# 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?

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 $20 \cdot 144 = 2880$.

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

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

• why to multiple by 6? – Jack Aug 15 '15 at 12:10
• 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 – karmalu Aug 15 '15 at 12:18

This is the solution $$($$page $$2)$$

This is the solution $$($$page $$1)$$

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