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It might not be the lack of hats in such a situation that's hazardous, but the lack of clothing well south of the hat region. Or if not hazardous, perhaps adventurous. e ...xactly!
I have a feeling the men without hats are happy to be partying in a dark room. After all, they can dance if they want to, and they can leave their friends behind.
It does indeed sound like a Berlin nightclub, but the calculation was not gender specific, and it could therefore also be applied to any dark room where people remove their hats on entering.
I'm a game developer, and a strikingly similar scenario - and result - came up awhile back when doing a deep dive on some item drop rate adjustments. Imagine you have a monster that drops an item when defeated at a rate of 1 in 100 times and then you defeat 100 of that monster. What's the chance you've gotten at least 1 of that item? Due to the "at least 1", this is easier to count the inverted result of "how many times did you fail to get the item" and repeat 100 times. So: (1-1/100)^100. And then invert that result: 1 - (1-1/100)^100. Giving a result of approximately ~63.4% chance of getting at least once. Generalizing this as n instead of 100, and then letting an n approach infinity, we get the result: Lim n->inf [1 - (1 - 1/n)^n] = 1 - (1 / e)
I remember mentally calculating critical chance for multiple critical chances way back in the Warcraft 3 days and my rough math always seemed to end up near either 33% or 66%
1/e also shows up in another famous math problem, which I'll poorly paraphrase: When dating, what % of the total pool should you check out before committing to one? The answer is 37% of the pool, 1/e.
11:00 just a note: I think here either the sum should be marked with i < j (not i =/= j), or, if written in this way, there should be 1/2 in front of it. You don't want to repeat elements: A1 intersection A2 and then A2 intersection A1. But the result is correct, the non-repeating sum over i < j is equal to n choose 2.
Came here to say this. The n choose 2 fixes this, since it ignores permutations of the choosing, which is why even though the left hand side is wrong, the right hand side is correct. This also means the higher order indices in the summation later on for the inclusion exclusion formula need to be written as i < j < k ... for all free indices from the set, or put another way 1 =< i < j < k ... =< n.
In my head I thought "its definitely something like e, or 1/e", and imagine my surprise when I saw the result! Although, not much of a surprise, whenever probability is involved, e will show up.
Yeah, for the mathematically initiated, it may be far less surprising. but it’s so amusing to take a silly word problem like that regarding hats, and the answer is 1/e 😂
I remember finishing our subfactorials a year ago and I loved them so much. I made a scratch project to plug a number in to give the sub factorial of the number.
I think that if we asked 10000 people the question (for some arbitrary number of hats, like 20) to give a percentage from 0-100 that no-one gets their hat back there would be quite a peak at 37%. Yeah, there would likely be one at 73% too.
As a fascinating exercise, consider the following: Suppose you help n people receive their own hats by randomly distributing all of them. Then, those with incorrect hats return them to you to randomly distribute again among those still missing hats. What is the expected number of iterations it will take for everyone to have their own hat? (I more or less have a proof for this, but i also have homework due today 😢)
Fixed points in this sense are cycles of length 1. The obvious generalisation is to permutations with other minimum (and indeed maximum) cycle lengths. These would be practical things to know.
So, if you need to calculate a subfactorial for some reason and you wish to save a lot of time, just divide the factorial by the number e then round to the nearest integer. It works every time.
As is obvious from their accurate, lifelike portraits, the men in this diagram are misidentified. They are actually Man 47, Man 312, and Man 14,703. 1. 2. and 3 all died in the pneumonia epidemic of 1919.
e and pi are the most interesting numbers like “oh you have a weird value for this problem which no field of mathematics even comes close to?” Plug in e or 1/e or e^2 or the eth root of e or e^pi just keep plugging variations of e and pi and it’ll probably work and if it doesn’t even Euler can’t help you
this was satisfying bc when saw the approximation before it felt so random yet underwhelming (like what, you just multiply the factorial by an essentially constant factor?) but this explains neatly where it came from
That reminds me, not of hats but of doggies’ Rsoles: The doggies held a meeting, They came from near and far, Some came by motorcycle, Some by motorcar. Each doggy passed the entrance, Each doggy signed the book, Then each unshipped his Rsole And hung it on the hook. One dog was not invited, It sorely raised his ire, He ran into the meeting hall And loudly bellowed, "Fire." It threw them in confusion And without a second look, Each grabbed another's Rsole, From off another hook. And that's the reason why, sir, When walking down the street, And that's the reason why, sir, When doggies chance to meet, And that's the reason why, sir, On land or sea or foam, He will sniff another's Rsole To see if it's his own. In this case e = 1 explains all the sniffing! 😁
But they were men without hats And they were dancing And they were friends The men without hats never claimed they didn't "own" hats. They just didn't have any, currently, as is the case with our dancers
The chance of any person getting their own hat is 1/n. Thus the chance of every one not getting their hat is (1-1/n)^n. This has the well known limit of 1/e as n tends to infinity.
I can't believe I didn't see 1/e coming. I just thought "ooh, Taylor series for cosh(1) minus Taylor series for sinh(1)". Of course, that _is_ 1/e, isn't it? Reason you have to round n!/e to get !n is that you're only summing the terms of the series up to n, and if you add more terms after that to approach n1?e, you'll only have shrinking, alternating fractional terms starting with n!/(n + 1)!, which at largest will be 1/2, meaning the sum of all fractional terms going to infinity will be strictly between -0.5 and +0.5, meaning you just round to eliminate all the fractional terms, and of course eliminating all the fractional terms gives you all the integer terms, which is the truncated series formula for the subfactorial.
Seems like there's an easier way to count using a recursive definition: let Ai be the set which has *exactly one* fixed point at index i. This means that index i is a fixed point, and the rest is a derangement: !(n-1). Likewise for Bij (the set which has exactly two fixed points). So the number of derangements is going to have the form !n = n! - ((n choose 1) ⋅ !(n-1) + (n choose 2) ⋅ !(n-2) + ... + (n choose n-1) ⋅ !(n-(n-1)) + 1) Then I'm sure we can do some re-arranging.
1:32 oh god there was a puzzle like this in proffessor layton, you had to figure out how likely it was that 2 people got their hat but one person didnt. I guessed EVERY number from 1-100% only to realise it was 0%
Seems like a useful way to calculate the secret santa arrangements where no-one gets its own gift. Thanks for the video, I wasn't sure if it was a joke video at first, but it was pretty interesting. You earned a sub :D
The fact that !n is equal to the rounded n!/e is pretty deranged! 11:24 so it wasn't a fever dream! It does exist! I remember doing this in high school but then never saw it again and couldn't remember what it actually does or how it works. And confusingly in my language it would translate to "n over 2" which in English is used for fractions so I never knew how to look for it. I will have a nice sleep today, knowing you helped me solve one years-long mystery.
We can subfactorial if we want to We can leave integers behind Cause your numbers derange And if that seems strange Then you've factorials in mind We can dance We can dance Everybody look at your pants
The "key game" parties for tolerant couples were popular in the US in the 70s. The men put their car keys in a pot and each woman drew a key from the pot and drove home with this man to have a little fun together. Different use case, but the math is the same 😊
You can 1/e if you want to, you can leave notations behind. Because your friends double count and if they double count then they're no friends of mine.
e is also closely related to pi. e^pi*i=-1. Also if f(x)=f^4(x), f(x) could equal both e^x, or sin(x). Also, since you mentioned rounding, both pi and e round to 3.
Nice video, well explained. perhaps a more practical application would be that there are several workers assigned a role, you want to reassign their role such that each person has a different role. how many ways are there of doing this. Perhaps role could be changed to position if the application is sport, maybe like the total football of the Dutch team during the 70s.
Here’s a fun fact, integration from 0 to infinity of (x-1)^ne^-x is equal to !n. I don’t think I’m the first to discover this, basically, expand (x-1)^n with binomial theorem, change sum and integral and you get the n-th discrete difference of n! Which is also known as the subfactorial. I’ll leave my derivation for the derrangement formula here as well. Let p(N,k) denote the number of permutations of N elements fixing k. The sum from k=0 to N of P(N,k) is N!. Now note that P(N,k)=C(N,k)P(0,N-k) since fixing k elements is the same as fixing k and NOT fixing the remaining ones. Now you have the sun from k=0 to N of C(N,k)P(0,N-k)=N!. This we know that N! Is the “binomial transform” of the the derrangements, meaning we can invert this sum of write the N-th derrangement as the N-th difference of N! With a bit of tweaking you can use an argument like this to count any permutations of fixing r elements.
Well, that was fun!
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i agree, that was fun
I feel like "It's 1/e, isn't it?" is the "he is right behind me, isn't he?" of maths
Great analogy!
Euler is always there. Sometimes he's downstairs.
Was not expecting to see a 2021 Abu Dhabi reference here 💀
It's always an expression to do with pi, e, or ln of something it feels like in these higher level math videos
@@blasphemer_amonit's not a reference to "2021 Abu Dhabi", it is a reference to the thousands of badly written movie scripts with tiresome clichés
IT'S ALWAYS e LEAVE ME ALONE EULER
I love that even in your furious use of caps lock you still have the respect for mathematics to properly write e as lower case
My rage doesn't make me disregard how symbols work. I have standards.
This smaller case e makes it seem like you say e calmly, and shout the rest
Euler will chase you for your life
I love this reply lmao 🤣
“A dark room with several men who aren’t wearing any hats.”
Oh! The nightmares I’ve had about that!!
😂😂
Nightmares? I had fantasizes about that
That's Inception level 1
The deeper nightmare is none of them getting their hats back.
This is a branch of Applied Mathematics I never knew I needed.
Oh, _hats_ not wearing hats, right. That's what they aren't wearing.
It might not be the lack of hats in such a situation that's hazardous, but the lack of clothing well south of the hat region. Or if not hazardous, perhaps adventurous.
e ...xactly!
I have a feeling the men without hats are happy to be partying in a dark room. After all, they can dance if they want to, and they can leave their friends behind.
True, but one has to worry about the hats going wherever they want to, a place the men may never find
And, if they don't dance, well they're no friends of mine
It will be definitely a SAFETY DANCE! 😉🤗😄 ua-cam.com/video/1p_BvaHsgGg/v-deo.html
I'm glad these men are so safety-conscious while dancing.
This is Weird : I was just doing some economic modelling and this result popped out. This explains the entropy of a market, as it estimates states.
That sounds very interesting, will have to research
You might find it useful that derrangements can be represented by an integral: int_{0}^{/infty} (e^(-x))(x-1)^ndx
@@joshuaiosevich3727 That's true. But I was finding the ratio by construction - ie a fractal result. This explained why the result converged on this.
@@joshuaiosevich3727 linear algeba ruins everything
@@815TypeSirius I'm afraid I'm too dense to get what you're saying.
as a programmer, seeing "!n" just makes me think "logical not n" which evaluates as either 0 or 1 depending if its non-zero lmao
bitwise not also works
I'd be more inclined to think bitwise not as I usually write logical not as ¬ but yeah this would be confusing 😭
@@cosmnik472 For that ~ is used like this ~n
or strict n in haskell
Those who have programmed in some home computer BASICs will be thinking "pling n".
0:38 this is getting very spicy
several men without hats
Spicy indeed, in a world of hatless men, where do we find God?
@@WrathofMaththe God is a man without a hat
Did they leave their friends behind?
@@tweer64 If their friends don't dance, well they're ... no friends of mine.
A dark room with many men in it... Sounds like a Berlin nightclub.
And then e showed up.
@@empathogen75 only if it was one of Those clubs, with hatless men.
It does indeed sound like a Berlin nightclub, but the calculation was not gender specific, and it could therefore also be applied to any dark room where people remove their hats on entering.
I feel like there's a Flight of the Conchords song about this kind of situation ...
@@coyetsyes, obviously. Did you miss the joke?
I'm a game developer, and a strikingly similar scenario - and result - came up awhile back when doing a deep dive on some item drop rate adjustments.
Imagine you have a monster that drops an item when defeated at a rate of 1 in 100 times and then you defeat 100 of that monster. What's the chance you've gotten at least 1 of that item?
Due to the "at least 1", this is easier to count the inverted result of "how many times did you fail to get the item" and repeat 100 times. So: (1-1/100)^100.
And then invert that result: 1 - (1-1/100)^100. Giving a result of approximately ~63.4% chance of getting at least once.
Generalizing this as n instead of 100, and then letting an n approach infinity, we get the result: Lim n->inf [1 - (1 - 1/n)^n] = 1 - (1 / e)
I remember mentally calculating critical chance for multiple critical chances way back in the Warcraft 3 days and my rough math always seemed to end up near either 33% or 66%
Geometric distribution is a bitch.
Brings a whole new meaning to "Statements dreamed up by the utterly Deranged"
(from the "stop doing math" meme)
this is literally how i process every combinatorics problem hoping all the terms cancel out when they dont
Finding this guy in a math video is a fever dream
I wasn't wearing socks and my toes blew up.
Sorry 😞
a small price for science
@@WrathofMath np, that solved having to trim toenails.
Unfortunate
Your fault for having toes
that's quite a deranged equation.
You could say that
1/e also shows up in another famous math problem, which I'll poorly paraphrase: When dating, what % of the total pool should you check out before committing to one? The answer is 37% of the pool, 1/e.
With how big dating pools are now due to modern transportation and dating apps, I still have a significant amount of work to do 🤣
It also marks the point where the graph y = x^x stops decreasing and starts to increase
Your pace and depth are perfect. I would not attempt those formulas on my own, but you made perfect sense if them. Thank you so very much!
11:00 just a note: I think here either the sum should be marked with i < j (not i =/= j), or, if written in this way, there should be 1/2 in front of it. You don't want to repeat elements: A1 intersection A2 and then A2 intersection A1. But the result is correct, the non-repeating sum over i < j is equal to n choose 2.
Came here to say this. The n choose 2 fixes this, since it ignores permutations of the choosing, which is why even though the left hand side is wrong, the right hand side is correct. This also means the higher order indices in the summation later on for the inclusion exclusion formula need to be written as i < j < k ... for all free indices from the set, or put another way 1 =< i < j < k ... =< n.
"A derangement" is such a hilarious term for something in math. I love it.
In my head I thought "its definitely something like e, or 1/e", and imagine my surprise when I saw the result! Although, not much of a surprise, whenever probability is involved, e will show up.
Yeah, for the mathematically initiated, it may be far less surprising. but it’s so amusing to take a silly word problem like that regarding hats, and the answer is 1/e 😂
What I love about mathematicians is that they'll mention Greek gods like it's 1200 BC.
When the Wrath of Zeus meets the Wrath of Math.
"Our fight will be legendary!"
I’m excited to integrate a “deranged“ math lesson into my sons home-study curriculum!
poor son😂
Oh cool. I love taking an integral of the subfactorial function.
i'm excited to show this to my math teacher
i was scared for just a second when you started drawing that hat rack
I was wondering how he would manage to hang the hats on it.
6:39 WHAT YOU'RE LETTING AI?!?!?!
AI's taking everyone's jobs, not even the set of all permutations where i is fixed is safe....
gotta do what you gotta do
The night theme of Hateno village made this an emotional hat story for me...
Takes me back to sophomore year of college
5:34 Missed opportunity to say it would blow our hats off. Booo!
I will not be tempted by your cheap puns!
This music is making me think I'm watching a Zullie the Witch video
5:38 "it will blow your socks off"
What a relief that it won't blow my _hat_ off 😌
The hats are long gone at this point 😂
Great explanation, you show how it’s calculated, how it’s relevant, and the end result is actually simple to calculate.
Thank you!
I appreciate the gentle transition to combinatorics via hats and Zeus's wrath :))
I will never write a "3" or a "2" as legibly as this man did @ 2:30. 😢
I was cooking with those
I did once, and then never again. 😢
However; not equal looked indistinguishable from +/- @11:30 :)
I remember finishing our subfactorials a year ago and I loved them so much. I made a scratch project to plug a number in to give the sub factorial of the number.
Super fun!
@ Hell yeah!
that's called an admin command
I think that if we asked 10000 people the question (for some arbitrary number of hats, like 20) to give a percentage from 0-100 that no-one gets their hat back there would be quite a peak at 37%.
Yeah, there would likely be one at 73% too.
Omfg it's really everywhere
Why would there be a peak at 73% as well?
@@Kapomafioso Second most popular "random" number from 0 to 100.
As a fascinating exercise, consider the following:
Suppose you help n people receive their own hats by randomly distributing all of them. Then, those with incorrect hats return them to you to randomly distribute again among those still missing hats. What is the expected number of iterations it will take for everyone to have their own hat?
(I more or less have a proof for this, but i also have homework due today 😢)
I bet the proof wouldn't fit in the margin, either.
is the expectation not just 1/n! ?
Enjoyed the SM64 music 😊
Love how this is done sooo much easier with like a for loop and if statement in programming
That was well worth the wait. Thanks for such a fun explanation.
Fixed points in this sense are cycles of length 1. The obvious generalisation is to permutations with other minimum (and indeed maximum) cycle lengths. These would be practical things to know.
10:21 I hear faint sounds of skyward sword. Haha
So, if you need to calculate a subfactorial for some reason and you wish to save a lot of time, just divide the factorial by the number e then round to the nearest integer. It works every time.
This was really fun, and the notation wasn’t too difficult. Thanks!
1/e is so hilarious, I'm amazed there isn't a sitcom about it.
Same!
Honestly amazed by how you could confidently go deep into a topic in such an entertaining way. Might be slightly jealous...
Can we talk about the beautiful Hateno Village music behind this?
As is obvious from their accurate, lifelike portraits, the men in this diagram are misidentified. They are actually Man 47, Man 312, and Man 14,703.
1. 2. and 3 all died in the pneumonia epidemic of 1919.
"it's the on left now"
5:33 Now I just want to know how likely is it that none of the viewers get both their socks back?
e and pi are the most interesting numbers like “oh you have a weird value for this problem which no field of mathematics even comes close to?” Plug in e or 1/e or e^2 or the eth root of e or e^pi just keep plugging variations of e and pi and it’ll probably work and if it doesn’t even Euler can’t help you
4:23 I hear your Easter Egg of putting the Select File Theme from SM64, as Pannenkoek does! Very clever!
Thanks for the presentation! This is a clssical gem of discrete mathematics.
this was satisfying bc when saw the approximation before it felt so random yet underwhelming (like what, you just multiply the factorial by an essentially constant factor?)
but this explains neatly where it came from
That reminds me, not of hats but of doggies’ Rsoles:
The doggies held a meeting,
They came from near and far,
Some came by motorcycle,
Some by motorcar.
Each doggy passed the entrance,
Each doggy signed the book,
Then each unshipped his Rsole
And hung it on the hook.
One dog was not invited,
It sorely raised his ire,
He ran into the meeting hall
And loudly bellowed, "Fire."
It threw them in confusion
And without a second look,
Each grabbed another's Rsole,
From off another hook.
And that's the reason why, sir, When walking down the street, And that's the reason why, sir, When doggies chance to meet, And that's the reason why, sir, On land or sea or foam, He will sniff another's Rsole To see if it's his own.
In this case e = 1 explains all the sniffing! 😁
Clearly, they were not doing the Safety Dance.
(Gen X earworm, activate!)
Well, then they're no friends of mine!
But they were men without hats
And they were dancing
And they were friends
The men without hats never claimed they didn't "own" hats. They just didn't have any, currently, as is the case with our dancers
I had exactly the same thing in my head 😂
i was trying to work out a comment like this…well played!
thank you for the pannenkoek2012 music
What a fantastic video, such an enjoyable watch, the mario 64 music was just the cherry on top
Thanks so much! Trying to get the music at the right volume, I think I got it just about right this time.
the moment he said "the probability will surprise you" I thought "probably 1/e or sth like that"...
3:47 ÐE SM64 MUSIC?!?!?!
The chance of any person getting their own hat is 1/n. Thus the chance of every one not getting their hat is (1-1/n)^n.
This has the well known limit of 1/e as n tends to infinity.
the file select theme in the background makes this so beautiful
It's a classic!
A wonderful and beautiful result, thanks.
So !5 is just 5! Divided by e and then rounded to the nearest whole number? Nice, knowledge acquired
So !5 would be 44?
Yup!
1:03 there's hawk tuah🗣️🗣️
HAWK TU-*fucking spits*
Bro
You are NOT funny
you should uncomment yourcomment now
I fw with you vro, keep going
oh hey, alternating harmonic series, love to see it
This entire video is a single unacknowledged safety dance joke.
This video amazed my mom as I watched this since she never heard of subfactorials before
Hot example, honestly. This is also the most lucid presentation of inclusion-exclusion I have ever seen. Well done.
Thanks a lot!
0:25 yes, it means 1 factorial times n = n
I can't believe I didn't see 1/e coming. I just thought "ooh, Taylor series for cosh(1) minus Taylor series for sinh(1)". Of course, that _is_ 1/e, isn't it?
Reason you have to round n!/e to get !n is that you're only summing the terms of the series up to n, and if you add more terms after that to approach n1?e, you'll only have shrinking, alternating fractional terms starting with n!/(n + 1)!, which at largest will be 1/2, meaning the sum of all fractional terms going to infinity will be strictly between -0.5 and +0.5, meaning you just round to eliminate all the fractional terms, and of course eliminating all the fractional terms gives you all the integer terms, which is the truncated series formula for the subfactorial.
Very interesting. So basically it goes up to n!/n! and the rest are n!/(n+i)!, I>0, which gets rounded away.
Seems like there's an easier way to count using a recursive definition:
let Ai be the set which has *exactly one* fixed point at index i. This means that index i is a fixed point, and the rest is a derangement: !(n-1).
Likewise for Bij (the set which has exactly two fixed points).
So the number of derangements is going to have the form !n = n! - ((n choose 1) ⋅ !(n-1) + (n choose 2) ⋅ !(n-2) + ... + (n choose n-1) ⋅ !(n-(n-1)) + 1)
Then I'm sure we can do some re-arranging.
1:32 oh god there was a puzzle like this in proffessor layton, you had to figure out how likely it was that 2 people got their hat but one person didnt. I guessed EVERY number from 1-100% only to realise it was 0%
You had me at Zelda music
And also Super Mario 64 music too.
This dude needs to add a songs used list in the video description
Love the Zelda background music!
Seems like a useful way to calculate the secret santa arrangements where no-one gets its own gift.
Thanks for the video, I wasn't sure if it was a joke video at first, but it was pretty interesting.
You earned a sub :D
omg the panennkoek music
I kept waiting for the hilarity, and was left with the impression that you math types are easily amused!
the real question is why do they care so much about their hats, like should that really the biggest of their concerns?
They are really expensive hats and they won’t be seeing each other for a long while so it is pretty important to them that they get the right hat
The fact that !n is equal to the rounded n!/e is pretty deranged!
11:24 so it wasn't a fever dream! It does exist! I remember doing this in high school but then never saw it again and couldn't remember what it actually does or how it works. And confusingly in my language it would translate to "n over 2" which in English is used for fractions so I never knew how to look for it. I will have a nice sleep today, knowing you helped me solve one years-long mystery.
Beautiful explanation. Thank you.
thank you SO MUCH for the Men Without Hats reference. I will do the Safety Dance in your honor, and with the light off indeed
It's here!
We can subfactorial if we want to
We can leave integers behind
Cause your numbers derange
And if that seems strange
Then you've factorials in mind
We can dance
We can dance
Everybody look at your pants
We had to prove something with derangements without the subfactorial ended up just becoming deranged 😂😂
Nice. You can instead of divising by e multiply by the precomputed 1/e or e^-1
I now wanna question what n!n is, is it n! × n, or n × !n...
parentheses are your friends
Id say n × !n bc you put numbers before variables like 2x
!n!
stuff like this is why im not better at math, that’s very ambiguous
well, clearly, the bodmas order wasn't enough 😢😂
The "key game" parties for tolerant couples were popular in the US in the 70s.
The men put their car keys in a pot and each woman drew a key from the pot and drove home with this man to have a little fun together.
Different use case, but the math is the same 😊
Should use the upside down exclamation mark in this case
12:45 writing (-1)^(j-1) feels illegal
Great expanation of the inclusion/exclusion proncople
As a programist I can say that !n means just "not n", like "anything but n"
You can 1/e if you want to, you can leave notations behind. Because your friends double count and if they double count then they're no friends of mine.
20:05 I was just waiting for it (it was worth it)
e is also closely related to pi. e^pi*i=-1. Also if f(x)=f^4(x), f(x) could equal both e^x, or sin(x). Also, since you mentioned rounding, both pi and e round to 3.
Or you know, cos(x) and basically any sine function of the form a*sin(x+b) or a*cos(x)+b*sin(x)
We call it Chaotic Permutations in Brazil
Nice video, well explained. perhaps a more practical application would be that there are several workers assigned a role, you want to reassign their role such that each person has a different role. how many ways are there of doing this. Perhaps role could be changed to position if the application is sport, maybe like the total football of the Dutch team during the 70s.
I'm already laughing and the joke hasn't even happened yet
Here’s a fun fact, integration from 0 to infinity of (x-1)^ne^-x is equal to !n. I don’t think I’m the first to discover this, basically, expand (x-1)^n with binomial theorem, change sum and integral and you get the n-th discrete difference of n! Which is also known as the subfactorial.
I’ll leave my derivation for the derrangement formula here as well. Let p(N,k) denote the number of permutations of N elements fixing k. The sum from k=0 to N of P(N,k) is N!. Now note that P(N,k)=C(N,k)P(0,N-k) since fixing k elements is the same as fixing k and NOT fixing the remaining ones. Now you have the sun from k=0 to N of C(N,k)P(0,N-k)=N!. This we know that N! Is the “binomial transform” of the the derrangements, meaning we can invert this sum of write the N-th derrangement as the N-th difference of N!
With a bit of tweaking you can use an argument like this to count any permutations of fixing r elements.
It's so fitting that we count derangements on the left. There are coincidences in nature.
This seems like something that may be non-polynomial.
Me: There's no way there's an explicit formula for this
Also me after watching the video: Oh right I did this in combinatorics like 7 years ago
My favorite part about it is that the sequence appears more slightly strange (the terms don’t all end in 000…)