Precalculus (2018 edition)
Sal explains the combination formula.
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- 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?(121 votes)
- 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.(310 votes)
- 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.(23 votes)
- 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(19 votes)
- At7:08, 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?(26 votes)
- (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).(22 votes)
- 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!(34 votes)
- That is correct! A true combination lock would be at most 𝑛! times weaker than a permutation lock!(12 votes)
- I don't really get how the combination formula works.(7 votes)
- 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.(65 votes)
- 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.(3 votes)
- 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.(40 votes)
- 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.(18 votes)
- That's very helpful. Many thanks as I was struggling with it too.(1 vote)
- At4:10, why is ABCD and DABC count as the same combination?(2 votes)
- 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?(2 votes)
- 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.(9 votes)
- I'm still confused why we multiply k! with (n-k)! I'm confused about how to distinguish permutations in word problems. WHat are some key words to look for?(2 votes)
- The first time you're exposed to permutations and combinations it takes a little bit to get your brain around it, so I think it never hurts to do as many examples. But each incremental example I'm gonna review what we've done before, but hopefully go a little bit further. So let's just take another example. This is in the same vein. In videos after this I'll start using other examples, other than just people sitting in chairs, but let's just stick with it for now. Let's say we have six people again. Person A, B, C, D, E, and F, so we have six people. Now let's put them into four chairs. We can go through this fairly quickly, one, two, three, four chairs. We've seen this show multiple times. How many ways, how many permutations are there of putting these six people into four chairs? Well the first chair, if we seat them in order, we might as well, we can say well there'd be six possibilities here. For each of those six possibilities there would be five possibilities of who we'd put here, because one person's already sitting down. For each of these 30 possibilities of seating these first two people, there'd be four possibilities of who we put in chair number three. Then for each of these, what is this, 120 possibilities there would be three possibilities of who we put in chair four. This six times five times four times three is the number of permutations. We've seen in one of the early videos on permutations that, when we talk about the permutation formula, one way to write this, if we wanted to write in terms of factorial, we could write this as six factorial, six factorial which is going to be equal to six times five times four times three times two times one, but we wanna get rid of the two times one. We're gonna divide that, we're gonna divide that. Now what's two times one? Two times one is two factorial. Where do we get that? We wanted the first four, the first four factors of six factorial. That's where the four came from. We wanted the first four factors, so the way we got two is we said six minus four. Six minus four, that's gonna give us the number that we wanna get rid of. We wanted to get rid of two or the factors we wanna get rid of. That's going to give us two factorial. If we do six minus four factorial. That's going to give us two factorial, which is two times one. Then these cancel out, and we are all set. This is one way, I put in the particular numbers here, but this is a review of the permutations formula, where people say, "Hey, if I'm saying n, "if I'm taking n things, then I want to figure out "how many permutations are there "of putting them into let's say k spots, "it's going to be equal to n factorial "over n minus k factorial." That's exactly what we did over here, where six is n and k, or four is k. Four is k. Actually let me color code the whole thing so that we see the parallel. All of that is review. Then we went into the world of combinations. In the world of combinations, we said permutations make a difference between who's sitting in what chair. For example, in the permutations world, and this is all review, we've covered this in the combinations video, in the permutations world A, B, C, D, and D, A, B, C, these would be two different permutations that's being counted in whatever number this is. This is what? This is 30 times twelve. This is equal to 360. Each of these, this is one permutation, this is another permutation, and if we keep doing it we would count up to 360. But we learned in combinations, when we're thinking about combinations, let me write combinations. If we're saying n choose, n choose k, or how many combinations are there? If we take k things, and we just wanna figure out how many combinatio- If we start with n, if we have pool of n things, and we wanna say how many combinations of k things are there, then we would count these as the same combination. What we really wanna do is we wanna take the number of permutations there are. We wanna take the number of permutations there are, which is equal to n factorial over n minus k factorial, over n minus k factorial, and we wanna divide by the number of ways that you can arrange four people. Once again, this takes- I remember the first time I learned it took my brain a little while. If it's taking you a little while to think about it, not a big deal. It can be confusing at first, but hopefully if you keep thinking about it hopefully you will see clarity at some moment. What we wanna do is we wanna divide by all of the ways that you can arrange four things. Cuz once again, in the permutations it's counting all of the different arrangements of four things, but we don't wanna count all of those different arrangements of four things. We wanna just say they're all one combination. We wanna divide by the number of ways to arrange four things. Or the number of ways to arrange k things. Let me write this down. What is the number of ways, number of ways, to arrange k things, k things, in k spots. I encourage you to pause the video, because this actually a review from the first permutation video. If you have k spots, let me do it so if this is the first spot, the second spot, third spot, and then you're gonna go all the way to the kth spot. For the first spot there could be k possibilities. There's k things that could take the first spot. For each of those k possibilities, how many things can be in the second spot? It's gonna be k minus one, because you already put, you already put something in the first spot. Then over here, what's it gonna be? K minus two all the way to the last spot, there's only one thing that can be put in the last spot. What is this thing here? K times k minus one times k minus two times k minus three all the way down to one. This is just equal to k factorial. The number of ways to arrange k things in k spots, k factorial. The number of ways to arrange four things in four spots, that's four factorial. The number of ways to arrange three things in three spots, it's three factorial. We could just divide this. We could just divide this by k factorial. This would get us, this would get us, n factorial divided by k factorial, k factorial times, times n minus k factorial, n minus k, n minus k, I'll put the factorial right over there. This right over here is the formula. This right over here is the formula for combinations. Sometimes this is also called the binomial coefficient. People always call this n choose k. They'll also write it like this, n choose k, especially when they're thinking in terms of binomial coefficients. I got into kind of an abstract tangent here, when I started getting into the general formula. Let's go back to our example. In our example we saw that there was 360 ways of seating six people into four chairs. What if we didn't care about who's sitting in which chairs and just wanna say, "How many ways are there "to choose four people "from a pool of six?" That would be that would be how many ways are there. That would be six. How many combinations if I'm starting with a pool of six, how many combinations are there? How many combinations are there for selecting four? Another way of thinking about it is how many ways are there to, from a pool of six items, people in this example, how many ways are there to choose four of them. That is going to be, we could do it- I'll apply the formula first, and then I'll reason through it. And like I always say, I'm not a huge fan of the formula. Every time I revisit it after a few years, actually just rethink about it, as opposed to memorizing it, because memorizing is a good way to not understand what's actually going on, but if we just apply the formula here, I really want you to understand what's happening in the formula, it would be six factorial over four factorial, over four factorial, times six minus four factorial, six, oops let me just, This is six minus four factorial, so this part right over here, six minus four fa- Let me write it out because I know this can be a little bit confusing the first time you see it. So six minus four factorial, factorial, which is equal to, which is equal to six factorial over four factorial, over four factorial, times this thing right over here is two factorial, times two factorial, which is going to be equal to, we can just write out the factorials, six times five times four times three times two times one, over four. Four time three times two times one times, times two times one. Of course that's going to cancel with that. Then the one doesn't really change the value, so let me get rid of this one here. Let's see, this three can cancel with this three. This four could cancel with this four. Then it's six divided by two is going to be three. So we are just left with three times five. We are left with, we are left with, there's fifteen combinations. There's 360 permutations for putting six people into four chairs, but there's only 15 combinations, because we're no longer counting all of the different arrangements for the same four people in the four chairs. We're saying, "Hey if it's the same four people, "that is now one combination." You can see how many ways are there to arrange four people into four chairs? That's the four factorial part right over here. The four factorial part right over here, which is four times three times two times one, which is 24. We essentially just took the 360 divided by 24 to get 15. Once again, I don't think I can stress this enough. I wanna make it clear where this is coming from. This right over here, let me circle, this piece right over here is the number of permutations. This is really just so you can get to six times five times four times three, which is exactly what we did up here, where we reasoned through it. Then we just wanna divide by the number of ways you can arrange four items in four spaces.