subfactorial & derangement, an explicit approach

#XmasIsReal #derangement

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blackpenredpen | 曹老師

[i_am_anxious]

Oh my gosh, you are making such unique intros and thumbnails, u r being so creative

[GourangaPL]

you said "but unfortunately i need more space", combining it with the sentence you always say, it would be "i don't like to be on the bottom, i like to be on the top, but unfortunately i need more space"...

[jay_sensz]

The Taylor series for e^-1 converges very quickly. That means that n doesn't have to be very large at all for it to be a good approximation. It seems that the following expression holds for all n>0:
!n = floor(n!/e + 0.5)

[NuptialFailures]

I really love your factorial family videos. I think that it would be interesting to see videos on the factorial (and the other types) of the imaginary number.

[davidrheault]

I just received my copy of the T-shirt, 3 weeks for delivery in canada.
A note for the video, !n is EXACTLY n!/e when you use the round function, I mean instead of doing 'sup factoreo of 4 by recursive formula, we can do
!4 = round (4!/e) = 9,
|3 = round(3!/e) = 2
!2 = round(2!/e) = 1
!1 = round(1!/e) = round(0.37) = 0

we can compute the digits of e by going high enough with the recursive formula, I did it high enough to compute 1000 digits of e :) Basically, you just need to divide the factoreo by the derangements

[nicolassamanez]

yo, i actually know how to do this

if you look at a three-way venn diagram, you can draw it by starting your way inside-out; draw a reuleaux triangle, then complete the 2-way intersections, then finally the circles.

you can generalize this for any n-gon. start out with a reuleaux (for odd n) or reuleaux-esque (in case of even n–just make the sides circle arcs) polygon, and work your way up from there, drawing in the intersections in increasing order, finishing with the circles.

so many years of doodling in math class finally paid off, it seems

[chinmayasahu]

I feel so dumb and so smart at the same time

[kiiometric]

I love your videos! I think you're underrated in youtube, you should get much more likes for that amazing stuff

[ChrisAsHell]

But is impossible that just two out of three get their gifts back... If they did, the third one also did

[HerbertLandei]

How about a "deranged" version of the choose operator? dechoose(3, 2) would mean "all possibilities for two people to choose from 3 gifts without getting their own"

[zubmit]

Love your videos and I'm not even studying math.

[bjarnivalur]

Hay! Do you make your videos for your students or just for fun? Or is there some other secret purpose behind them?

[codyhintz]

Toward the end you mention for big enough n you can use !n(about)=n!e^(-1) so how big does n have to be for this to apply?

[ffggddss]

Except for n = 0, you can write
!n = {n!/e}
where {x} is the *nearest* integer to x. In fact, the difference goes as
!n – {n!/e} ~ (–1)ⁿ/(n+2)
in the sense that the ratio of those two sides 1 as n  ∞; and that it converges to 1 faster than any other term of the form a/(n+b).

Fred

PS. Somehow I missed this when it came out over a year & a half ago, but I'm glad I found it now. Next, I will check out its predecessor – the recursive approach.

[angelmendez-rivera]

This is good stuff! Now to generalize this using the Incomplete Gamma function, or just good old Gamma function?

[yrcmurthy]

Sir, what is the relationship between subfactorial with hyperfactorial and Double factorial.

[sugarfrosted]

I was wondering if this divided factorial and it does.

[clubstepdj]

I still don't get it
Is it ok if the 3rd person gets his own gift (gift C) as a gift?

Is it "all 3 of us cannot get the same gift" or "only i can't get my own gift, screw with others"?

[przemysawkwiatkowski]

Actually how good is the approximation n!/e ?

[askyle]

At the last bit i thought you were also gonna plug the Stirling approximation ._.