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

It is currently 20 Sep 2014, 16:09

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 subordinates does Marcia have?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 05 Nov 2012
Posts: 151
Followers: 1

Kudos [?]: 9 [0], given: 56

Re: perms [#permalink] New post 26 Dec 2013, 20:40
200<2^n-n-1<500

I did get this equation but how do you solve it guys? GMAT may not be testing it but I am interested to know the process of solving this equation. This one is easy by plugging, but what if the equation is 289700<2^n-n-1<526100 :roll: :evil: :oops:
Kaplan GMAT Prep Discount CodesKnewton GMAT Discount CodesGMAT Pill GMAT Discount Codes
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23494
Followers: 3505

Kudos [?]: 26532 [0], given: 2712

Re: perms [#permalink] New post 27 Dec 2013, 02:43
Expert's post
Amateur wrote:
200<2^n-n-1<500

I did get this equation but how do you solve it guys? GMAT may not be testing it but I am interested to know the process of solving this equation. This one is easy by plugging, but what if the equation is 289700<2^n-n-1<526100 :roll: :evil: :oops:


You won't be asked to solve 289700<2^n-n-1<526100 on the GMAT.
_________________

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

Intern
Intern
avatar
Joined: 10 Jan 2014
Posts: 24
Followers: 0

Kudos [?]: 1 [0], given: 6

Re: perms [#permalink] New post 16 Feb 2014, 12:43
Bunuel wrote:
BarneyStinson wrote:
How many subordinates does Marcia Have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.


This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.


Little correction:
For (2) we have \{s_1,s_2,s_3,...s_n\}. Each subordinate (s_1,s_2,s_3,...s_n) has TWO options: either to be included in the list or not. Hence total # of lists - 2^n, correct. But this number will include n lists with 1 subordinate as well 1 empty list.

As the lists should contain at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.

Lists with 1 subordinate - n: \{s_1,0,0,0...0\}, \{0,s_2,0,0,...0\}, \{0,0,s_3,0,...0\}, ... \{0,0,0...s_n\}.
List with 0 subordinate - 1: \{0,0,0,...0\}

So we'll get 200<2^n-n-1<500, --> n=8. Sufficient.

For (1):
C^2_n=28 --> \frac{n(n-1)}{2!}=28 --> n(n-1)=56 --> n=8. Sufficient.

Answer: D.


I'm sorry for bringing this old post up again but I'm afraid I don't quite understand your solution.

For statement 1, why did you use 2^n here and not a factorial approach? I just don't understand the logic behind it. For example, if Marcia had 4 different subordinates - e.g Anna, Bea, Christian, Doug (ABCD) - then there would be 4!/(2!2!) ways to create a list with two employee names on it. if the number of different two-name lists is to be between 200 and 500, then Marcia could have either 21 employees (21!/(2!19!)=210) or 32 employees (32!/2!30!=496). Since we do not know how many different employee names she puts on those lists, we can't really tell how many different subordinates she has.

I know I made a mistake somewhere, but I don't know where or why my approach/ logic is wrong. Could you please enlighten me? :)
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23494
Followers: 3505

Kudos [?]: 26532 [0], given: 2712

Re: perms [#permalink] New post 17 Feb 2014, 06:43
Expert's post
damamikus wrote:
Bunuel wrote:
BarneyStinson wrote:
How many subordinates does Marcia Have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.


This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.


Little correction:
For (2) we have \{s_1,s_2,s_3,...s_n\}. Each subordinate (s_1,s_2,s_3,...s_n) has TWO options: either to be included in the list or not. Hence total # of lists - 2^n, correct. But this number will include n lists with 1 subordinate as well 1 empty list.

As the lists should contain at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.

Lists with 1 subordinate - n: \{s_1,0,0,0...0\}, \{0,s_2,0,0,...0\}, \{0,0,s_3,0,...0\}, ... \{0,0,0...s_n\}.
List with 0 subordinate - 1: \{0,0,0,...0\}

So we'll get 200<2^n-n-1<500, --> n=8. Sufficient.

For (1):
C^2_n=28 --> \frac{n(n-1)}{2!}=28 --> n(n-1)=56 --> n=8. Sufficient.

Answer: D.


I'm sorry for bringing this old post up again but I'm afraid I don't quite understand your solution.

For statement 1, why did you use 2^n here and not a factorial approach? I just don't understand the logic behind it. For example, if Marcia had 4 different subordinates - e.g Anna, Bea, Christian, Doug (ABCD) - then there would be 4!/(2!2!) ways to create a list with two employee names on it. if the number of different two-name lists is to be between 200 and 500, then Marcia could have either 21 employees (21!/(2!19!)=210) or 32 employees (32!/2!30!=496). Since we do not know how many different employee names she puts on those lists, we can't really tell how many different subordinates she has.

I know I made a mistake somewhere, but I don't know where or why my approach/ logic is wrong. Could you please enlighten me? :)


The lists should contain at least 2 subordinates, not exactly 2.
_________________

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

Intern
Intern
avatar
Joined: 20 Oct 2012
Posts: 7
Location: United States
Twilight: Sparkle
GMAT 1: 370 Q24 V17
Followers: 0

Kudos [?]: 5 [0], given: 124

Re: How many subordinates does Marcia have? [#permalink] New post 30 Apr 2014, 06:55
Hi Guys, :)
plz , I do not understand any of the answers , can anyone give a more simple explanation . Thank You in advance.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23494
Followers: 3505

Kudos [?]: 26532 [0], given: 2712

Re: How many subordinates does Marcia have? [#permalink] New post 30 Apr 2014, 06:57
Expert's post
gmatlover23 wrote:
Hi Guys, :)
plz , I do not understand any of the answers , can anyone give a more simple explanation . Thank You in advance.


It's a tough problem. Not every question has a silver bullet 20 sec solution.
_________________

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
avatar
Joined: 20 Oct 2013
Posts: 66
Followers: 0

Kudos [?]: 0 [0], given: 27

Re: perms [#permalink] New post 12 May 2014, 04:52
Bunuel wrote:
For (2) we have \{s_1,s_2,s_3,...s_n\}. Each subordinate (s_1,s_2,s_3,...s_n) has TWO options: either to be included in the list or not. Hence total # of lists - 2^n, correct. But this number will include n lists with 1 subordinate as well 1 empty list.

As the lists should contain at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.

Lists with 1 subordinate - n: \{s_1,0,0,0...0\}, \{0,s_2,0,0,...0\}, \{0,0,s_3,0,...0\}, ... \{0,0,0...s_n\}.
List with 0 subordinate - 1: \{0,0,0,...0\}

So we'll get 200<2^n-n-1<500, --> n=8. Sufficient.

For (1):
C^2_n=28 --> \frac{n(n-1)}{2!}=28 --> n(n-1)=56 --> n=8. Sufficient.

Answer: D.


Dear Bunnel

How can you even understand what the qs is saying... i couldnt understand what statement 1 was saying? what list are they talking about? :(
_________________

Hope to clear it this time!!
GMAT 1: 540
Preparing again :(

Senior Manager
Senior Manager
avatar
Joined: 28 Apr 2014
Posts: 291
Followers: 0

Kudos [?]: 23 [0], given: 46

GMAT ToolKit User
How many subordinates does Marcia have? [#permalink] New post 06 Jul 2014, 22:49
BarneyStinson wrote:
This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.


Just out of interest , how will the above understanding change if we have say 'a' open slots and 'b' candidates ( b>a) ?. Will it be bCa (2^a) ?

Bunuel any pointers for more information on this concept will be much appreciated
Intern
Intern
avatar
Status: Preparing for GMAT
Joined: 10 Dec 2013
Posts: 25
Location: India
Concentration: Marketing, Leadership
GMAT 1: 530 Q46 V18
WE: Other (Entertainment and Sports)
Followers: 0

Kudos [?]: 5 [0], given: 61

Re: How many subordinates does Marcia have? [#permalink] New post 26 Jul 2014, 12:14
From 1, how did we get n(n-1)/2! ?? Shouldn't it be n!/(n-2)!*2 ?? Can someone please explain!
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23494
Followers: 3505

Kudos [?]: 26532 [0], given: 2712

Re: How many subordinates does Marcia have? [#permalink] New post 26 Jul 2014, 12:19
Expert's post
suhaschan wrote:
From 1, how did we get n(n-1)/2! ?? Shouldn't it be n!/(n-2)!*2 ?? Can someone please explain!


C^2_n;

\frac{n!}{(n-2)!*2!};

\frac{(n-2)!*(n-1)*n}{(n-2)!*2!};

\frac{n(n-1)}{2!}.

Does this make sense?
_________________

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 subordinates does Marcia have?   [#permalink] 26 Jul 2014, 12:19
    Similar topics Author Replies Last post
Similar
Topics:
How many prime factors does N have? shivanigs 7 06 Oct 2012, 00:31
ps:How many roots does equation have? pmal04 3 21 Jun 2009, 13:33
How many faces does the solid have? sunilgupta 8 27 Jul 2007, 04:26
How many factors does 36^2 have GMATT73 12 18 Dec 2005, 02:06
5 How many terminating zeroes does 200! have? TS 5 15 Apr 2005, 05:42
Display posts from previous: Sort by

How many subordinates does Marcia have?

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   [ 30 posts ] 



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®.