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Good set of PS 2

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Re: Good set of PS 2  [#permalink]

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New post 13 Aug 2010, 12:07
Hi Bunuel

for the time and work question in this problem set can you pls explain this step in a bit more detail:

"Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"

How did you figure out the speed increased by 1.6?

For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work---

your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks.
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New post 13 Aug 2010, 13:50
1
gmat1011 wrote:
Hi Bunuel

for the time and work question in this problem set can you pls explain this step in a bit more detail:

"Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"

How did you figure out the speed increased by 1.6?

For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work---

your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks.


Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance.

\(Time*Rate=Distance\) <--> \(Time*Rate=Job \ done\).

We know that 10 people need 55 days to complete the job --> let the combined rate of 10 people be x --> \(Time*Rate=x*55=Job \ done\);
Now, if combined rate of 10 people is \(x\), then combined rate of 16 man will be \(1.6x\) (1.6 times more than x as 16 is 1.6 times more than 10), so \(new \ time*new \ rate=same \ job\) --> \(t_2*1.6x=x*55\) --> \(t_2=\frac{55}{1.6}\approx{34.8}\) (as rate increased 1.6 times then time needed to do the same job will decrease 1.6 times).

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.
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Re: Good set of PS 2  [#permalink]

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New post 30 Sep 2010, 15:59
TehJay wrote:
I don't know if it's the broken English, but question 9 doesn't make any sense. 15 people applied to both college X and college Y. Those 15 people are composed of 20% of the people who applied to college X and 25% of the people who applied to college Y. So .2x + .25y = 15.

What if 50 people applied to college X and 20 people applied to college Y? Then you have (.2)(50) + (.25)(20) = 10 + 5 = 15. But 70 isn't an answer choice.


9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y?
(A) 135
(B) 120
(C) 115
(D) 105
(E) 90

Question means the following:
15 people applied to both college X and Y, these 15 people represent 20% of all applicants who applied to X and 25% of all applicants who applied to Y.


\(0.2*x=15\) --> \(x=75\) --> # of people who applied only to X is: \(75-15=60\);
\(0.25*y=15\) --> \(y=60\) --> # of people who applied only to Y is: \(60-15=45\);

# of people who applied only to X OR Y is: \(60+45=105\).

Answer: D.

Hope it's clear.
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Re: Good set of PS 2  [#permalink]

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New post 02 Oct 2010, 03:03
Bunuel wrote:
ANSWERS (OA):

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
(A) 28
(B) 32
(C) 48
(D) 60
(E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately:
1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24
Or
2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8.
Total=24+8=32.

"Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32"

Answer: B.


Bunuel,

For Q1, Can u please elaborate more on this >>
Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

I was able to get answer by first method as u have mentioned. But, I am failing to get the computation mentioned in another method.

Thanks!
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Re: Good set of PS 2  [#permalink]

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New post 02 Oct 2010, 05:27
samark wrote:
Bunuel wrote:
ANSWERS (OA):

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
(A) 28
(B) 32
(C) 48
(D) 60
(E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately:
1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24
Or
2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8.
Total=24+8=32.

"Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32"

Answer: B.


Bunuel,

For Q1, Can u please elaborate more on this >>
Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

I was able to get answer by first method as u have mentioned. But, I am failing to get the computation mentioned in another method.

Thanks!


Total # of arrangements:
Drivers seat: 2 (either mother or father);
Front seat: 4 (any of 4 family members left);
Back seat: 3! (arranging other 3 family members on the back seat);
So. total # of arrangements is 2*4*3!=48.

# of arrangements with sisters sitting together:
Sisters can sit together only on the back seat either by the left window or by the right window - 2, and either {S1,S2} or {S2,S1} - 2 --> 2*2=4;
Drivers seat: 2 (either mother or father);
Front seat: 2 (5 - 2 sisters on back seat - 1 driver = 2);
Back seat with sisters: 1 (the last family member left);
So, # of arrangements with sisters sitting together is 4*2*2*1=16.

48-16=32.

Hope it's clear.
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Re: Good set of PS 2  [#permalink]

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New post 09 Jan 2011, 12:22
144144 wrote:
can someone plz try and show me question 2 in the C way?

1 mean 1C9.
thanks.


2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?
(A) 271/900
(B) 27/100
(C) 7/25
(D) 1/9
(E) 1/10

There are total 900 3 digit numbers;

There are 8*9*9=648 (you can write this as 8C1*9C1*9C1 if you like) 3-digit numbers without 7 in its digits (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25.

Answer: C.
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Re: Good set of PS 2  [#permalink]

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New post 09 Jan 2011, 16:10
pesfunk wrote:
Perfect Answer :)

Economist wrote:
Bunuel wrote:
Please find below new set of PS problems:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Nobody attempted 4, let me give a try.
total work = 110*10 = 1100 man days
now, from day 1 to day 55, 10 men worked = 550 man days of work was done.
from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6.
C.



I found one more solution to this problem by Bunnel but the answer is 5... that is B..
But here its mentioned C...


4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

This one was solved incorrectly:
Days to finish the job for 10 people 110 days.
On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people).
Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.
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Re: Good set of PS 2  [#permalink]

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New post 09 Jan 2011, 16:17
jullysabat wrote:
Economist wrote:
Bunuel wrote:
Please find below new set of PS problems:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Nobody attempted 4, let me give a try.
total work = 110*10 = 1100 man days
now, from day 1 to day 55, 10 men worked = 550 man days of work was done.
from day 61 to day 100, 16 men worked = 640 man days of work was done.

so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6.
C.



I found one more solution to this problem by Bunnel but the answer is 5... that is B..
But here its mentioned C...


4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

This one was solved incorrectly:
Days to finish the job for 10 people 110 days.
On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people).
Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.


Solution provided by Economist is not correct (the one with answer C). OA's and solutions are given in my post on the 2nd page. OA for the quoted question is indeed B:

Given: 10-man crew needs 110 days to complete the construction.

"On the 61-st day, after 5 days of rain ..." --> as it was raining for 5 days then they must have bee working for 55 days thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 men).

Then contractor "hired 6 more people" --> speed of construction increased 1.6 times, so the new 16-man crew needed 55/1.6=~34.4 days to complete the construction, but after they were hired job was done in 100-60=40 days --> so 5 days rained. (They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.)

Answer: B.
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New post 16 Jan 2011, 02:41
praveenvino wrote:
9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y?
(A) 135
(B) 120
(C) 115
(D) 105
(E) 90

I am getting a single equation like .2X + 0.25Y = 15. I feel like missing something. Can someone explain what it could be?


OA-s and solutions are given in my post on 2nd page. Solution for this question:

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y?
(A) 135
(B) 120
(C) 115
(D) 105
(E) 90

Question means the following:
15 people applied to both college X and Y, these 15 people represent 20% of all applicants who applied to X and 25% of all applicants who applied to Y.


\(0.2*x=15\) --> \(x=75\) --> # of people who applied only to X is: \(75-15=60\);
\(0.25*y=15\) --> \(y=60\) --> # of people who applied only to Y is: \(60-15=45\);

# of people who applied only to X OR Y is: \(60+45=105\).

Answer: D.

Hope it's clear.
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Re: Good set of PS 2  [#permalink]

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New post 20 Jul 2012, 08:39
[quote="Bunuel"]Please find below new set of PS problems:



5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45


R= 0,12, then:
\(\frac{R}{S}= \frac{12}{100}\)
\(\frac{R}{S}= \frac{3}{25}\)
\(25R = 3S\)
Since 45 is the only number which has 3 and 5 as his primes (E) should be the answer.
Please provide OA.
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New post 20 Jul 2012, 08:54
Stiv wrote:
Bunuel wrote:
Please find below new set of PS problems:



5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45


R= 0,12, then:
\(\frac{R}{S}= \frac{12}{100}\)
\(\frac{R}{S}= \frac{3}{25}\)
\(25R = 3S\)
Since 45 is the only number which has 3 and 5 as his primes (E) should be the answer.
Please provide OA.


Please read the thread.

OA-s and solutions are given in my post on 2nd page.


5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

\(s\) divided by \(t\) yields the remainder of \(r\) can always be expressed as: \(\frac{s}{t}=q+\frac{r}{t}\) (which is the same as \(s=qt+r\)), where \(q\) is the quotient and \(r\) is the remainder.

Given that \(\frac{s}{t}=64.12=64\frac{12}{100}=64\frac{3}{25}=64+\frac{3}{25}\), so according to the above \(\frac{r}{t}=\frac{3}{25}\), which means that \(r\) must be a multiple of 3. Only option E offers answer which is a multiple of 3

Answer: E.
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Re: Good set of PS 2  [#permalink]

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New post 27 Jul 2012, 00:10
For those who like visualizing, here is a solution for Question 7 using graphs.

Please, refer to the attached drawing, in which the three graphs \(y=1/x,\) \(y=2x,\) and \(y=x^2\) are depicted for \(x>0\).
The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions.

So, the correct orderings are:
If \(x\) between 0 and A: \(x^2<2x<1/x\)
If \(x\) between A and B: \(x^2<1/x<2x\)
If \(x\) between B and C: \(1/x<x^2<2x\)
If \(x\) greater than C: \(1/x<2x<x^2\)

We can see that only the first two of the above options are listed as answers (I and II).

Answer: D.
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New post 24 Sep 2012, 13:02
pleonasm wrote:
1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
(A) 28
(B) 32
(C) 48
(D) 60
(E) 120


Good challenging question.

Driving spot - only two variations. 1) Mother drives 2) Father drives

The permutations come up in the spots for the other 4 seats.

So, 2 default values * 4P4 = total 2*4! = 48 permutations

Total # ways - # ways together = # ways can't sit together

The two daughters can only sit together when they are together in the back seat.
So that's:
12B
21B
B12
B21

...where 1 and 2 represent daughter #1 and #2. "B" represents the boy child.

so that's 4 ways we counted.

4 ways * variations on other 3 seats
4 ways * (default mother/father drives and variations on other 2 seats)
4 ways * (2 * 2P2)
4 ways * 2 * 2! = 16

Total ways = 48
Variations together = 16
Variations not together = 48 - 16 = 32
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New post 17 Mar 2013, 02:42
Bunuel wrote:
ANSWERS (OA):


6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

700-65 = 635

Answer: E.



Looks like the above analysis is flawed.

Ways to select 2 men out of 8 is \(8C2\)
Ways to select 3 women out of 5 is \(5C3\)
Ways to select the remaining 1 person out of the remaining 8 = 8

Total number of ways of forming a committee as above = 28 * 10 *8 = 2240 - (1)

Now for the restriction that 2 men cannot serve together consider the ways they can serve together= 1*10*8=80 - (2)

(1) - (2) = 2160 ways looks like the right answer.
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New post 17 Mar 2013, 02:50
SravnaTestPrep wrote:
Bunuel wrote:
ANSWERS (OA):


6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

700-65 = 635

Answer: E.



Looks like the above analysis is flawed.

Ways to select 2 men out of 8 is \(8C2\)
Ways to select 3 women out of 5 is \(5C3\)
Ways to select the remaining 1 person out of the remaining 8 = 8


Total number of ways of forming a committee as above = 28 * 10 *8 = 2240 - (1)

Now for the restriction that 2 men cannot serve together consider the ways they can serve together= 1*10*8=80 - (2)

(1) - (2) = 2160 ways looks like the right answer.


OA for this question is E.

2240 has duplications in it.
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Re: Good set of PS 2  [#permalink]

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New post 17 Mar 2013, 02:53
Bunuel wrote:
SravnaTestPrep wrote:
Bunuel wrote:
ANSWERS (OA):


6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

700-65 = 635

Answer: E.



Looks like the above analysis is flawed.

Ways to select 2 men out of 8 is \(8C2\)
Ways to select 3 women out of 5 is \(5C3\)
Ways to select the remaining 1 person out of the remaining 8 = 8


Total number of ways of forming a committee as above = 28 * 10 *8 = 2240 - (1)

Now for the restriction that 2 men cannot serve together consider the ways they can serve together= 1*10*8=80 - (2)

(1) - (2) = 2160 ways looks like the right answer.


OA for this question is E.

2240 has duplications in it.


I don't think they do. Why don't you try on a smaller example?
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New post 17 Mar 2013, 03:06
SravnaTestPrep wrote:
Bunuel wrote:
SravnaTestPrep wrote:
Looks like the above analysis is flawed.

Ways to select 2 men out of 8 is \(8C2\)
Ways to select 3 women out of 5 is \(5C3\)
Ways to select the remaining 1 person out of the remaining 8 = 8


Total number of ways of forming a committee as above = 28 * 10 *8 = 2240 - (1)

Now for the restriction that 2 men cannot serve together consider the ways they can serve together= 1*10*8=80 - (2)

(1) - (2) = 2160 ways looks like the right answer.


OA for this question is E.

2240 has duplications in it.


I don't think they do. Why don't you try on a smaller example?


Actually there is no need, but just to illustrate.

Two sets {A, B, C} and {a, b, c}. How many ways to select a committee of 3 so that there is at least one out of each set.

My way (correct one): \(C^1_3*C^2_3+C^2_3*C^1_3=3*3+3*3=18\). We could do this in another way: \(C^3_6-2=18\).

Your way (wrong one): \(C^1_3*C^1_3*4=36\).

Let's check:
Aab;
Aac;
Abc;
Bab;
Bac;
Bbc;
Cab;
Cac;
Cbc;

ABa;
ABb;
ABc;
ACa;
ACb;
ACc;
BCa;
BCb;
BCc.

Total of 18 ways.

Hope it's clear.
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New post 17 Mar 2013, 04:35
Dear Bunuel,

Thanks for taking the trouble to list the set of possibilities for an example. I really appreciate it. But I think there is a basic point that is being missed here. In the above example let us say two of the possibilities are AaB and BaA given my approach. The point is that we are implicitly assuming importance to A in the first possibility and to B in the second possibility as we first satisfy the basic requirement of 1 man and 1 woman. "A" being picked to satisfy the basic requirement tells that A needs to be there and is more important than B. And vice versa. In my thinking that implicit importance and ordering makes the two possibilities different. So my reading would be in such questions the uniqueness of the committee is not only with respect to who constitute it but also the way it is selected.

An example to illustrate my point would be : In how ways a class of 10 can be formed from a pool of 100 applicants so that 8 are selected on merit.
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New post 17 Mar 2013, 05:07
SravnaTestPrep wrote:
Dear Bunuel,

Thanks for taking the trouble to list the set of possibilities for an example. I really appreciate it. But I think there is a basic point that is being missed here. In the above example let us say two of the possibilities are AaB and BaA given my approach. The point is that we are implicitly assuming importance to A in the first possibility and to B in the second possibility as we first satisfy the basic requirement of 1 man and 1 woman. "A" being picked to satisfy the basic requirement tells that A needs to be there and is more important than B. And vice versa. In my thinking that implicit importance and ordering makes the two possibilities different. So my reading would be in such questions the uniqueness of the committee is not only with respect to who constitute it but also the way it is selected.

An example to illustrate my point would be : In how ways a class of 10 can be formed from a pool of 100 applicants so that 8 are selected on merit.


Not sure I understand what you are trying to say here. Anyway, {AaB} and {BaA} are the same two groups.

So, the correct answer is E.
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Re: Good set of PS 2  [#permalink]

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New post 17 Mar 2013, 05:57
Dear Bunuel,

Consider this:

In how ways 3 scholars be selected from a pool of 10 scholars so that 2 are selected on merit.

Let us say one possible set of scholars is: A B C with A and B being selected on merit and C on other considerations. We have the same set which is : C B A but Now C and B being selected on merit and A on other considerations.

Don't you think the two groups are unique even though constituted by the same people?

Is not your question,

"How many 3 member committees be formed from 6 people so that each has at least one man and one woman?"

asking the same?

Here "one man and one woman" is analogous to "merit" in the above example. The point is you are not going to select "one man and one woman" randomly. This is not an abstract problem and so the solution should be based on the practical facts. Since whether ordering is important or not for a particular problem is crucial to using the right formula we need to be right in making that decision.

The way one would think, the selection of a committee is done in the above case would be not by the roll of dice but by some practical considerations. So to first satisfy the essential requirement that I select a man and a woman I go by some practical considerations. The fact that I first select "A" over "B" says that I prefer "A" in some way over "B" or "A" is in some way more important than "B". So to me the selection AaB is different from the selection BaA since order, though not physical, is implied. It is like in the first case "A" is selected on merit and "B" on merit in the second case.
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Re: Good set of PS 2 &nbs [#permalink] 17 Mar 2013, 05:57

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