Need help in this Permutation problem!

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paritosh_b: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Jan 17, 2010 9:51 am

by paritosh_b » Thu Nov 04, 2010 8:33 am

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?

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Source: Beat The GMAT — Problem Solving | Thu Nov 04, 2010 8:33 am

limestone: Master | Next Rank: 500 Posts; Posts: 307; Joined: Sun Jul 11, 2010 7:52 pmFollowed by: 1 member; Thanked: 36 times; GMAT Score:640

by limestone » Thu Nov 04, 2010 10:41 am

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.

Last edited by limestone on Thu Nov 04, 2010 6:01 pm, edited 1 time in total.

"There is nothing either good or bad - but thinking makes it so" - Shakespeare.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Nov 04, 2010 12:49 pm

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.

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paritosh_b: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Jan 17, 2010 9:51 am

by paritosh_b » Thu Nov 04, 2010 6:17 pm

Got it.Thanks

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Shawshank: Master | Next Rank: 500 Posts; Posts: 305; Joined: Mon Jul 27, 2009 5:38 am; Thanked: 10 times

by Shawshank » Thu Nov 04, 2010 7:16 pm

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 Redemtion -- Hope is still alive ...

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Nov 04, 2010 7:21 pm

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.

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