Need help in this Permutation problem!

Hi,

There are 7 letters here : R,A,I,N,B,O,W.
3 must be put in order: A,I,O => Call this Group 1
4 can have random positions: R,N,B,W => Call this Group 2

Let's call I is a letter in Group 1, O is a letter in Group 2:
The word consists of 7 letters : three I and four O.
Total of ways to form 7-letter word from three I and four O: 7!/ (3!4!) = 35

For group 1: A,I,O must be in exact order, so number of way is : 1
For group 2: R,N,B,W can be put randomly, so number of ways is: 4! = 24

Number of ways that satifies the above condition : 35*24 = 840.

Hope this helps.

paritosh_b wrote:Q]
Word:RAINBOW
How many words are there in which 'A' is always before 'I' and 'I' is always before 'O' in the word 'RAINBOW'?
Ans:840.

I am not able to solve this prob.Can someone please help me out?

Here's another approach:

We have 7 total positions.
From these 7 positions, any combination of 3 will give us an acceptable way to place A-I-O: 7C3 = 35.

The number of ways to arrange the remaining 4 letters = 4! = 24.

Now we multiply the number of choices we have for A-I-O with the number of choices we have for the remaining 4 letters:
35*24 = 840.

Got it.Thanks

HI,

Am a bit confused.
O followed this approach.

I considered "AIO" as 1 Letter.
Thus the tal number of letters to be rerranged was 4+1=5

So the total numner of ways - 5! = 120,

What is wrong in this apporach ????

Shawshank wrote:HI,

Am a bit confused.
O followed this approach.

I considered "AIO" as 1 Letter.
Thus the tal number of letters to be rerranged was 4+1=5

So the total numner of ways - 5! = 120,

What is wrong in this apporach ????

The problem states only that A has to come before I, which has to come before O; it doesn't state that the 3 letters have to be adjacent to each other. By considering AIO as a unit, you're overly restricting the number of possible arrangements.