A row of seats in a movie hall contains 10 seats. 3 Girls : GMAT Problem Solving (PS)
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# A row of seats in a movie hall contains 10 seats. 3 Girls

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A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]

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17 Aug 2010, 01:21
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A row of seats in a movie hall contains 10 seats. 3 Girls and 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

[Reveal] Spoiler:
7/15

Last edited by harshjoshi91 on 17 Aug 2010, 03:36, edited 1 time in total.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Aug 2010, 02:05
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Hi,

Let us first figure out the total # of ways in which the girls do not sit together.

Total # of ways for the 7 boys to stand = 7!

After these boys have stood in their positions, there are 8 positions for the 3 girls to occupy.

# of ways for the girls to choose 3 positions = 8C3

# of ways of arranging themselves = 3!

Hence, total # of ways for the girls = 8C3 times 3!

Overall, # of ways when no two girls are together = 7! times 8C3 times 3!

n(S) = n(Sample space for these 10 students) = 10!

Hence, probability for no two girls to sit together = 7! * 8C3 * 3! / 10!

= 8C3 * 3! / (10 * 9 * 8)

= (8 * 7 * 6 ) / (10 * 9 * 8)

= (7*6) / (10*9)

=7/15

What is the OA, please. The answer gives me the feeling that there is an easier way to approach this problem.

Thanks.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Aug 2010, 12:41
I guess we can use the method below...

But why cant we solve it by using this method?

1- P(2GrlsTogether) - P(3GrlsTogether)

*****************
Kudos me if you liked this post
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Aug 2010, 13:51
4gmatmumbai wrote:
Hi,

Let us first figure out the total # of ways in which the girls do not sit together.

Total # of ways for the 7 boys to stand = 7!

After these boys have stood in their positions, there are 8 positions for the 3 girls to occupy.

# of ways for the girls to choose 3 positions = 8C3

# of ways of arranging themselves = 3!

Hence, total # of ways for the girls = 8C3 times 3!

Overall, # of ways when no two girls are together = 7! times 8C3 times 3!

n(S) = n(Sample space for these 10 students) = 10!

Hence, probability for no two girls to sit together = 7! * 8C3 * 3! / 10!

= 8C3 * 3! / (10 * 9 * 8)

= (8 * 7 * 6 ) / (10 * 9 * 8)

= (7*6) / (10*9)

=7/15

What is the OA, please. The answer gives me the feeling that there is an easier way to approach this problem.

Thanks.

Well done!

Please give the OA with explanation it should really have the easier approach
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Aug 2010, 16:17
Pavle wrote:
4gmatmumbai wrote:
Hi,

Let us first figure out the total # of ways in which the girls do not sit together.

Total # of ways for the 7 boys to stand = 7!

After these boys have stood in their positions, there are 8 positions for the 3 girls to occupy.

# of ways for the girls to choose 3 positions = 8C3

# of ways of arranging themselves = 3!

Hence, total # of ways for the girls = 8C3 times 3!

Overall, # of ways when no two girls are together = 7! times 8C3 times 3!

n(S) = n(Sample space for these 10 students) = 10!

Hence, probability for no two girls to sit together = 7! * 8C3 * 3! / 10!

= 8C3 * 3! / (10 * 9 * 8)

= (8 * 7 * 6 ) / (10 * 9 * 8)

= (7*6) / (10*9)

=7/15

What is the OA, please. The answer gives me the feeling that there is an easier way to approach this problem.

Thanks.

Well done!

Please give the OA with explanation it should really have the easier approach

No, after giving some thought i think it's the best approach. Callculating probabilities with 3 girls together and then 2 girls together is a lot more time-consuming than that.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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19 Aug 2010, 09:39
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Are you sure?

I tell you why:
First, we have to identify the total number of combinations of seats in which the three girls can occupy.
So, 10C3 = 120 possible combinations.

Now, we have to identify the combinations in which there are not two girls together.

Only to offer a better explanation, I will write the combinations:

Seat N°: 1 2 3 4 5 6 7 8 9 10
G B G B G B B B B B
B G B G B G B B B B
B B G B G B G B B B
B B B G B G B G B B
B B B B G B G B G B
B B B B B G B G B G

Based on this, there are only 6 scenarios in which there are not two girls together.
Obviously, it is not necessary to write this combinations, you only have to see that when the third girl is in the 10th position, the first girl is in the 6th. But the first girl cannot be in the 7th, because there is not the 11th position for the third girl.
(I think that there is a way to solve this part with combinatronics, but I didn't find it)

So, we have this 6 scenarios and the total number of events (120).
We divide: 6 / 120 = 1/20

What do you think?

I think I deserve kudos
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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20 Aug 2010, 08:04
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Expert's post
metallicafan wrote:
Are you sure?

I tell you why:
First, we have to identify the total number of combinations of seats in which the three girls can occupy.
So, 10C3 = 120 possible combinations.

Now, we have to identify the combinations in which there are not two girls together.

Only to offer a better explanation, I will write the combinations:

Seat N°: 1 2 3 4 5 6 7 8 9 10
G B G B G B B B B B
B G B G B G B B B B
B B G B G B G B B B
B B B G B G B G B B
B B B B G B G B G B
B B B B B G B G B G

Based on this, there are only 6 scenarios in which there are not two girls together.
Obviously, it is not necessary to write this combinations, you only have to see that when the third girl is in the 10th position, the first girl is in the 6th. But the first girl cannot be in the 7th, because there is not the 11th position for the third girl.
(I think that there is a way to solve this part with combinatronics, but I didn't find it)

So, we have this 6 scenarios and the total number of events (120).
We divide: 6 / 120 = 1/20

What do you think?

I think I deserve kudos

This approach is not correct. There are more cases possible, for no girls to sit together, for example:

G B B G B G B B B B
Or:
G B G B B B B B B G
...

Answer given by 4gmatmumbai is correct.

A row of seats in a movie hall contains 10 seats. 3 Girls & 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

Consider the following:
*B*B*B*B*B*B*B*

Now, if girls will occupy the places of 8 stars no girl will sit together.

# of ways 3 girls can occupy the places of these 8 stars is $$C^3_8$$;
# of ways 3 girls can be arranged on these places is $$3!$$;
# of ways 7 boys can be arranged is $$7!$$.

So total # of ways to arrange 3 Girls and 7 boys so that no girls are together is $$C^3_8*3!*7!$$;
Total # of ways to arrange 10 children is $$10!$$.

So $$P=\frac{C^3_8*3!*7!}{10!}=\frac{7}{15}$$.

Hope it's clear.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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07 Oct 2010, 06:04
*B*B*B*B*B*B*B*

by considering this we are getting total of 15 seats , but actually there are only 10 seats?
can any one help me to resolve my confusion?
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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07 Oct 2010, 06:42
anilnandyala wrote:
*B*B*B*B*B*B*B*

by considering this we are getting total of 15 seats , but actually there are only 10 seats?
can any one help me to resolve my confusion?

*B*B*B*B*B*B*B* represent 7 boys (B) and 8 empty slots (*) between them. Then 3 girls are placed in some particular 3 slots out of 8 (thus ensuring that the girls are separated by boys), so in the end there are 7+3=10 objects not 15.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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23 Jul 2011, 23:22
Hi,

If we had 2 girls instead of 3 girls , we could have applied 1 - P(2 girls together) ?

For eg:

if we have 2 girls and 7 boys, then the answer would have been :

1 - (8! 2! / 9!)

8! for grouping 7 boys and 1 group of 2 girls together

Is this correct?

Thanks.
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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23 Dec 2011, 05:03
Could you plz help me why my approach is wrong?

P(not 2 girls sit together)=1-P(2 girls sit together or 3 girls sit together)
If we name girls as X, Y, and Z then the above equation:

p(not 2 girls sit together)=1-p(XY or XZ or ZY or XYZ)

$$P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}$$

$$P(xyz)=\frac{8!*3!}{10!}$$

So, P(not 2 girls sit together)=$$1-\frac{9!*2+9!*2+9!*2+8!*3!}{10!}=1-\frac{2}{3}=\frac{1}{3}$$

What's wrong???
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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31 Dec 2011, 00:53
Using the following slot:
_B_B_B_B_B_B_B_

Boys can be arranged in 7! ways
3 Girls can occupy 8 "_" of the slot. i. e. 8p3
Total possible arrangement = 10!

(8p3*7!)/10!= 7/15
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Jul 2012, 00:53
I think 7/15 is wrong answer as you cant have 8 spaces. Total no of seats are fixed in this case there wont be any casw when *B*B*B*B*B*B*B* would be possible.

This will be only possible when seats are unlimited or it was asked like IN HOW MANY WAYS BOYS AND GIRLS CAN SEAT IN A ROW

but here its given that 10 seats are there..

Right approach would be total- (2 together)- (3 together)
total=10!
2 together= G1G2 G3 B1 B2 B3 B4 B5 OR G3G2 G3 B1 B2 B3 B4 B5 OR G1G3 G3 B1 B2 B3 B4 B5
= 3* ( Gx Gy B1 B2 B3 B4 B5)
= 3*( 7!- 2*6!) (2*6! is for cases when Gx Gy are together)
= 3!*6!5!

3 Together = G1G2G3 B1 B2 B3 B4 B5
= 6!*3!
= (10!-6!*6!-6!3!)
= 6!(10*9*8*7- 6*5!- 3!)
=6!(5*3*8*7*6 - 6*5!- 6)
= 6!*6( 840- 120- 1)
=6!*6(719)
So prob = 6!*6*719/10!
= 6*719/10*9*8*7
=719/840
Ans= 719/840

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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Jul 2012, 00:54
7/15 is wrong by just sensing the question....50% cases girls will not sit together..I think prob would be more of no two girl seating together

rt??
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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17 Jul 2012, 01:16
jaigoyal wrote:
I think 7/15 is wrong answer as you cant have 8 spaces. Total no of seats are fixed in this case there wont be any casw when *B*B*B*B*B*B*B* would be possible.

This will be only possible when seats are unlimited or it was asked like IN HOW MANY WAYS BOYS AND GIRLS CAN SEAT IN A ROW

but here its given that 10 seats are there..

Right approach would be total- (2 together)- (3 together)
total=10!
2 together= G1G2 G3 B1 B2 B3 B4 B5 OR G3G2 G3 B1 B2 B3 B4 B5 OR G1G3 G3 B1 B2 B3 B4 B5
= 3* ( Gx Gy B1 B2 B3 B4 B5)
= 3*( 7!- 2*6!) (2*6! is for cases when Gx Gy are together)
= 3!*6!5!

3 Together = G1G2G3 B1 B2 B3 B4 B5
= 6!*3!
= (10!-6!*6!-6!3!)
= 6!(10*9*8*7- 6*5!- 3!)
=6!(5*3*8*7*6 - 6*5!- 6)
= 6!*6( 840- 120- 1)
=6!*6(719)
So prob = 6!*6*719/10!
= 6*719/10*9*8*7
=719/840
Ans= 719/840

OA for this question is 7/15.

It seems that you didn't understand the solution above.

Each arrangement given by placing the girls in the empty slots (in *B*B*B*B*B*B*B* ) will still occupy only 10 seats (naturally) since there are 10 people. For example:

GBGBGBBBBB or BGBGBBBBBG ...
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]

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28 Aug 2012, 02:16
Why this approach is wrong? I am new to this forum. Plz excuse me for any violations in posting a topic.

Could you plz help me why my approach is wrong?

P(not 2 girls sit together)=1-P(2 girls sit together or 3 girls sit together)
If we name girls as X, Y, and Z then the above equation:

p(not 2 girls sit together)=1-p(XY or XZ or ZY or XYZ)

P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}

P(xyz)=\frac{8!*3!}{10!}

So, P(not 2 girls sit together)=1-\frac{9!*2+9!*2+9!*2+8!*3!}{10!}=1-\frac{2}{3}=\frac{1}{3}

What's wrong???
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Re: PS: Probability : 3Girls 7 Boys [#permalink]

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28 Aug 2012, 05:51
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shahideh wrote:
Could you plz help me why my approach is wrong?

P(not 2 girls sit together)=1-P(2 girls sit together or 3 girls sit together)
If we name girls as X, Y, and Z then the above equation:

p(not 2 girls sit together)=1-p(XY or XZ or ZY or XYZ)

$$P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}$$

$$P(xyz)=\frac{8!*3!}{10!}$$

So, P(not 2 girls sit together)=$$1-\frac{9!*2+9!*2+9!*2+8!*3!}{10!}=1-\frac{2}{3}=\frac{1}{3}$$

What's wrong???

You didn't distinguish between the cases of exactly two girls sitting together and one sitting alone and the case when all three girls sit together.
$$P(xy)=P(xz)=P(zy)= \frac{9!*2}{10!}$$ each of these also includes some cases when all three girls sit together.
Now, it gets even more complicated. When considering xy sitting together, it might happen that z joined them, so you also had the arrangement xyz or zxy, but since you took only 2! (just switching between x and y only), you subtracted only 4 out of the full 3!=6 possibilities. Which means that you subtracted $$3\cdot\frac{4}{6}=2$$ or twice more the probability P(xyz).

To your result, you should add back twice P(xyz). You will get $$\frac{1}{3}+2\cdot\frac{2\cdot3}{9\cdot10}=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}.$$
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]

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28 Aug 2012, 07:18
harshjoshi91 wrote:
A row of seats in a movie hall contains 10 seats. 3 Girls and 7 boys need to occupy those seats. What is the probability that no two girls will sit together?

[Reveal] Spoiler:
7/15

Here is a solution based on direct calculations of probabilities:

The probability of two girls sitting together is $$9\cdot\frac{3}{10}\cdot\frac{2}{9}=\frac{3}{5}.$$
When sitting 2 girls together - GG - 3 out of 10 choices for the first girl, then 2 out of 9 for the second girl.
The factor 9 accounts for the 9 possibilities where to place the two girls in the row of 10 seats:

GG********
*GG*******
...
********GG

The probability of all three girls sitting together is $$8\cdot\frac{3}{10}\cdot\frac{2}{9}\cdot\frac{1}{8}=\frac{1}{15}.$$
When sitting all 3 girls together - GGG - 3 out of 10 choices for the first girl, then 2 out of 9 for the second girl, and 1 out of 8 for the third girl.
The factor 8 accounts for the 8 possibilities where to place the three girls in the row of 10 seats:

GGG*******
*GGG******
...
*******GGG

Now we should be careful: when computing the probability for 2 girls sitting together, we didn't say anything about the third girl.
Those sitting arrangements include also cases of all three girls sitting together, and maybe more than once.
In fact, we can determine that we included all the cases when 3 girls sit together twice.
For any arrangement of two girls together, say GG, and the third girl g sitting with them, we also considered the arrangements gGG and GGg when placed on the same three seats, like
GGg*******
gGG*******
...
***GGg****
***gGG****
...

Therefore, for the final probability we get $$1-\frac{3}{5}+\frac{1}{15}=\frac{7}{15}.$$
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]

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28 Aug 2012, 07:57
very diffcult question..i must say..above 700 level..

i did it wrong..after bunuel explanation i got that where i was wrong..thank u bunuel..
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]

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25 Sep 2012, 05:37
Total no. of seats In the theater = 10
Total no of possible arrangements = 10 Factorial

No. of Boys = 7

No. of Girls = 3

The 7 boys can be arranged in 7 seats in 7 Factorial different ways ...

so we have B B B B B B B As our arrangement so far ..

In order for the girls to sit in such a way that they are NEVER together we can have each girl occupy any position from Position no. 1 ie before the first Boy and position no. 8 ie after the last boy ... Therefore they can have 8 different positions , but we have only 3 girls ...Therefore the Permutations for arranging 3 girls in 8 different places are P ( 8,3) which is = 336 ..

We have 336 possible arrangements for GIRLS such that no 2 are together , and we have 7 factorial arrangements for boys...

Therefore Probabability = (336 x 7 Factorial) / (10 Factorial ie. total no. of possible arrangements)

This is equal to 7/15 ..
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls   [#permalink] 25 Sep 2012, 05:37

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