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

It is currently 14 Feb 2016, 05:29
GMAT Club Tests

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

Math: Combinatorics

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3573
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 457

Kudos [?]: 2614 [0], given: 359

GMAT ToolKit User Top 10 in overall
Re: Math: Combinatorics [#permalink] New post 04 Dec 2011, 06:17
Expert's post
...k objects taken from a set of n distinct objects....

so k is always less or equal N by definition.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Manager
avatar
Joined: 02 Feb 2012
Posts: 195
Location: United States
Schools: HBS '19
GPA: 3.08
Followers: 1

Kudos [?]: 4 [0], given: 104

CAT Tests
Re: Math: Combinatorics [#permalink] New post 14 Mar 2012, 15:34
Thank you for this post! Very helpful!
Manager
Manager
avatar
Joined: 30 May 2008
Posts: 76
Followers: 0

Kudos [?]: 48 [0], given: 26

Re: Math: Combinatorics [#permalink] New post 11 Apr 2012, 22:29
walker wrote:
The topic is not finished.

COMBINATORICS

created by: walker
edited by: bb, Bunuel

--------------------------------------------------------
Image


This post is a part of [GMAT MATH BOOK]



--------------------------------------------------------

Circular arrangements

Let's say we have 6 distinct objects, how many relatively different arrangements do we have if those objects should be placed in a circle.

Image

The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
\(R = \frac{n!}{n} = (n-1)!\)

Tips and Tricks

Any problem in Combinatorics is a counting problem. Therefore, a key to solution is a way how to count the number of arrangements. It sounds obvious but a lot of people begin approaching to a problem with thoughts like "Should I apply C- or P- formula here?". Don't fall in this trap: define how you are going to count arrangements first, realize that your way is right and you don't miss something important, and only then use C- or P- formula if you need them.


Resources

Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]

--------------------------------------------------------

[Reveal] Spoiler: Images


Attachment:
Math_Combinatorics_6balls.png
Attachment:
Math_Combinatorics_6balls_b.png
Attachment:
Math_Combinatorics_6balls_l.png
Attachment:
Math_Comb_Round_t.png



having a hard time understand the formula under circular arrangement :(
Intern
Intern
avatar
Joined: 25 Jun 2012
Posts: 36
Followers: 0

Kudos [?]: 27 [0], given: 4

Re: Math: Combinatorics [#permalink] New post 04 Dec 2012, 19:26
2
This post was
BOOKMARKED
just wanted to throw these videos from khan academy out there, super helpful in dumbing down the concepts and holding your hand through it

combinations:

http://www.khanacademy.org/math/probabi ... inations_1

permutations:

http://www.khanacademy.org/math/probabi ... utations_1

These should be required watching, MGMAT does a thorough job of confusing my ass in this area.
Intern
Intern
User avatar
Joined: 01 Sep 2012
Posts: 12
Location: United States
Followers: 0

Kudos [?]: 5 [0], given: 19

Re: Math: Combinatorics [#permalink] New post 16 Mar 2013, 23:07
Great post., Many thanks!!
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 31356
Followers: 5367

Kudos [?]: 62588 [0], given: 9457

Re: Math: Combinatorics [#permalink] New post 10 Jul 2013, 23:07
Expert's post
Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 34 [0], given: 23

Re: Math: Combinatorics [#permalink] New post 07 Sep 2013, 19:03
Hi - I'm a little confused by the last part of the explanation above:

"here is a basket with 4 blue balls, 3 red balls and 2 yellow balls. The balls of the same color are identical. How many ways can a child pick balls out of this basket ?

Solution
We combine techniques from the 2 cases above. Consider the blue balls as one entity and instead of asking the yes/no question, ask the question how many ways to choose the balls ? This is answered using the second case, and the answer is (4+1). Combining this with the ways to choose red and yellow balls, the overall ways to choose balls would be (4+1)*(3+1)*(2+1)."

What is the + 1 for and how can we just multiply the 3 sets - 4*3*2? Can't the child pick all 4 blue and no red etc?
Expert Post
MBA Section Director
User avatar
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 2385
Location: India
City: Pune
GPA: 3.4
WE: Business Development (Manufacturing)
Followers: 290

Kudos [?]: 2045 [0], given: 1630

GMAT ToolKit User Top 10 in overall
Re: Math: Combinatorics [#permalink] New post 08 Sep 2013, 18:39
Expert's post
russ9 wrote:
What is the + 1 for and how can we just multiply the 3 sets - 4*3*2? Can't the child pick all 4 blue and no red etc?


+1 is for NOT choosing any of the n identical balls.

Hope that helps :)
Attachments

Untitled123.png
Untitled123.png [ 10.2 KiB | Viewed 3294 times ]


_________________

MBA Admission Help Desk

Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 34 [0], given: 23

Re: Math: Combinatorics [#permalink] New post 06 Apr 2014, 12:42
walker wrote:

Combination

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(C^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects ((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter.

\(C^n_k = \frac{n!}{k!(n-k)!}\)



Permutation

A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(P^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the order of remained (n-k) objects doesn't matter.

\(P^n_k = \frac{n!}{(n-k)!}\)

If we exclude order of chosen objects from permutation formula, we will get combination formula:

\(\frac{P^n_k}{k!} = C^n_k\)



Hello friends - this page seems to be the best find yet. I've been using Manhattan gmat books and their content on Combinatorics and Probability is pretty lacking so please excuse my rudimentary questions.

One topic that i'm having a very hard time grasping is the "ordered vs. unordered" set. Seems like Permutation is ordered and Combination is unordered. I have absolutely no clue what that means and more importantly, I can't connect the dots to figure out why it affects the equations the way it does?

Any help would be appreciated.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 31356
Followers: 5367

Kudos [?]: 62588 [0], given: 9457

Re: Math: Combinatorics [#permalink] New post 06 Apr 2014, 12:56
Expert's post
1
This post was
BOOKMARKED
russ9 wrote:
walker wrote:

Combination

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(C^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects ((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter.

\(C^n_k = \frac{n!}{k!(n-k)!}\)



Permutation

A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(P^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the order of remained (n-k) objects doesn't matter.

\(P^n_k = \frac{n!}{(n-k)!}\)

If we exclude order of chosen objects from permutation formula, we will get combination formula:

\(\frac{P^n_k}{k!} = C^n_k\)



Hello friends - this page seems to be the best find yet. I've been using Manhattan gmat books and their content on Combinatorics and Probability is pretty lacking so please excuse my rudimentary questions.

One topic that i'm having a very hard time grasping is the "ordered vs. unordered" set. Seems like Permutation is ordered and Combination is unordered. I have absolutely no clue what that means and more importantly, I can't connect the dots to figure out why it affects the equations the way it does?

Any help would be appreciated.


In how many way we can choose two letters out of {a, b, c}, of the order of the letters does not matter?

{a, b}
{a, c}
{b, c}

So, in 3 ways: \(C^2_3=3\).

In how many way we can choose two letters out of {a, b, c}, of the order of the letters matters?

{a, b}
{b, a}
{a, c}
{c, a}
{b, c}
{c, b}

So, in 6 ways: \(P^2_3=6\).

I'd advice to go through easier questions on the topic to understand better. Check here: ds-question-directory-by-topic-difficulty-128728.html and here: gmat-ps-question-directory-by-topic-difficulty-127957.html

Hope it helps.
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 34 [0], given: 23

Re: Math: Combinatorics [#permalink] New post 06 Apr 2014, 13:08
Bunuel wrote:
russ9 wrote:
walker wrote:

Combination

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(C^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects ((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter.

\(C^n_k = \frac{n!}{k!(n-k)!}\)



Permutation

A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

\(P^n_k\)

knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination:

1. The total number of arrangements of n distinct objects is n!
2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the order of remained (n-k) objects doesn't matter.

\(P^n_k = \frac{n!}{(n-k)!}\)

If we exclude order of chosen objects from permutation formula, we will get combination formula:

\(\frac{P^n_k}{k!} = C^n_k\)



Hello friends - this page seems to be the best find yet. I've been using Manhattan gmat books and their content on Combinatorics and Probability is pretty lacking so please excuse my rudimentary questions.

One topic that i'm having a very hard time grasping is the "ordered vs. unordered" set. Seems like Permutation is ordered and Combination is unordered. I have absolutely no clue what that means and more importantly, I can't connect the dots to figure out why it affects the equations the way it does?

Any help would be appreciated.


In how many way we can choose two letters out of {a, b, c}, of the order of the letters does not matter?

{a, b}
{a, c}
{b, c}

So, in 3 ways: \(C^2_3=3\).

In how many way we can choose two letters out of {a, b, c}, of the order of the letters matters?

{a, b}
{b, a}
{a, c}
{c, a}
{b, c}
{c, b}

So, in 6 ways: \(P^2_3=6\).

I'd advice to go through easier questions on the topic to understand better. Check here: ds-question-directory-by-topic-difficulty-128728.html and here: gmat-ps-question-directory-by-topic-difficulty-127957.html

Hope it helps.


Thanks for the tips. I'll work on doing the easier questions. Cheers.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 8242
Followers: 420

Kudos [?]: 111 [0], given: 0

Top 10 in overall
Re: Math: Combinatorics [#permalink] New post 20 Apr 2015, 11:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Manager
Manager
avatar
Joined: 25 Nov 2014
Posts: 118
WE: Engineering (Manufacturing)
Followers: 1

Kudos [?]: 20 [0], given: 42

GMAT ToolKit User Reviews Badge
Re: Math: Combinatorics [#permalink] New post 07 Aug 2015, 22:27
can u please share something like this on "Probability" as well?????
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 31356
Followers: 5367

Kudos [?]: 62588 [0], given: 9457

Re: Math: Combinatorics [#permalink] New post 16 Aug 2015, 09:09
Expert's post
Manager
Manager
User avatar
Joined: 18 Aug 2014
Posts: 133
Location: Hong Kong
Followers: 1

Kudos [?]: 58 [0], given: 36

Re: Math: Combinatorics [#permalink] New post 27 Oct 2015, 12:04
One question..when we say combinations are unordered.. and take the example of lottery..6 numbers chosen from 49.

We have 49! / 6! 43!

But I would say that the numbers are ordered..or order matters.. it is different to have 123456 vs 213456

Are there some examples for combinations vs permutations ?
Re: Math: Combinatorics   [#permalink] 27 Oct 2015, 12:04

Go to page   Previous    1   2   [ 35 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
1 Combinatorics: Anagram ahmed1313 1 23 Jul 2015, 02:20
4 Combinatorics saif123 1 19 Aug 2013, 12:47
1 Experts publish their posts in the topic PS Combinatorics Cityboy 5 24 Apr 2013, 13:11
7 Experts publish their posts in the topic Combinatorics Summary fxsunny 7 29 Jan 2012, 22:28
Probability and Combinatorics gmatbull 3 20 Apr 2010, 09:02
Display posts from previous: Sort by

Math: Combinatorics

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.