[Discrete Mathematics] Derangements

We talk about derangements.
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In this video we look at derangements. That is, the number of ways we can return items so that nobody gets their own item back.
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[@yaweli2968]

Wow, good job. I have struggling with combinatorics, cuz I missed tons of classes. With you, i will catch up. Thanks.

[@GerardFunClub]

I'm majoring in statistics and this video has helped me a lot, thank you, sir!

[@proggenius2024]

It's cool to learn from someone who knows what he's teaching inside out!

[@abdullah-ayy]

I don't get why would anyone dislike this video?
He's better than most of my proffs

[@Indhu9205]

Thank you so much for uploading all these awesome clearly explained videos!!!I'm preparing for an exam tmr and you have helped me more than my prof 😊😊😊😊😊

[@darianharrison4836]

Thank you very much for these coursers, they are very helpful

[@antmanistheman]

I'm a bit confused it seems that if you multiplied the total number of derangements by the total number of ways in which to arrange student their would be more way to derange them then arrange them I don't think youre wrong I just know theres something I don't understand. This question pertains to the problem about 17:00 minutes in concerning the desks.

[@yoshiroryuuma1319]

The formula is correct; however his explanation might not be very clear which leads you to confusion. The first 12! means assignment without doubt, but it means the assignment for the first class (either geometry or biology in this case); one instance included in 12!, let's say(suppose "1,2,3, ... , 12" means each student's name), is "1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12", based on which we can do derangement for the second class. Without the first 12!, we can't even take a referrence. Hope my poor English can enlighten you a little bit.

[@Ashish-zs5by]

@@yoshiroryuuma1319 thank you so much..
I was also confused in this..
Now it is clear

[@ArunKumar-ds6di]

@@yoshiroryuuma1319 Thank you very much

[@captainfoodman]

great video!!
but I didn't understand the Ms.pezzulo question. where the question states she teaches both classes. so in the first class the no of ways she can assign desks to 12 students so that they don't get their own is D12. but then for the second class, all these new arrangements must also be avoided. so during the second class it will be (12!-D12). then shouldn't we multiply D12(12!-D12) for the final answer.
and this question didn't tell us to do two different things such as "assigning to 12 students and then rearranging them" right?? it goes like 'how many ways can she assign it to them such that they don't get their own desks.'
And if I'm wrong, then why did they give the information that she teaches two classes one after the other.

[@hengyilam7586]

love your work！

[@Selim_Hasan_Raj]

these videos are awesome. thank you so much.

[@bradimi3776]

For dn = n!/e, is it true that if n is even, then you would always take the ceiling of n!/e, and if n is odd, then you would always take the floor of n!/e?

[@adityasinghverma4765]

Thanks a ton man!! :) You're awesome!

[@jafrancaibanez5041]

Good job, all I have to say!

[@rymlallaouh2574]

Thank you sir

[@davemartin157]

At time 4:19, you say S4 is uncounted due to multiplication by 0!. However, isn't 0! = 1! = 1?

[@razeer1232]

Yes. Maybe he made a mistake because !0 = 1 and !1 = 0.

[@jasonspence]

Huh... that's clever to choose 6 of the 12 to do the derangement... I instead calculated it out as inclusion-exclusion for exactly 6 of the criteria (i.e. E6 = 12![ 1/6! - nCr(7,1)/7! + nCr(8,2)/8! - ... + nCr(12,6)/12! ] ) and it comes to the same number!

[@kenan5798]

Thank you very much !

[@Jordie389]

very helpful

[@manishvarma7379]

I was pretty surprised when you wrote 10,11,12 in hexadecimal 🤣🤣🤣

[@kshitijingle9738]

can u make a video for this:-
Dn= nCn-r * Dr

[@burninredcrab10]

When you factored out n! at 10:30, I think the last term should just be 1/1 or 1. Not 1/n! since you factored it out.

[@jorgemd]

AbsoluteZero He factored out n!. If you factor out an n! from (n!/n!) you get (1/n!). Think of it this way, (n!/n!)=1 and therefore if we factor an n! from 1 we get (1/n!).

[@burninredcrab10]

Oh right, thats actually such a simple explanation, thank you!

[@souravganguly5680]

e^x =1+x+1/(2!)x^2+1/(3!)x^3.... if x=1; e=1+1/(1!)+1/(2!)+... So how come dn=n!/e?

[@sambit370]

how to solve dearangement problem having identical object... such as dearangement of "varoon"

[@Trevtutor]

Best to use inclusion-exclusion for that.

[@Pride1283]

Why does S0 need to be included for this example...?

[@TheTrevLife]

+Pride1283 Our answer would be negative, which makes no sense.

[@maxrush206]

16:35 actually it's the name of my band

[@suvanshsharma505]

it was so boring...I couldn't even watch it completely

[@simayturgut]

bad 4 u