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Seating - Probability Question

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Seating - Probability Question

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?
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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.
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by ramyaravindran » Thu Jan 28, 2010 1:53 pm
I took the same approach that you took and got stuck.
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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!
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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.
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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
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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???
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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
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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).
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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!!
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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)
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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.
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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
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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
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