINBOW be arranged so that only 2 vowels always remain together? 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?
 A: 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 $20 \cdot 144 = 2880$.
A: 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
A: 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
A:  This is the solution $($page $2)$
A:  This is the solution $($page $1)$
