Problem on permutation

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Pinku: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Thu Jul 22, 2010 3:15 pm

by Pinku » Fri Aug 06, 2010 4:05 pm

There are 7 boys and 3 girls in a class. In how many ways can they be seated in a row so that no two of the 3 girls are next to each other?
a) 336 b)56* 7! c)168*6! d)336*7! e)672 *7!

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Source: Beat The GMAT — Problem Solving | Fri Aug 06, 2010 4:05 pm

fitzgerald23: Master | Next Rank: 500 Posts; Posts: 219; Joined: Mon Mar 08, 2010 8:51 pmFollowed by: 5 members; Thanked: 62 times; GMAT Score:750

by fitzgerald23 » Fri Aug 06, 2010 8:08 pm

Ways to seat all 10 people = 10!
Ways that the girls could be together= 8!
Ways to arrange 3 girls = 3!

Total ways to seat without girls sitting next to each other =

10!- 8!3! = 8!((10 * 9)- ( 1 * 6))= 8!(90-6)= 8!(84) = 7!(8 * 84)= 7!*672

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muralithe1: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Fri Mar 05, 2010 9:50 pm; Thanked: 1 times

by muralithe1 » Sun Aug 08, 2010 4:50 pm

Here is my analysis.

10! = 9!3C2(two grils in one set and one gril individual) - 8!3(all girls treated as one unit) + x (set in which no two girls sit next to each other)...

So i got x = 39*8!.

Please correct my analysis.

Thanks a lot.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sun Aug 08, 2010 6:44 pm

Solution:.
Let the 7 boys be seated in a row.
There will be 8 places between them in which the 3 girls can be arranged.
This can be done in 8P3 ways.
Also the boys can be arranged among themselves in 7! Ways.
So total number of possible arrangements is 8P3*7! which is 336*7! ways.

Correct answer is d)

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muralithe1: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Fri Mar 05, 2010 9:50 pm; Thanked: 1 times

by muralithe1 » Sun Aug 08, 2010 6:57 pm

Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...

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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 » Sun Aug 08, 2010 7:32 pm

muralithe1 wrote:Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...

I was asked by PM to discuss how you could approach this problem using the methodology that I used in the following thread: https://www.beatthegmat.com/permuatation ... tml#283495. Here goes:

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10!

Bad arrangements:

Number of ways to arrange 3 girls together = 3!*8! = 6*8!

Number of ways to arrange AB, C, D, E, F, G, H, I, J = 9!
Number of ways to arrange ABC, D, E, F, G, H, I, J and CAB, D, E, F, G, H, I, J = 2*8!
So number of arrangements with AB together but not next to C = 9! - 2*8! = 8!*(9-2) = 7*8!
Since AB can be replaced with BA, AC, CA, BC, and CB -- for a total of 6 variations -- number of arrangements with 2 girls together but not next to 3rd girl = 6*(7 *8!) = 42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

For this problem, Rahul's method is much more efficient!

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aarati: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Thu Aug 05, 2010 3:15 am

by aarati » Mon Aug 09, 2010 12:06 am

GMATGuruNY wrote:
muralithe1 wrote:Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...
I was asked by PM to discuss how you could approach this problem using the methodology that I used in the following thread: https://www.beatthegmat.com/permuatation ... tml#283495. Here goes:

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10!

Bad arrangements:

Number of ways to arrange 3 girls together = 3!*8! = 6*8!

Number of ways to arrange AB, C, D, E, F, G, H, I, J = 9!
Number of ways to arrange ABC, D, E, F, G, H, I, J and CAB, D, E, F, G, H, I, J = 2*8!
So number of arrangements with AB together but not next to C = 9! - 2*8! = 8!*(9-2) = 7*8!
Since AB can be replaced with BA, AC, CA, BC, and CB -- for a total of 6 variations -- number of arrangements with 2 girls together but not next to 3rd girl = 6*(7 *8!) = 42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

For this problem, Rahul's method is much more efficient!

thanks guru,but i did n't understood...can u explain me little bit clear.......

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mainhoon: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sun Aug 08, 2010 8:36 pm

by mainhoon » Mon Aug 09, 2010 3:03 am

so then again what is the answer, i see an answer of 336 7! and 672 7!

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Mon Aug 09, 2010 3:31 am

mainhoon wrote:so then again what is the answer, i see an answer of 336 7! and 672 7!

336 * 7!

Check Rahul's post.

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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 » Mon Aug 09, 2010 5:39 am

thanks guru,but i did n't understood...can u explain me little bit clear.......

Let's break it down step by step.

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10! (There are 10! ways to arrange 10 elements.)

From the 10! total arrangements, we need to subtract the bad arrangements. There are two types of bad arrangements in this problem:

1) The 3 girls ABC all sit together
2) 2 girls sit together but not next to the 3rd girl

Number of ways to arrange 3 girls together:
If we think of the 3 girls as block ABC, we're arranging 8 elements (block ABC and the 7 boys). Number of ways to arrange 8 elements = 8!
ABC itself can be arranged in different orders. Number of ways to arrange ABC = 3!
Multiplying, total arrangements with the 3 girls together = 3!*8! = 6*8!

Number of ways to arrange 2 girls together but not next to the 3rd girl:
Let's start with girls AB.
Number of ways to arrange AB and the 8 other children = 9! (There are 9! ways to arrange 9 elements.)
We need to subtract all the arrangements that include ABC together and CAB together because we counted these above when we counted the number of ways to arrange all 3 girls together.
Number of ways to arrange ABC and the 7 other children is 8!
Number of ways to arrange CAB and the 7 other children is 8!
Subtracting, the number of arrangements with AB together but not next to C = 9!- 8!-8!= 8!(9-1-1)=7*8!
Since AB can be reversed to BA, the number of arrangements with BA together but not next to C = 7*8!
So bad arrangements with AB or BA next to each other but not next to C = 7*8! + 7*8! = 14*8!
Each pair of girls will give us another 14*8! bad arrangements. From ABC we can make 3*2/1*2 = 3 pairs.
So total arrangements with 2 girls together but not next to the 3rd girl = 3*(14*8!)=42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = Total arrangements - Bad arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

Hope this helps!

Good = Total - Bad can be an efficient way to solve many problems. For the problem above, however, determining the number of bad arrangements is complex. I suggest that you use Rahul's method, which is much more efficient!

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