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Re: Math: Combinatorics
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04 Dec 2011, 07:17
.. .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
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14 Mar 2012, 16:34
Thank you for this post! Very helpful!



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Re: Math: Combinatorics
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11 Apr 2012, 23:29
walker wrote: The topic is not finished.COMBINATORICScreated by: walkeredited by: bb, Bunuel This post is a part of [ GMAT MATH BOOK]  Circular arrangementsLet'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: \(R = \frac{n!}{n} = (n1)!\)Tips and TricksAny 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. ResourcesWalker's post with Combinatorics/probability problems: [ Combinatorics/probability Problems] 
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



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Re: Math: Combinatorics
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04 Dec 2012, 20:26
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_1permutations: http://www.khanacademy.org/math/probabi ... utations_1These should be required watching, MGMAT does a thorough job of confusing my ass in this area.



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Re: Math: Combinatorics
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17 Mar 2013, 00:07
Great post., Many thanks!!



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Re: Math: Combinatorics
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Re: Math: Combinatorics
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07 Sep 2013, 20:03
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
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08 Sep 2013, 19:39
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
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Re: Math: Combinatorics
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06 Apr 2014, 13:42
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 (nk) objects ((nk)!) as the order of chosen k objects and remained (nk) objects doesn't matter.
\(C^n_k = \frac{n!}{k!(nk)!}\)
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 (nk) objects ((nk)!) as the order of remained (nk) objects doesn't matter.
\(P^n_k = \frac{n!}{(nk)!}\)
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
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06 Apr 2014, 13:56
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 (nk) objects ((nk)!) as the order of chosen k objects and remained (nk) objects doesn't matter.
\(C^n_k = \frac{n!}{k!(nk)!}\)
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 (nk) objects ((nk)!) as the order of remained (nk) objects doesn't matter.
\(P^n_k = \frac{n!}{(nk)!}\)
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: dsquestiondirectorybytopicdifficulty128728.html and here: gmatpsquestiondirectorybytopicdifficulty127957.htmlHope it helps.
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Re: Math: Combinatorics
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06 Apr 2014, 14:08
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 (nk) objects ((nk)!) as the order of chosen k objects and remained (nk) objects doesn't matter.
\(C^n_k = \frac{n!}{k!(nk)!}\)
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 (nk) objects ((nk)!) as the order of remained (nk) objects doesn't matter.
\(P^n_k = \frac{n!}{(nk)!}\)
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: dsquestiondirectorybytopicdifficulty128728.html and here: gmatpsquestiondirectorybytopicdifficulty127957.htmlHope it helps. Thanks for the tips. I'll work on doing the easier questions. Cheers.



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Re: Math: Combinatorics
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07 Aug 2015, 23:27
can u please share something like this on "Probability" as well?????



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Re: Math: Combinatorics
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16 Aug 2015, 10:09



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Re: Math: Combinatorics
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27 Oct 2015, 13:04
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 ?



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Re: Math: Combinatorics
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05 Jan 2017, 18:44
Circular arrangements having hard time understanding how did you arrive at the formula of dividing n! by n. Can someone please elaborate?
Thanks Kunal



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Re: Math: Combinatorics
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06 Jan 2017, 02:40



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Re: Math: Combinatorics
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06 Jan 2017, 17:45
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 = (n1)!



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Re: Math: Combinatorics
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16 May 2017, 14:17
In a class of 10 students, in how many ways can you form a academic committee, if the committee must contain a minimum of 2 students ?[/color][/i]
Total number of ways to form committees (case 1) = 2^10 Number of ways to form of size 0 = 1 Number of ways to form of size 1 = 10 (each student picked one at a time) So no of ways = 1024 1 10 = 1013[/quote]
can we solve this problem with formula C?



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Re: Math: Combinatorics
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02 Aug 2018, 14:04
Bunuel wrote: 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 \((n1)!\). 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: \(\frac{n!}{n} = (n1)!\). Check other Arrangements in a Row and around a Table questions in our Special Questions Directory. Hi Bunuel Could you clarify, if we need to know for the GMAT what would be: The number of circular permutations of the elements in a MULTISET (a set with n indistinguishable objects, ie. {a,a,a,b,c,d}) And if so, how should we approach such problems, is there a formula? Similar to the one below: The number of CIRCULAR permutations of the elements in a SET of DISTINCT elements is: \(\frac{n!}{(nr)!r}\) Thanks in advance.




Re: Math: Combinatorics &nbs
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