Find all School-related info fast with the new School-Specific MBA Forum

It is currently 29 Jul 2014, 03:29

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many different 5-person teams can be formed from a group

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 22 Dec 2009
Posts: 40
Followers: 0

Kudos [?]: 7 [0], given: 13

How many different 5-person teams can be formed from a group [#permalink] New post 22 Jun 2010, 18:14
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

87% (01:53) correct 13% (01:17) wrong based on 193 sessions
How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.


The answer is D and I know how to figure out now but is there any trick to know each question is sufficient without actual compute? cuz its time consuming until I found out e.g. 1) is 9!/5!4!

thanks in advance
[Reveal] Spoiler: OA
5 KUDOS received
Manager
Manager
avatar
Joined: 03 May 2010
Posts: 89
WE 1: 2 yrs - Oilfield Service
Followers: 11

Kudos [?]: 57 [5] , given: 7

GMAT Tests User Reviews Badge
Re: Manhattan Q/DS [#permalink] New post 22 Jun 2010, 20:31
5
This post received
KUDOS
We need a unique value.

1. (x+2) C 5 = 126. There is only one possible value for x+2 that would yield a value of 126. Don't bother trying to find out what it is. Remember, the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr.

2. (x+1) C 3 = 56 Again, you should be able to see that there can be only one value of x+1 that would yield a value of 56. Why bother finding out what the value is? As long as we have an equation in one variable, we can find a value.
Intern
Intern
avatar
Joined: 22 Dec 2009
Posts: 40
Followers: 0

Kudos [?]: 7 [0], given: 13

Re: Manhattan Q/DS [#permalink] New post 22 Jun 2010, 21:27
HI AbhayPrasanna.. Thanks for the reply...

I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...

How do you know (x+2) C 5 = 126 is computable
Manager
Manager
User avatar
Joined: 28 Feb 2010
Posts: 176
WE 1: 3 (Mining Operations)
Followers: 3

Kudos [?]: 20 [0], given: 33

Re: Manhattan Q/DS [#permalink] New post 24 Jun 2010, 01:50
I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value.
But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.

@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.
_________________

Regards,
Invincible...:)
"The way to succeed is to double your error rate."
"Most people who succeed in the face of seemingly impossible conditions are people who simply don't know how to quit."

Expert Post
8 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18803
Followers: 3262

Kudos [?]: 22681 [8] , given: 2635

Re: Manhattan Q/DS [#permalink] New post 25 Jun 2010, 04:41
8
This post received
KUDOS
Expert's post
gmatJP wrote:
HI AbhayPrasanna.. Thanks for the reply...

I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...

How do you know (x+2) C 5 = 126 is computable


praveenism wrote:
I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value.
But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.

@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.


The point here is the following:
Suppose we are told that there are 10 ways to choose x people out of 5. What is x? C^x_5=10 --> \frac{5!}{x!(5-x)!}=10 --> x!(5-x)!=12 --> x=3 or x=2. So we cannot determine single numerical value of x. Note that in some cases we'll be able to find x, as there will be only one solution for it, but generally when we are told that there are n ways to choose x out of m there will be (in most cases) two solutions of x possible.

But if we are told that there are 10 ways to choose 2 out of x, then there will be only one value of x possible --> C^2_x=10 --> \frac{x!}{2!(x-2)!}=10 --> \frac{x(x-1)}{2!}=10 --> x(x-1)=20 --> x=5.

In our original question, statement (1) says that there are 126 ways to choose 5 out of x+2 --> there will be only one value possible for x+2, so we can find x. Sufficient.

Just to show how it can be done: C^5_{(x+2)}=126 --> (x-2)(x-1)x(x+1)(x+2)=5!*126=120*126=(8*5*3)*(9*7*2)=5*6*7*8*9 --> x=7. Basically we have that the product of five consecutive integers ((x-2)(x-1)x(x+1)(x+2)) equal to some number (5!*126) --> only one such sequence is possible, hence even though we have the equation of 5th degree it will have only one positive integer solution.

Statement (2) says that there are 56 ways to choose 3 out of x+1 --> there will be only one value possible for x+1, so we can find x. Sufficient.

C^3_{(x+1)}=56 --> (x-1)x(x+1)=3!*56=6*7*8 --> x=7. Again we have that the product of three consecutive integers ((x-1)x(x+1)) equal to some number (3!*56) --> only one such sequence is possible, hence even though we have the equation of 3rd degree it will have only one positive integer solution.

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

2 KUDOS received
Kaplan GMAT Instructor
User avatar
Joined: 21 Jun 2010
Posts: 75
Location: Toronto
Followers: 20

Kudos [?]: 97 [2] , given: 2

Re: Manhattan Q/DS [#permalink] New post 25 Jun 2010, 22:29
2
This post received
KUDOS
Bunuel supplied an awesome and exhaustive mathematical algebraic explanation. Perhaps it will be of benefit to review the concept verbally as well.

8C5 = 8C3 because everytime we pull a subgroup of 5 objects from the bigger group of 8, we can see that we are also "setting aside" a subgroup of 3. Likewise, everytime we pull out a subgroup of 3, we also set aside a subgrup of 5. So, the number of ways we can pull out 5 object subgroups must be the same as the number of ways we can pull out 3 object subgroups.

But if there are 126 ways to pull 5 objects from a big group "x + 2", then "x+2" must be just one value. If it were not, then it would imply that increasing or decresing the size of the big group doesn't necessarily affect how many ways you can pull out a smaller subgroup--surely an absurd conclusion. Absurd because clearly there are more ways to pull 5 objects out of a set of 100 than out of a set of, say, 10.
_________________

Kaplan Teacher in Toronto
http://www.kaptest.com/GMAT

Prepare with Kaplan and save $150 on a course!

Image

Kaplan Reviews

Intern
Intern
avatar
Joined: 03 Dec 2010
Posts: 22
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: How many different 5-person teams can be formed from a group [#permalink] New post 06 Apr 2012, 07:07
Hello to all,

Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,

3Cx+1 =56 and how, 5Cx+2 = 126 ?

Thnx
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18803
Followers: 3262

Kudos [?]: 22681 [0], given: 2635

Re: How many different 5-person teams can be formed from a group [#permalink] New post 06 Apr 2012, 07:28
Expert's post
priyalr wrote:
Hello to all,

Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,

3Cx+1 =56 and how, 5Cx+2 = 126 ?

Thnx


# of ways to pick k objects out of n distinct objects is C^k_n=\frac{n!}{(n-k)!*k!}.

# of ways to pick 3 people out of x+1 people is C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}. Now, since (x+1)!=(x-2)!*(x-1)*x*(x+1) then (x-2)! get reduced and we'll have: \frac{(x-1)*x*(x+1)}{3!}. We are told that this equals to 56: \frac{(x-1)*x*(x+1)}{3!}=56 --> (x-1)x(x+1)=3!*56=6*7*8 --> x=7.

You can apply similar logic to C^5_{(x+2)}=126.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 07 Sep 2011
Posts: 65
Location: United States
Concentration: Strategy, International Business
GMAT 1: 640 Q39 V38
WE: General Management (Real Estate)
Followers: 3

Kudos [?]: 26 [0], given: 3

Re: How many different 5-person teams can be formed from a group [#permalink] New post 20 Aug 2012, 02:39
Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)



Bunuel wrote:
priyalr wrote:
Hello to all,

Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,

3Cx+1 =56 and how, 5Cx+2 = 126 ?

Thnx


# of ways to pick k objects out of n distinct objects is C^k_n=\frac{n!}{(n-k)!*k!}.

# of ways to pick 3 people out of x+1 people is C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}. Now, since (x+1)!=(x-2)!*(x-1)*x*(x+1) then (x-2)! get reduced and we'll have: \frac{(x-1)*x*(x+1)}{3!}. We are told that this equals to 56: \frac{(x-1)*x*(x+1)}{3!}=56 --> (x-1)x(x+1)=3!*56=6*7*8 --> x=7.

You can apply similar logic to C^5_{(x+2)}=126.

Hope it's clear.
Manager
Manager
avatar
Joined: 05 Mar 2012
Posts: 67
Schools: Tepper '15 (WL)
Followers: 0

Kudos [?]: 3 [0], given: 7

Re: How many different 5-person teams can be formed from a group [#permalink] New post 22 Aug 2012, 13:23
manjeet1972 wrote:
Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)



[/quote]

Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.

Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.

So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.
Manager
Manager
User avatar
Joined: 07 Sep 2011
Posts: 65
Location: United States
Concentration: Strategy, International Business
GMAT 1: 640 Q39 V38
WE: General Management (Real Estate)
Followers: 3

Kudos [?]: 26 [0], given: 3

Re: How many different 5-person teams can be formed from a group [#permalink] New post 22 Aug 2012, 21:38
Good explanation. Through practice I will learn.

Injuin wrote:
manjeet1972 wrote:
Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1)
(x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3)
(x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)





Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.

Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.

So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.[/quote]
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18803
Followers: 3262

Kudos [?]: 22681 [0], given: 2635

Re: How many different 5-person teams can be formed from a group [#permalink] New post 30 Jun 2013, 23:46
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: How many different 5-person teams can be formed from a group   [#permalink] 30 Jun 2013, 23:46
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic How many 5 person committees chosen at random from a group c mumbijoh 6 03 Dec 2013, 08:41
2 Experts publish their posts in the topic How many different groups of 3 people can be formed from a niheil 5 10 Oct 2010, 13:19
In how many ways can a group of 8 be divided into 4 teams of leeye84 4 24 Aug 2007, 08:04
1 In how many ways can a group of 8 be divided into 4 teams of mitul 9 22 Dec 2006, 21:37
How many different words can be formed with the letters of anirban16 2 19 Mar 2005, 08:16
Display posts from previous: Sort by

How many different 5-person teams can be formed from a group

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.