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### Course: Precalculus>Unit 8

Lesson 4: Combinations

# Combination formula

Learn the difference between permutations and combinations, and how to calculate them using factorials. This video also discusses binomial coefficients and the formula for combinations, nCk. Solidify your understanding with an example: how would one seat six people in four chairs?

## Want to join the conversation?

• I really don't understand the difference between permutations and combinations? For permutations is it just "choosing" three people from six, and for combinations, just determining the number of "combinations" that they can be placed in to fill the chairs?
• If the order doesn't matter, it is a combination.
If the order does matter, it is a permutation.

A permutation is an ordered combination. In this case, it doesn't matter what order the people are placed in to fill the chairs, it just matters which people you chose.
• Does anyone have a good mnemonic for remembering the difference between permutations and combinations? I try to think of p for place, so in permutations, order matters. That's not very catchy. I'm sure someone has a better one.
• Combos are (C)areless about the order
Permutations take great (P)ains to keep track of the order

ps. i just came up with i so its not very catchy
• So, if you really think about it, a combination lock should actually be called a permutation lock, since in the case of the locks, order does matter!
• That is correct! A true combination lock would be at most 𝑛! times weaker than a permutation lock!
• I don't really get how the combination formula works.
• Ok, let's start by an example. If there are 3 chairs and 5 people, how many permutations are there? Well, for the first chair, 5 people can sit on it. On the second chair (5-1) people can sit on the chair. On the third chair (5-2) people can sit on the chair. So the total number of permutations of people that can sit on the chair is 5*(5-1)*(5-2)=5*4*3=60. We can make a general formula based on this logic. For n people sitting on k chairs, the number of possibilities is equal to n*(n-1)*(n-2)*...1 divided by the number of extra ways if we had enough people per chair. So the formula for the number of permutations is n!/((n-k)!.

The number of combinations is the number of ways to arrange the people on the chairs when the order does not matter. In our example, let the 5 people be A, B, C, D, and E. So some of the permutations would be ABC, ACB, BAC, BCA, CAB and CBA. If we didn't care about these specific orders and only cared that they were on the chairs then we could group these people as one combination. So the number of combinations is equal to the number of permutations divided by the size of the groups(which in this case is 6). The group size can be calculated by permuting over the number of chairs which is equal to the factorial of the number of chairs(k!). So the formula for calculating the number of combinations is the number of permutations/k!. the number of permutations is equal to n!/(n-k)! so the number of combinations is equal to (n!/(n-k)!)/k! which is the same thing as n!/(k!*(n-k)!).

So the number of combinations in our example is equal to 5!/(3!*(5-3)!) >=120/(6*2) => 120/12 => 10.
I hope I didn't confuse you.
• At , Sal says that n!/(n-k)!/k! equals n!/k!(n-k)!
Now I'm sure he's right but why is that?
I mean a/b/c would be (a*c)/b so why is this different?
• (a/b)/c = ((a/b)(1/c)) / (c * (1/c)) This is just multiplying the top and bottom by (1/c). The bottom then cancels out to 1 and the top is a/(b*c).
• The part that confused me about the combinations formula is what the multiplication of k! in the denominator is doing to the formula. It's deceiving because the k! is actually DIVIDING the entire permutation equation. This reminded me of the principle of Inclusion/Exclusion which is basically about subtracting over counted elements. In light of this, it becomes clear that the permutation formula is counting one combination k! times, k = the number of chairs or spots (4 in the video). That means ABCD is 1 COMBINATION but it has 4! PERMUTATIONS (ABCD, ADCB, DCBA...etc). But were not counting permutations only COMBINATIONS, thus all we want to count is the FIRST PERMUTATION of the four letters. Essentially, what the k! is doing by dividing the TOTAL PERMUTATIONS, is it is canceling all the additional permutations that were counted MORE THAN ONCE in the permutations formula. Leaving us with the FIRST PERMUTATION that was counted or ONE COMBINATION of those 4 letters. I hope this helped someone. I've been struggling for over a week with this.
• That's very helpful. Many thanks as I was struggling with it too.
• So if there are 6 people, but 0 chairs, I would compute 6 choose 0. That gives me 6!/0!(6-0!), which gives me 6!/1 x 6!. That is equal to 1. I don't understand how I can arrange 6 people into four chairs in ONE way. I would assume it would be ZERO, or undefined.
• Six people into 0 chairs means all 6 people are standing, which is the only possibility. So you have 1, not 0 or undefined possibility.
• At , why is ABCD and DABC count as the same combination?
• The difference between permutation and combination is that in permutation, order matters. In combination, order doesn't matter.
(1 vote)
• what would be the number of combinations for a 4 digit password if you take numbers from 0 to 9 and digits can be repeated?
• 10,000 combinations.
First method: If you count from 0001 to 9999, that's 9999 numbers. Then you add 0000, which makes it 10,000.
Second method: 4 digits means each digit can contain 0-9 (10 combinations). The first digit has 10 combinations, the second 10, the third 10, the fourth 10. So 10*10*10*10=10,000.
• Ok so let me get this straight.....

Let's say we have 6C4. The formula gives us:

6!/(4!(6-4)!)

Here 6! is the number of permutations of 6 objects taken 6 at a time, and we divide that by (6-4)!=2! for the number of permutations of 6 objects taken 4 at a time (2! eliminates the last two possibilities multiplied while taking 6 from 6).

We further divide the entire thing by 4! to eliminate the possibilities of the order of arranging the 4 objects taken from the original 6, thus giving us a combination.

Is that it?
• Yes, that is correct!

The formula for combinations, also known as binomial coefficients, is represented as nCr, where n is the total number of objects and r is the number of objects to be chosen. The formula for nCr is:

``nCr = n! / (r! * (n-r)!)``

In your example, you have 6 objects and you want to choose 4 of them. So, using the formula, you get:

``6C4 = 6! / (4! * (6-4)!)``

``= (654321) / ((4321) * (2*1))``

``= (65) / (21)``

``= 15``

So, the number of combinations of 6 objects taken 4 at a time is 15.
(1 vote)