Combination formula | Probability and combinatorics | Probability and Statistics | Khan Academy

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[JoelChristophel]

8:26 "Memorizing is a good way to not understand what is going on." Nailed it.

[NestyPeach]

I got it! Thank you (I didnt even cry this time!!!)

[qweify]

And as Moses spoke these words at 10:30 the lord opened up a path through the fog in my brain with a strong east wind. The wind blew all that night, leaving only clarity.

[LisaHall]

I'm currently studying for my GED and it feels hopeless. But people like you are helping me to understand this stuff and gives me hope that I WILL pass that test. THANK YOU so freaking much!! :D

[skilz]

I haven't had a probability or statistics course in almost 25 years. This was a great refresher. When it comes to mathematics I don't forget too much, but this was one of those things that would allude me since I don't use it all too often. It's good to be reminded at times and this was a very simple and elegant explanation for basic Combinatorics. It also shows the power of using factorials along with its relationship to the binomial coefficient.

[yakunsharma]

The day I get a job - my first donation will be to Khan Academy. I cannot appreciate you enough!

[andrewsmith]

Excellent explanations. I have a degree in economics and the highest UK high school maths qualifications, but I never really "got" combinations and permutations until now. I'd just learnt the formulas by rote, but now I really understand. Onwards to a MSc in Statistics in a year!

[monoinluv]

I have an assignment over this topic and you're making it much easier to understand than lecture and discussion!

[Cheva-aree]

OMG, i understand it already... This is much better than listening to my teacher's nonsense talking... By the way, im gonna make this into a report//' "D

[leahjaelgreen]

I absolutely love your video, I have been struggling with this. You actually made it make sense to me. i’ve been taking an online class, so I’m teaching it to myself. Please don’t quit making videos I’m coming back to you! Thank you 🙏

[kyandrew]

I have an exam in 10 minutes and I understood a lot more from this compared to reading the textbook

[caht]

guy has a soothing voice, we need all math teachers to be like him

[Snoo]

I actually understood this, I am not studying probability or statistics I was actually searching for some other algebra stuff but I found the concept interesting and now I know how to calculate combinations which is cool

[juanmoralesdracus]

Let's hope this complements your work (amazing, by the way). This helped me join both ideas:
- Permutations: if we have 6 elements (A, B, C, D, E, F), picked in groups of 4, we can get 6*5*4*3 = 360 different groups of them (permutations formula, already demonstrated by you), which look like ABCD, ABCE, BCFE, ...
- Combinations: from those 360 permutations, how many of them are included in the same combination? In combinatorial world order of elements doesn't matter, so ABCD = ACBD = ADCB = ... So, in other words, if we now pick 4 elements (in this case, A, B, C and D, because they form ABCD, which is the group we are studying as an example), in groups of 4, how many permutations of them can we get?
Since this is a permutation and we know the formula (4 elements in groups of 4), we know we get 4*3*2*1 = 4! = 24 permutations.

This means ABCD = ACBD = ... for a total of 24 permutations which are inside the same combination (A, B, C and D in whatever order). So, if 24 permutations = 1 combination, we see that nº permutations = nº combinations * 24, then nº combinations = nº permutations/24 = 360/24 = 15 combinations. This, of course, also works for another 4 letters (BCDE = BDCE = ... 24 permutations for 1 combination of letters B, C, D and E, and so on).

Recalling: if we have 360 permutations of our elements and we can make groups of 24 of them which belong to the same combination, then we only have 360/24 = 15 combinations

Generalizing:
For n elements, taken in groups of k elements, there are n!/(n-k)! permutations.
Now we take k elements (which make a group in the previous step), and take them in groups of k elements, to have k!/(k-k)!= k!/0! = k! permutations that belong to the same combination.
Finally, we divide n!/(n-k)! permutations by k! of them which are grouped in each combination and have n!/[k!(n-k)!) combinations of n elements taken in groups of k elements.

I hope this serves you all, for me it has clarified everything finally. Thanks!

[chengli]

After months of struggling, this is the clearest and easiest explanation of permutations and combinations I've come across. Thank you! You are truly one of the greats of our time

[zhaofengzheng]

Why am I paying thousands of dollars tuition fee?

[devkathuria]

if u didn't get it, then start listening from 10:30 again and think Peace.

[tuomokriikkula]

Khan Academy is smart for hiring fluent English speakers. This is easy to follow all across the globe.

[oddyellow]

I was really stressed because of permutation and combination. Never would've thought that it's not that hard to understand. Thank you for giving us a great explaination.

[sevenaries]

8:35 School System: *shocked Pikachu face*