Introduction to combinations | Probability and Statistics | Khan Academy

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[DakoGamerZ]

Imagine going to sleep every night, resting and knowing you have helped so many people out there to learn! Hats off to you man

[salimalajmi]

This was very helpful! I'm 34 and last year I decided to go back to university. this semester I'm taking statistics intro and I have never seen or heared of permutations and combinations. This video explained it all in less than 10 min. Thanks Khan (or whoever is talking in the video).

[paulybanez]

dude thank you so much, we don't have online classes and our school just sent us worksheets to do.Thanks a bunch man

[lynnkonyali]

thanks for not making me feel stupid anymore

[lcfsoft]

He's used all three by now "have sitted", "have sat" and "have sitten".

[noopurmehrotra]

In case someone is wondering "what possibly could those 20 combinations be?" well here they are
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF

[UltimateBargains]

So, is a combination lock misnamed? Is the true name of a combination lock a permutation lock?

[abiyermias]

Now, I understand the formula for combinations. Thank you!

[lokeshgoyal]

Those who have difficulty imagining this concept logically, Let me put in simple words !
First you calculate permutation for given question, as you know permutation generates all the ways we can arrange elements, including duplicates!
Now to calculate combinations, all we have to remove duplicates.

For question A, B, C, D, E, F and total spots available is 3, then permutation would be 120
And if we want to calculate combinations, all we have to remove the duplicates from our permuations.

So if we calculate, how many ways we can arrange 3 people at 3 spots, the answer would be 6. Meaning, that each 3 letters produces 6 different mutations, for which we should count 1 in case of calculating combinations. So if we divide total permutations by toatl dupllicates generated by each 3 letters at a time. The answer would be 20.


Conclusion : Permutation tells us all the ways including duplicates as well (according to combinations point of view)
So to calcuate combination all we have to remove is the duplicates generated by n spots permuation.

[haridaskr]

Well explained! Thanks! All this while I didn't know the concept of combinations

[mx]

omg tysm im in 7th and i have a quiz tmr first period and I have no idea what to do and this helped a lot so ty omg

[hugodivi]

Really well explained, it all came together at the end, THANK YOU!

[TasyaAdzkiya]

So, the reason why we divide the number of arrangements/permutations by the number of arrangements in a combination group is that because now that we don't care about the order, we consider the (in this case) 6 arrangements in a combination group of (in this case) 3 letters as one whole entity/one group that contain 3 letters in it because that's all that matters now that we don't care about the order.

[Snoo]

Wow, I now know how to calculate possible combinations of stuff

[jossmits]

super explanation. no formulas needed. back to basics. another hat off for you professor.

[mansimarkaur]

You explain the origin of the concepts so well which is something often neglected! Thank you!!

[johanliebert]

You are a true genius, I like the way you write, the way you speak and illustrate things in a simple way, thanks a lot !!

[SouL-obwx]

Very helpful. I've been racking my brain going through a lot of different lectures trying to understand the difference between the two but this video cleared it up. Thank you

[roseb]

I know this video is old but im doing somthing that may involve revisting this concept and i realize that I may not be 100 percent sure i understand it. Is another way to think of combinations as from all the possible ways to rearrange in this video the example is 3 from 5 how many sets can i make ( were each set will have different letters) but first one must find out the how many groups of 3 count as 1 set. ( which is the same as how many different groups of 3 one can make from 3 things) then by dividing the number of permutation by this number one can determine how many sets they can make?

[patrickstar]

How would you go about setting up a problem for seating these people (A - F) for more than 6 seats? Or is this addressed in a separate video?