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COMBINATION PERMUTATION QUESTION HELP!!

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Joined: 23 Dec 2009
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14 Jan 2010, 03:55
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I HAVE SOME DOUBT IN SOME QUESTION I HOPE THAT SOMEONE CAN HELP ME HERE IN GMATCLUB..
1.
fritz is taking an examination that consist of two parts A and B with the following instructions:
Part A must be completed before starting part B
Part A contains 3 questions and a student must answer two
Part B contains 4 questions and a student must answer two.

in how many ways can the test be completed
A 12
B 15
C 36
D 72
E 90

can you solve for me this question? i have tried to solve it but I dont understand why in the solution they use permutation instead of combinations. thank you for your help!

2.
the university of Maryland, university of Vermont and Emory University have each 4 soccer players. if a team of 9 is to be formed with an equal number of players from each university, how many numbers of ways can the selections be done?
A 3
B 4
C 12
D 16
E 25

I cannot find the right answer. my answer will be 4^3 because I multiply 4C3*4C3*4C3..
can u explain me why it isnt correct.

3
Each employee at a certain bank is either a clerk or an agent or both. of every 3 agents one is also a clerk. of every two clerks, one is also an agent. what is the probability that an employee randomly selected from a bank is boh an agent and a clerk?
A 1/2
B 1/3
C 1/4
D 1/5
E 2/5

I cant solve this question, the book solution are wording and not clear.

and i want to add just a question.
when i face with a probability problems like in a box there 32 blu marbles and 12 red marbles... which is the probability that picking two marbles are both red? in these case when someone picks at the same time more than one marbles what do i have to do to solve it? thanks

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Manager
Joined: 06 Jan 2010
Posts: 62
Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 09:12
2
1.
fritz is taking an examination that consist of two parts A and B with the following instructions:
Part A must be completed before starting part B
Part A contains 3 questions and a student must answer two
Part B contains 4 questions and a student must answer two.

for part a; he must choose 2 of 3 questions. and these 2 can be in more than one order
so 3c2*2!

for part b; similarly he must choose 2 out of 4 question and can answer those 2 questions in more than 1 order
so
4c2*2!

so total ways is 3c2*2*4c2*2 = 36
Manager
Joined: 06 Jan 2010
Posts: 62
Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 09:13
1
the university of Maryland, university of Vermont and Emory University have each 4 soccer players. if a team of 9 is to be formed with an equal number of players from each university, how many numbers of ways can the selections be done?
A 3
B 4
C 12
D 16
E 25

I cannot find the right answer. my answer will be 4^3 because I multiply 4C3*4C3*4C3..
can u explain me why it isnt correct.

i think you're right
even i get 4c3*4c3*4c3 = 64
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 09:24
3
Each employee at a certain bank is either a clerk or an agent or both. of every 3 agents one is also a clerk. of every two clerks, one is also an agent. what is the probability that an employee randomly selected from a bank is boh an agent and a clerk?
A 1/2
B 1/3
C 1/4
D 1/5
E 2/5

This can be done using Baye's theorem on conditional probability
Conditional probability is when the probability of something occuring is given assuming another condition has already occured

In this case, given the employee is a clerk, there's 1 in 2 chance that he's also an agent
and
Given the employee is an agent, there's a 1 in 3 chance that he's also a clerk
It is important to note that these 2 sentences are not reciprocal

Baye's theorem says that P(A/B) = P(AnB)/(P(B)
Put another way
if you know that event B has occured, what is the probability that A has occured
you're looking for all those events in set b, where a has also occured
or you're looking for all those events where b and a have occured
but instead of dividing it by total no of events, you divide by total number of events in set B

so from the question we know
P(clerk/agent) = 1/3
P(agent/clerk) = 1/2

from baye's theorem
p(clerk/agent) = p(clerk n agent)/p(agent)
==>
p(agent) = p(clerk n agent)/p(clerk/agent);
substituting vales
p(agent) = 3 p(clerk n agent)
==>
solving for p(clerk) in a similar manner, we get
p(clerk) = 2 p(clerk n agent)

now
p(agent) + p(clerk) - p(agent n clerk) = 1
substituting values in terms of p(agent n clerk) we get p(agent n clerk ) = 1/4

hope that helps
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 10:10
thank you so much for your usefull explanation, expecially in the last question.
I have use combinations too in the first problems but the book in the solution use permutations and i dont understand why... is it corret using combinations also for you?

thanks
Manager
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Posts: 62
Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 13:49
3p2 is the same as 3c2*2!

the thing to understand about combinations is that it is just a selection where the order doesn't matter
permutation is selecting and then ordering
so 3c2 * 2 is selecting(3c2) and then ordering(2)
hope that makes sense.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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14 Jan 2010, 13:55
at the same time would still be the usual way of solving the problem

ways of pick 2 red marbles 12c2
total ways of picking marbles (12+ 32) c2

so probability is 12c2/44 c2

the time when you've to solve the problem differently is when they say : the marble drawn is replaced

in those cases you'll solve it as as 12/44*12/44
Intern
Joined: 27 Sep 2010
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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27 Sep 2010, 15:48
questions 1 and 3 have been answered.

Here is my thought for question 2:

For each school you have 4C3= 4!/3! = 4
because you have 3 schools: 3*4 = 12 possibilities.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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27 Sep 2010, 16:08

Each university has exactly 1 student to leave out. This can be done in 4 ways. There are 3 universities, each can leave out in 4 ways. So total ways is 4^3 or 64
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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27 Sep 2010, 16:28
Keep in mind it is a combination problem, not a permutation one.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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27 Sep 2010, 16:29
lucalelli88 wrote:
I HAVE SOME DOUBT IN SOME QUESTION I HOPE THAT SOMEONE CAN HELP ME HERE IN GMATCLUB..
1.
fritz is taking an examination that consist of two parts A and B with the following instructions:
Part A must be completed before starting part B
Part A contains 3 questions and a student must answer two
Part B contains 4 questions and a student must answer two.

in how many ways can the test be completed
A 12
B 15
C 36
D 72
E 90

can you solve for me this question? i have tried to solve it but I dont understand why in the solution they use permutation instead of combinations. thank you for your help!

I'm curious where these questions are from, because the wording, in particular in the first question, is poor. It isn't clear in the first question whether order matters; if the student answers question 1 first, then question 3, is that different from answering question 3 first, then question 1? There's no way to decide based on the wording; the question 'In how many ways can the test be completed?' is open to different legitimate interpretations. So I don't find this a good practice problem.

lucalelli88 wrote:
3
Each employee at a certain bank is either a clerk or an agent or both. of every 3 agents one is also a clerk. of every two clerks, one is also an agent. what is the probability that an employee randomly selected from a bank is boh an agent and a clerk?
A 1/2
B 1/3
C 1/4
D 1/5
E 2/5

I cant solve this question, the book solution are wording and not clear.

Question 3 is similar to some questions I saw on a recent real GMAT - an overlapping sets question which provides ratios between the two overlapping groups. Here, using the ratios given, if we have 1 person who is both an agent and a clerk, we have 2 who are only agents and 1 who is only a clerk. So for every 4 people, there is 1 who is an agent and a clerk, and the answer is 1/4. There's no need to use Baye's Theorem here.
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Intern
Joined: 27 Sep 2010
Posts: 3
Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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27 Sep 2010, 16:37
IanStewart wrote:
Question 3 is similar to some questions I saw on a recent real GMAT - an overlapping sets question which provides ratios between the two overlapping groups. Here, using the ratios given, if we have 1 person who is both an agent and a clerk, we have 2 who are only agents and 1 who is only a clerk. So for every 4 people, there is 1 who is an agent and a clerk, and the answer is 1/4. There's no need to use Baye's Theorem here.

a Venn diagram can help in this kind of case.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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11 Oct 2010, 03:28
lucalelli88 wrote:
I HAVE SOME DOUBT IN SOME QUESTION I HOPE THAT SOMEONE CAN HELP ME HERE IN GMATCLUB..
1.
fritz is taking an examination that consist of two parts A and B with the following instructions:
Part A must be completed before starting part B
Part A contains 3 questions and a student must answer two
Part B contains 4 questions and a student must answer two.

in how many ways can the test be completed
A 12
B 15
C 36
D 72
E 90

This question doesn't really make sense.
Assuming that order doesn't matter when answering questions within Part A or Part B:
3C2 * 4C2 = 3 * 6 = 18.

Assuming order does matter when answering questions within each part. I'm not sure why this would occur given the information in the problem statement.
3P2 * 4P2 = 6 * 12 = 36.

lucalelli88 wrote:
2.
the university of Maryland, university of Vermont and Emory University have each 4 soccer players. if a team of 9 is to be formed with an equal number of players from each university, how many numbers of ways can the selections be done?
A 3
B 4
C 12
D 16
E 25

"the university of Maryland, university of Vermont and Emory University have each 4 soccer players."
3 bags of 4: M1,M2,M3,M4 V1,V2,V3,V4 E1,E2,E3,E4

"if a team of 9 is to be formed"
combo box arrangement
(_)(_)(_)(_)(_)(_)(_)(_)(_)/9! <--------the next clause indicates that this is not the correct setup for the problem.

"with an equal number of players from each university"
Oops. Rewrite the combo box arrangement.
divide the combo box arrangement into 3 groups
(_)(_)(_)/3! Maryland
(_)(_)(_)/3! Vermont
(_)(_)(_)/3! Emory

"how many numbers of ways can the selections be done?"
Fill in the boxes.
(4)(3)(2)/3! * (4)(3)(2)/3! * (4)(3)(2)/3! = 4*4*4 = 64

lucalelli88 wrote:
3
Each employee at a certain bank is either a clerk or an agent or both. of every 3 agents one is also a clerk. of every two clerks, one is also an agent. what is the probability that an employee randomly selected from a bank is boh an agent and a clerk?
A 1/2
B 1/3
C 1/4
D 1/5
E 2/5

"Each employee at a certain bank is either a clerk or an agent or both. "
C,A,CA

"of every 3 agents one is also a clerk."
1: A
2: A
3: AC

"of every two clerks, one is also an agent."
1: A
2: A
3: AC
4: C
Total = 4

"what is the probability that an employee randomly selected from a bank is boh an agent and a clerk?"
Employee #3 / Total = 1/4

lucalelli88 wrote:
and i want to add just a question.
when i face with a probability problems like in a box there 32 blu marbles and 12 red marbles... which is the probability that picking two marbles are both red? in these case when someone picks at the same time more than one marbles what do i have to do to solve it? thanks

Probability Table: Create and work backward.
# of reds
0:
1:
2:
------------------------------
Total =

******************************************

Total = Bag of 44, pick 2 = 44C2 = 44*43/2 = 22*43 = 946

2 reds: 12 reds, pick 2 = 12C2 = 12*11/2 = 6*11 = 66

1 red: 12 reds, pick 2 * 32 blues = 12C1 * 32 = 12 * 32 = 384

0 reds: Total up the column = 946 - 66 - 384 = 496.
Or, 32 blues, pick 2 = 32C2 = 32*31/2 = 16*36 = 496.

******************************************

# of reds
0: 496
1: 384
2: 66
------------------------------
Total = 946

******************************************

Use the table to answer any probability question.
P(red = 0) = 496/946
P(red = 1) = 384/946
P(red = 2) = 66/946
P(no reds or all reds) = (496+66)/946
P(less than 2 reds) = (384+496)/946
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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25 May 2011, 06:30
I assume you guys are using the NOVA book. This problem is tricky and I've noticed that most of the questions in this chapter are worded in ways that could be considered "open to interpretation"

The key to this problem is that there are no dependencies between the different groups choices. You are NOT being asked to find all the possible unique 9 person teams given the problems constraints. This is not expressed very well in the wording. What the question writer is trying to convey is "What is the total number of choices that could be made?(Considering each university chooses independently)"

Think of it using this narrative. You are on the selection committee for Emory university and you are told to select 3 of your 4 players. How many choices do you have? The answer is 4!/3! --> 24/6 --> 4 choices. Now at the other two universities each of those selection committees also have 4 choices. If you want to know what the total number of choices that can be made is you simply add those three numbers together 4+4+4=12.

Its easy to fall into a trap where you think the answer is 64. You arrive at this answer by building a tree . This leads you to the conclusion that that there are 4^3 nodes at the lowest order. The problem is that to use this tree you need to make an assumption that there is a dependency between the choices(i.e. that Emory's selection is somehow linked to the University of Vermont's, and so forth). Since the problem never explicitly establishes this dependency(by saying something like Emory must make their choice before University of Vermont can make theirs) because of this we must assume that the choices are independent.

The bottom line is this question is vaguely worded and reinforces the consensus that this entire book could use a good proof-reading. The problems you will encounter on the GRE/GMAT can still be tricky but they will have gone through a review process that eliminates any possibility of interpretation.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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25 May 2011, 08:52
nsfyn55 wrote:

Think of it using this narrative. You are on the selection committee for Emory university and you are told to select 3 of your 4 players. How many choices do you have? The answer is 4!/3! --> 24/6 --> 4 choices. Now at the other two universities each of those selection committees also have 4 choices. If you want to know what the total number of choices that can be made is you simply add those three numbers together 4+4+4=12.

I agree that the wording of most of these questions is pretty hopeless, but no matter how I read the soccer question, I can't see any interpretation that would make addition the correct thing to do. We add in counting problems when we have different ways of arriving at the result we're interested in - that is, when there are several cases to consider. That's not what is happening in this question; here we have choices from different universities, and when we make a sequence of choices, we need to multiply. Mind you, the question "how many numbers of ways can the selections be done" doesn't make any sense to begin with, so there isn't much point debating what it might mean.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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25 May 2011, 09:43
Thats what I think you missed. This problem does not indicate or even imply a sequence of choices. It says "how many numbers of ways can the selections be made". Technically, this sentence is grammatically correct in terms of pluralization and the keywords are "numbers" and "selections" -- > plural. This indicates that we are dealing with the counts of each university's selections independently. Each university has 4 choices. 4+4+4=12.

Sorry I'm with NOVA guy on this one. If the problem said what are the total number of unique teams that can be formed if you must select 3 from each group then the answer is 64.
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25 May 2011, 12:55
nsfyn55 wrote:
Thats what I think you missed.

I don't believe I missed a thing. The question asks "how many numbers of ways can the selections be made". This is a pedantic point, but suppose we take that literally, and decipher what it might mean. If there are 100 people in a room, and I ask the question "how many numbers of people are there in the room?" the answer is one: there is one number of people in the room, and that number is equal 100. It's an almost nonsensical question to ask, and while it may be 'technically... grammatically correct', it's mathematically meaningless.

If you're adding the number of teams from each university, you're answering the question 'how many different 4-person teams could arise from this selection process'. Then it's correct to add, but that's not at all what the question asks.

Mind you, this isn't a particularly productive discussion, because the wording of the questions in the original post is so bad that it's just a guessing game how to interpret them. Were you to see this question on the GMAT, it would surely be set up in such a way that you would multiply your choices from each university, so that's the version of the question that might be worthwhile to study.
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25 May 2011, 13:22
You can call it unproductive all you want.

What the question is saying in a less than perfectly worded though still correct way is...

1. A selection is picking 3 from a pool of 4 - 4 Combinations
2. There are three institutions making selections - 3 institutions
3. How many selections are there in total. - 12 total

Adding wording, guessing intent, and speculating at what you probably will see on the test do not change the fact that the answer to this problem as its worded is 12.
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Re: COMBINATION PERMUTATION QUESTION HELP!!  [#permalink]

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25 May 2011, 14:06
nsfyn55 wrote:
You can call it unproductive all you want.

What the question is saying in a less than perfectly worded though still correct way is...

1. A selection is picking 3 from a pool of 4 - 4 Combinations
2. There are three institutions making selections - 3 institutions
3. How many selections are there in total. - 12 total

Adding wording, guessing intent, and speculating at what you probably will see on the test do not change the fact that the answer to this problem as its worded is 12.

As I explained above, as the question is worded the answer is 1. I still don't understand, from the wording of the question, how you're arriving at your interpretation, but I've wasted far too many hours in the past debating the interpretation of questions that were hopelessly worded to begin with, and I don't care to do it again, so if you'd like to think the answer is 12 to the question, I won't try any further to persuade you otherwise; I'll just say, in case it's of any benefit to other test takers reading this thread, that under the only interpretation I find reasonable here, the answer ought to be 4^3 = 64.

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