Seating - Probability Question - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

Seating - Probability Question

Expert replies

ramyaravindran: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Mon Apr 28, 2008 5:33 am; Thanked: 2 times

by ramyaravindran » Thu Jan 28, 2010 12:57 pm

There are 6 guys and 4 girls.what is the probability of seating all the guys and girls in such a way that no two girls sit together?
I dont remember the answers right now.. but i'm sure all the options ranged from 14440 to 64440. Can anyone help me solve this one?

Like this post

Source: Beat The GMAT — Problem Solving | Thu Jan 28, 2010 12:57 pm

dmitriyaleyev: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Thu May 21, 2009 5:53 pm; Thanked: 1 times

by dmitriyaleyev » Thu Jan 28, 2010 1:43 pm

I am not great at perm q's, but this is how I would attempt to solve it...

1. number of combinations of a group of 4 girls and 6 separate guys = >> 10!/7!3! = 120
2. ----------------------------------------------- 3 girls, 1 separate, and 6 separate guys = >> 10!/8!2! = 45
3. ----------------------------------------------- 2 girls, a group of 2 girls and 6 separate guys =>>> 10!/8!2! = 45
and then we have 2 ways to arrange 120 >>>>>> 240
2 ways to arrange 45 >>>>>>>>>>>>>>>>>>>>90
8 (?) ways to arrange 45 for number 3 >>>>>>>>360

Then I am stuck....
Please help.

Like this post

ramyaravindran: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Mon Apr 28, 2008 5:33 am; Thanked: 2 times

by ramyaravindran » Thu Jan 28, 2010 1:53 pm

I took the same approach that you took and got stuck.

Like this post

ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pmFollowed by: 1 member; Thanked: 26 times

by ace_gre » Thu Jan 28, 2010 3:26 pm

Since girls cannot be seated together, lets seat the boys first. 6 boys can be seated in 6! ways.
Seating 6 boys creates 7 spaces between them for seating 4 girls.

_B1_B2_B3_B4_B5_B6_

So 4 girls can be seated in any of the 7 spaces==> 7P4 = 7!/ 3!

Total # of ways = 6! * 7! / 3!

Like this post

dmitriyaleyev: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Thu May 21, 2009 5:53 pm; Thanked: 1 times

by dmitriyaleyev » Thu Jan 28, 2010 4:36 pm

Just an idea, that in this method you are completely dismissing probability of two guys sitting together.
it is possible that
B1_B2B3_B4_B5_ for example.
I dont think it is a right answer.

Like this post

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Thu Jan 28, 2010 4:47 pm

dmitriyaleyev wrote:Just an idea, that in this method you are completely dismissing probability of two guys sitting together.
it is possible that
B1_B2B3_B4_B5_ for example.
I dont think it is a right answer.

He is not, I will demonstrate How:
_B1_B2_B3_B4_B5_B6_ is the original arrangement

Now we are selecting 4 out of 7 Gaps for the girls to sit. If a gap is not selected essentially two of the boys will sit together

In the specific example you mentioned, the gaps selected are 2, 4, 5,6 ; it is only a case which is addressed by ace_gre's Model

Always borrow money from a pessimist, he doesn't expect to be paid back.

Like this post

dmitriyaleyev: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Thu May 21, 2009 5:53 pm; Thanked: 1 times

by dmitriyaleyev » Thu Jan 28, 2010 5:02 pm

Let me just get it straight:
6 boys can seat in 6! ways >>>> 720 ways
4 girls can sit in 7!/3! ways >>>> 840 ways ..... is it because the order does not matter? I counted it as 7!/4!3!....
so 720 x 840 = 604,800 ways???

Like this post

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Thu Jan 28, 2010 5:08 pm

dmitriyaleyev wrote:Let me just get it straight:
6 boys can seat in 6! ways >>>> 720 ways
4 girls can sit in 7!/3! ways >>>> 840 ways ..... is it because the order does not matter? I counted it as 7!/4!3!....
so 720 x 840 = 604,800 ways???

It is an arrangement problem and Order does matter.

We arrange 6 boys in 6! ways leaving gaps in between them so that girls can sit only in these gaps, thus making sure that no girl will sit together. now there are 7 gaps in the arrangement of boys since we are allowing girls only to sit in the gaps.
We have to arrange 4 girls in 7 places. that can be done in 7P3 ways

total ways = 6!*7P3 =6!*7!/3! indeed

Always borrow money from a pessimist, he doesn't expect to be paid back.

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Jan 28, 2010 5:32 pm

ramyaravindran wrote:There are 6 guys and 4 girls.what is the probability of seating all the guys and girls in such a way that no two girls sit together?
I dont remember the answers right now.. but i'm sure all the options ranged from 14440 to 64440. Can anyone help me solve this one?

Can you provide some more information about the question? Here's what we need to know:

1) are there only 10 seats available? When we talk about "seating", there's almost always a fixed number of seats; we can't solve unless we know how many there are.

2) are the seats in a straight line or a circle? The solution would be different in each case.

3) what's the source of the question? Always good to provide that info!

As it stands, this couldn't be a real GMAT question, since there are unresolved ambiguities (i.e. there's no one "correct" answer to this question).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Like this post

ramyaravindran: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Mon Apr 28, 2008 5:33 am; Thanked: 2 times

by ramyaravindran » Fri Jan 29, 2010 6:56 am

Stuart

Here are my responses to your question

1. The seating is fixed and we only have 10 seats.

2. The seating is in a row.

3. I saw this question on testmagic but I am not able to find the link now.

I would be curious to know what the solution would be if the seats are in a circle.

Thanks!!

Like this post

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Fri Jan 29, 2010 12:17 pm

ramyaravindran wrote:Stuart

Here are my responses to your question

1. The seating is fixed and we only have 10 seats.

2. The seating is in a row.

3. I saw this question on testmagic but I am not able to find the link now.

I would be curious to know what the solution would be if the seats are in a circle.

Thanks!!

1, 2 The solution given above takes care of these.

Circular permutation

here are 10 seats in a circle and we have to make 6 Boys and 4 girls sit

Well,

6 boys can sit in a circle with gaps in between them in 5! ways (circular permutation) and there are 6 gaps and 4 girls can be arranged 6 gaps in 6P4 ways

so total is 120*30 = 3600 (I do not think it is right but I am trying)

Always borrow money from a pessimist, he doesn't expect to be paid back.

Like this post

ace_gre: Senior | Next Rank: 100 Posts; Posts: 98; Joined: Mon Nov 23, 2009 2:30 pmFollowed by: 1 member; Thanked: 26 times

by ace_gre » Fri Jan 29, 2010 5:08 pm

I second Ajith's response for circular seating.. I would approach the question the same way.

Like this post

chintudave: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Thu Sep 25, 2008 6:10 pm; Thanked: 2 times

by chintudave » Fri Jan 29, 2010 6:25 pm

With all the ambiguities out of the way. This is my painful way. Experts please share some quicker way!

Since no two girls can sit together there are 30 unique options as listed below.

1-3-5-7
1-3-5-8
1-3-5-9
1-3-5-10
1-3-6-8
1-3-6-9
1-3-6-10
1-3-7-9
1-3-7-10
1-3-8-10
1-4-6-8
1-4-6-9
1-4-6-10
1-4-7-9
1-4-7-10
1-4-8-10
1-5-7-9
1-5-7-10
1-5-8-10
1-6-8-10
2-4-6-8
2-4-7-9
2-4-8-10
2-5-7-9
2-5-8-10
2-6-8-10
3-5-7-9
3-5-8-10
3-6-8-10
4-6-8-10

Now girls can sit in 4! ways and boys can sit in 6! ways in each option above. So total is 6! * 4! * 30 ways.

The total number of ways without restriction is 10!.

So the probability is ...

= 6!*4!*30/10!
= 30/210
=1/7.

Let me know if you'll think this is incorrect.

Regards
Chirag

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Feb 02, 2010 2:16 pm

So, rewriting the question as:

There are 10 chairs in a row. If 6 boys and 4 girls are to be randomly seated in those chairs, what's the probability that no two girls are seated together?

Probability = # desired outcomes / total # of possibilities

Let's start with the easy part (always a good plan on the GMAT): total # of possibilities

We have 10 people to arrange (order definitely matters). There are n! ways to arrange n distinct objects, so the denominator is 10!.

Now the fun part: the # of desired outcomes.

As has previously been explained (very well, in fact), if we don't want two girls together, the best way to start is to look at the boys and the gaps where there's room for girls:

_b_b_b_b_b_b_

We can see that we have 6 boys to arrange, so there are 6! possible sequences for them.

There are 7 possible gaps and 4 girls to place in those spots. When we're not filling all the spots, we use the regular permutations formula:

nPk = n!/(n-k)!

in which n is the total number of objects and k is the number that we're using.

Here, n=7 and k=4, so:

7P4 = 7!/(7-4)! = 7!/3! = 7*6*5*4

In this question we're arranging boys AND girls, i.e. we have MULTIPLE arrangements. Whenever we have MULTIPLE arrangements, we MULTIPLY the individual arrangements. Consequently, there are:

6!*7*6*5*4 total arrangements that fit our desired criteria.

Now our final answer:

Probability = # desired / total #

= (6!*7*6*5*4) / 10!

= 7*6*5*4/10*9*8*7

= 6*5*4/10*9*8

= 120/720

= 1/6

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review