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A row of seats in a movie hall contains 10 seats. 3 Girls

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

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New post 23 Sep 2014, 13:10
my solution:
(8P3 * 7P7) / 10!

Place the boys first. Now you have 8 empty slots (add one before and after the first and last boys) for three girls. That's 8P3.

Always divide by the total outcomes, which is 10P10, that is 10!

I think some people (from what I've read) are confused with the question. No two girls means not two, not three. Each has to sit alone.

Hope I helped!
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 29 Sep 2014, 22:18
Probability that no two girls will sit together= 1- (P(2 girls sit together)+P(3 girls sit together))
= 1- ( (3C2 + 3C3)/(10C3) )
= 1- ( 3/2+1/120)
= 1- (5/240)
= 235/240
Probability that no two girls will sit together= 47/48
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A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 01 Oct 2014, 11:21
I understood all the explanations by visualising the process. First the total number of ways which is obviously our denominator or 10!. It's 10! because you can't have repetition with humans.

Secondly, you place the 7 boys on the seats, missing one seat each time. There are 7! ways to do this, this goes in the numerator and is multiplied by 8P3 or 8C3 x 3!.

The only thing I'm still struggling to understand is why we do 8P3 for the 3 girls and 8 open seats once the 7 boys are seated. I.e. the *'s in most of people's explanations. Why are we using 8P3 or 8C3 x 3! (basically another way to say 8P3).

Why can we not just use 8C3? Why does the order matter where we place the girls in the stars?

EDIT: Nevermind, I think i've got it.
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 16 Jan 2016, 23:29
Bunuel wrote:
metallicafan wrote:
Are you sure?
My answer is 1/20.

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 8-)


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.


Hi Bunuel,

Just one thing i am getting confused about: the stem says that the row contains 10 seats. So, when we are saying that girls can occupy * number of seats i.e. 8 in number - are we not going beyond the confines of the question, since if that was correct the row should have at least 8+7 i.e. 15 seats.
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 16 Jan 2016, 23:42
WilDThiNg wrote:
Bunuel wrote:
metallicafan wrote:
Are you sure?
My answer is 1/20.

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 8-)


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.


Hi Bunuel,

Just one thing i am getting confused about: the stem says that the row contains 10 seats. So, when we are saying that girls can occupy * number of seats i.e. 8 in number - are we not going beyond the confines of the question, since if that was correct the row should have at least 8+7 i.e. 15 seats.


No. You are assuming that all the spots will be occupied at the same time.

If you are given, *B*B*B*B*B*B*B*

My arrangement can be *B*BGB*BGB*B*BG or GBGB*B*B*BGB*B*, so technically you only have 10 occupied seats and hence within the confines of the question.

Hope this helps.
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 17 Jan 2016, 00:10
[No. You are assuming that all the spots will be occupied at the same time.

If you are given, *B*B*B*B*B*B*B*

My arrangement can be *B*BGB*BGB*B*BG or GBGB*B*B*BGB*B*, so technically you only have 10 occupied seats and hence within the confines of the question.

Hope this helps.[/quote]

Thank you for such a prompt response. Agree that the number of seats occupied eventually would only be 10, however, when we are selecting 3 seats for Gs out of 8 remaining options - those 8 options need to exist, to begin with. Now, when Bs have already occupied 7 seats out of 10 (the number of seats that exist), are there not only 3 seats left to choose from.
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A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 17 Jan 2016, 00:12
WilDThiNg wrote:

Thank you for such a prompt response. Agree that the number of seats occupied eventually would only be 10, however, when we are selecting 3 seats for Gs out of 8 remaining options - those 8 options need to exist, to begin with. Now, when Bs have already occupied 7 seats out of 10 (the number of seats that exist), are there not only 3 seats left to choose from.


You are correct with your last statement that there are only 3 actual spots to work with. Not all '*' are available at the same time.
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A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 14 Nov 2016, 07:34
EvaJager wrote:
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 \(\frac{1}{3}+2\cdot\frac{2\cdot3}{9\cdot10}=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}.\)


Hi Evajager,

I used the same approach and ended with 1/3.

I understand your point, however my logic was:

G1G2..(boys)...G3 (let's take first GG as 1 person, we have 9 people now)

= 9! * 3! (Gs can be arranged inside the girls group) ---->> this also includes the cases when 3 girls are together, as you mentioned. Among the arrangements, there are the ones which G3 is next to G1 or G2. Thus, isn't total # of no 2 girls sitting together:

= 10! - 9! * 3!
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 29 May 2017, 13:37
zet wrote:
EvaJager wrote:
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 \(\frac{1}{3}+2\cdot\frac{2\cdot3}{9\cdot10}=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}.\)


Hi Evajager,

I used the same approach and ended with 1/3.

I understand your point, however my logic was:

G1G2..(boys)...G3 (let's take first GG as 1 person, we have 9 people now)

= 9! * 3! (Gs can be arranged inside the girls group) ---->> this also includes the cases when 3 girls are together, as you mentioned. Among the arrangements, there are the ones which G3 is next to G1 or G2. Thus, isn't total # of no 2 girls sitting together:

= 10! - 9! * 3!

Hi people Im new here so excuse me if Im doıng something wrong.

What if I just choose 3 spaces from 8 places 8C3 and divide it by 10C3 which is sitting girls in the normal seat number
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls  [#permalink]

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New post 01 May 2019, 02:13
Why do we consider the Boys' seating position when probability concerns only about the girls' position?
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Re: A row of seats in a movie hall contains 10 seats. 3 Girls   [#permalink] 01 May 2019, 02:13

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