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Re: Math: Combinatorics [#permalink]
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Kudos
Thank you for the posted information....much appreciated!!
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Re: Math: Combinatorics [#permalink]
Do it before this April bro!!!! Gotta brush my skills!!!!
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Re: Math: Combinatorics [#permalink]
I missed circular arrangement while brushing up my concepts. Thanks for the good post.
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Re: Math: Combinatorics [#permalink]
This is very helpful!
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Re: Math: Combinatorics [#permalink]
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Waiting for the updates..... You hit the nail on the head

"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
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Re: Math: Combinatorics [#permalink]
many many thanks for the whole math collection...waiting for update and new one...
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Re: Math: Combinatorics [#permalink]
Just read this post and the one about probability. A great help! Now I will try some application...
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Re: Math: Combinatorics [#permalink]
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Walker,
How come no updates on this chapter? If you're too busy to complete it, could you let me know what additional concepts to add? I could send a draft to you...?!
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Re: Math: Combinatorics [#permalink]
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gmatdelhi wrote:
Walker,
How come no updates on this chapter? If you're too busy to complete it, could you let me know what additional concepts to add? I could send a draft to you...?!


You can post here any concepts/examples you think aren't covered by my post. At least everyone who read this thread will see them. I will do my best to include them in the first post.
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Re: Math: Combinatorics [#permalink]
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Hi ,

I am just confused when to use permutation and when to use combination formula while solving problems. i mean what is the basic difference between Permutation and combination.
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Re: Math: Combinatorics [#permalink]
Good Info! Thank you for the posted!
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Re: Math: Combinatorics [#permalink]
Guys please help me,
How can I apply arrangements fromula for examples 2 and 3 in section enumeration?

if n=3 than N=could equal to 6 and 3 but not 4.
Please help.
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Re: Math: Combinatorics [#permalink]
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nice one shrouded.

I actually do the reverse. I use \(P = A^a * B^b * C^c\) for all the combination problems.

eg....for n identical, P is of the form = a*a* .........n times where a is prime number

so number of ways of selecting n identical -> n+1 using the formula of number of divisors.[ (a+1)*(b+1)*(c+1)]

for n non-identical P is of the form = a*b*c*...........n times

so number of ways of selecting n non-identical -> (1+1)*(1+1)...n times using the formula of number of divisors.

= 2^n

similarly if the number of identical balls are given then take \(P = A^a * B^b * C^c\) where a,b,and c are the number of identical balls of type A,B,and C respectively.

Then use the number of divisor formula = \((a+1)*(b+1)*(c+1)\)
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Re: Math: Combinatorics [#permalink]
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Kudos ... gr8 work... helped me alot in understanding the conce3pts of combinatorics.. as i m always confused with them... thnks for the notes...
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Re: Math: Combinatorics [#permalink]
Thanks walker for this post is really helpful
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Re: Math: Combinatorics [#permalink]
Great Post thanks!
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Re: Math: Combinatorics [#permalink]
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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:

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


Great Post.

I believe that the formulas can be used only when \(N\geq K\)
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