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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?

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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 $20 \cdot 144 = 2880$.

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  • $\begingroup$ Can you think of a shorter approach? $\endgroup$ – Aditya Agarwal Aug 15 '15 at 11:34
  • $\begingroup$ Not really, but I would like to see a simpler one. $\endgroup$ – Wojciech Karwacki Aug 15 '15 at 11:37
  • $\begingroup$ @AdityaAgarwal i think second approach would work fine with you $\endgroup$ – Jack Aug 15 '15 at 11:44
  • $\begingroup$ @WojciechKarwacki I think the second approach would be simpler $\endgroup$ – Jack Aug 15 '15 at 11:44
  • $\begingroup$ @WojciechKarwacki: See if you like my answer $\endgroup$ – true blue anil Aug 15 '15 at 18:30
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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

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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

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  • $\begingroup$ why to multiple by 6? $\endgroup$ – Jack Aug 15 '15 at 12:10
  • $\begingroup$ You can see the block of two vowels you choose as a single letter, so you have $6!$. You wrote $5!$ so you had to multiply by 6 $\endgroup$ – karmalu Aug 15 '15 at 12:18
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enter image description here This is the solution $($page $2)$

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enter image description here This is the solution $($page $1)$

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  • $\begingroup$ Please type your solution. Don't just give a link. $\endgroup$ – YuiTo Cheng Jan 31 '19 at 8:33

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