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Math: Combinatorics

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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.
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Re: Math: Combinatorics [#permalink] New post 14 Mar 2012, 15:34
Thank you for this post! Very helpful!
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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]



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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]

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[Reveal] Spoiler: Images


Attachment:
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Attachment:
Math_Combinatorics_6balls_b.png
Attachment:
Math_Combinatorics_6balls_l.png
Attachment:
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having a hard time understand the formula under circular arrangement :(
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Re: Math: Combinatorics [#permalink] New post 04 Dec 2012, 19:26
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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.
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Re: Math: Combinatorics [#permalink] New post 16 Mar 2013, 23:07
Great post., Many thanks!!
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Re: Math: Combinatorics [#permalink] New post 10 Jul 2013, 23:07
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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?
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Re: Math: Combinatorics [#permalink] New post 08 Sep 2013, 18:39
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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 :)
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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.
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Re: Math: Combinatorics [#permalink] New post 06 Apr 2014, 12:56
Expert's post
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.
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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.
Re: Math: Combinatorics   [#permalink] 06 Apr 2014, 13:08
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