Math: Combinatorics : GMAT Quantitative Section - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 17:59

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Math: Combinatorics

Author Message
TAGS:

### Hide Tags

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 545

Kudos [?]: 3554 [0], given: 360

### Show Tags

04 Dec 2011, 06:17
...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
Joined: 02 Feb 2012
Posts: 204
Location: United States
GMAT 1: 770 Q49 V47
GPA: 3.08
Followers: 1

Kudos [?]: 6 [0], given: 104

### Show Tags

14 Mar 2012, 15:34
Thank you for this post! Very helpful!
Manager
Joined: 30 May 2008
Posts: 76
Followers: 1

Kudos [?]: 93 [0], given: 26

### Show Tags

11 Apr 2012, 22:29
1
This post was
BOOKMARKED
walker wrote:
The topic is not finished.

COMBINATORICS

created by: walker
edited by: bb, Bunuel

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

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.

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
Joined: 25 Jun 2012
Posts: 36
Followers: 0

Kudos [?]: 29 [0], given: 4

### Show Tags

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:

permutations:

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

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

### Show Tags

16 Mar 2013, 23:07
Great post., Many thanks!!
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93038 [0], given: 10541

### Show Tags

10 Jul 2013, 23:07
Bumping for review*.

*New project from GMAT Club!!! Check HERE

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

Kudos [?]: 53 [0], given: 23

### Show Tags

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?
MBA Section Director
Status: On vacation...
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 3937
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Followers: 391

Kudos [?]: 2873 [0], given: 2159

### Show Tags

08 Sep 2013, 18:39
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 [ 10.2 KiB | Viewed 4089 times ]

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

Kudos [?]: 53 [0], given: 23

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93038 [0], given: 10541

### Show Tags

06 Apr 2014, 12:56
Expert's post
2
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.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 53 [0], given: 23

### Show Tags

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
Joined: 09 Sep 2013
Posts: 13432
Followers: 575

Kudos [?]: 163 [0], given: 0

### Show Tags

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.
_________________
Manager
Joined: 25 Nov 2014
Posts: 140
WE: Engineering (Manufacturing)
Followers: 1

Kudos [?]: 25 [0], given: 69

### Show Tags

07 Aug 2015, 22:27
can u please share something like this on "Probability" as well?????
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93038 [0], given: 10541

### Show Tags

16 Aug 2015, 09:09
Expert's post
1
This post was
BOOKMARKED
AA2014 wrote:
can u please share something like this on "Probability" as well?????

Check here: math-probability-87244.html
_________________
Manager
Joined: 18 Aug 2014
Posts: 132
Location: Hong Kong
Schools: Mannheim
Followers: 1

Kudos [?]: 66 [0], given: 36

### Show Tags

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 ?
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13432
Followers: 575

Kudos [?]: 163 [0], given: 0

### Show Tags

28 Oct 2016, 08:32
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.
_________________
Intern
Joined: 05 Jan 2017
Posts: 3
Followers: 0

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

### Show Tags

05 Jan 2017, 17:44
Circular arrangements having hard time understanding how did you arrive at the formula of dividing n! by n. Can someone please elaborate?

Thanks
Kunal
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93038 [1] , given: 10541

### Show Tags

06 Jan 2017, 01:40
1
KUDOS
Expert's post
kuvshah wrote:
Circular arrangements having hard time understanding how did you arrive at the formula of dividing n! by n. Can someone please elaborate?

Thanks
Kunal

The number of arrangements of n distinct objects in a row is given by $$n!$$.
The number of arrangements of n distinct objects in a circle is given by $$(n-1)!$$.

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:

$$\frac{n!}{n} = (n-1)!$$.

Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.
_________________
Intern
Joined: 05 Jan 2017
Posts: 3
Followers: 0

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

### Show Tags

06 Jan 2017, 16:45
Thanks for the reply. I got it now. I understood this from a youtube videos - the link I cannot post it yet as I am new to this forum. Since we have n! ways to arrange distinct objects in circle and there are n repeats (if we shift all object by one position, we will get because the same relative arrangement in a circle), the equation is n!/(repeat i.e. n). So n!/n = (n-1)!
Re: Math: Combinatorics   [#permalink] 06 Jan 2017, 16:45

Go to page   Previous    1   2   [ 39 posts ]

Similar topics Replies Last post
Similar
Topics:
1 Combinatorics: Anagram 1 23 Jul 2015, 02:20
4 Combinatorics 1 19 Aug 2013, 12:47
1 PS Combinatorics 5 24 Apr 2013, 13:11
7 Combinatorics Summary 7 29 Jan 2012, 22:28
Probability and Combinatorics 3 20 Apr 2010, 09:02
Display posts from previous: Sort by