Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 15 Aug 2010
Posts: 8

A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]
Show Tags
17 Aug 2010, 02:21
4
This post received KUDOS
11
This post was BOOKMARKED
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? ANSWER:
Last edited by harshjoshi91 on 17 Aug 2010, 04:36, edited 1 time in total.



Intern
Joined: 15 Aug 2010
Posts: 23
Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Aug 2010, 03:05
5
This post received KUDOS
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.
_________________
Naveenan Ramachandran 4GMAT  Mumbai



Intern
Joined: 15 Aug 2010
Posts: 8

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Aug 2010, 13:41
I am not sure about the answer.. 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



Manager
Joined: 29 Jul 2010
Posts: 124

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Aug 2010, 14:51
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



Senior Manager
Joined: 03 Nov 2005
Posts: 385
Location: Chicago, IL

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Aug 2010, 17:17
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 timeconsuming than that.
_________________
Hard work is the main determinant of success



Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1657
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
19 Aug 2010, 10:39
2
This post received KUDOS
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 = 1/20What do you think? I think I deserve kudos
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/myirlogbookdiary133264.html
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
20 Aug 2010, 09:04
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 = 1/20What 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 07 Feb 2010
Posts: 158

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
07 Oct 2010, 07:04
*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? thanks in advance



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
07 Oct 2010, 07:42



Intern
Joined: 28 Mar 2011
Posts: 34

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
24 Jul 2011, 00:22
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?
Please advise.
Thanks.



Manager
Joined: 25 May 2011
Posts: 152

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
23 Dec 2011, 06:03
Could you plz help me why my approach is wrong?
P(not 2 girls sit together)=1P(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)=1p(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???



Senior Manager
Joined: 13 May 2011
Posts: 307
WE 1: IT 1 Yr
WE 2: Supply Chain 5 Yrs

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
31 Dec 2011, 01:53
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



Intern
Joined: 17 Jul 2012
Posts: 2

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Jul 2012, 01:53
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
Give your comment



Intern
Joined: 17 Jul 2012
Posts: 2

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Jul 2012, 01:54
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??



Math Expert
Joined: 02 Sep 2009
Posts: 39744

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
17 Jul 2012, 02:16
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
Give your comment 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 ...
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 03 Oct 2010
Posts: 4

Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]
Show Tags
28 Aug 2012, 03:16
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)=1P(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)=1p(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???



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: PS: Probability : 3Girls 7 Boys [#permalink]
Show Tags
28 Aug 2012, 06:51
1
This post received KUDOS
shahideh wrote: Could you plz help me why my approach is wrong?
P(not 2 girls sit together)=1P(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)=1p(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 \(\frac{1}{3}+2\cdot\frac{2\cdot3}{9\cdot10}=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}.\)
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]
Show Tags
28 Aug 2012, 08:18
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? ANSWER: 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 \(1\frac{3}{5}+\frac{1}{15}=\frac{7}{15}.\)
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Senior Manager
Joined: 06 Aug 2011
Posts: 400

Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]
Show Tags
28 Aug 2012, 08:57
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..
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Senior Manager
Joined: 03 Sep 2012
Posts: 336
Location: United States
Concentration: Healthcare, Strategy
GPA: 3.88
WE: Medicine and Health (Health Care)

Re: A row of seats in a movie hall contains 10 seats. 3 Girls [#permalink]
Show Tags
25 Sep 2012, 06:37
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 ..
_________________
"When you want to succeed as bad as you want to breathe, then you’ll be successful.”  Eric Thomas




Re: A row of seats in a movie hall contains 10 seats. 3 Girls
[#permalink]
25 Sep 2012, 06:37



Go to page
1 2
Next
[ 30 posts ]




