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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.

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?

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?

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 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?

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.
_________________

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.
_________________

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:

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)!

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?

gmatclubot

Re: Math: Combinatorics
[#permalink]
16 May 2017, 14:17

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