Arrangements and Derangements with Repetition

[@ethancheung1676]

aren't you double counted? the first "false" derangement !5 (p1urple -> ***p1**) contains (p2**p1**), also the second "false" derangement !5 (purp2le -> p2****) also contains (p2**p1**). so I think you should not -!4 but +!4 to make up the count.

[@blackpenredpen]

Ethan Cheung When we "derange" the first p and fix the second p, it prevents the first p goes back to the first place. So the !5 wouldn't have a configuration with p goes first. Similar for the second !5. Thus we still have to subtract !4. I will pin this comment since I don't think I explained it well enough in the video regarding to that part and maybe others can add their explanations too. : )

[@waynemv]

I had the same question. I did some investigation myself with the help of Wolfram Alpha and a text editor (Notepad++) that counts matches using regular expressions. Here is what I learned. There are !6 = 265 derangements of six unique items. I used WolframAlpha to generate a list with all derangements of "Purple". (Since it was case sensitive, P was not considered the same as p.) I copied the list to my text editor and did minor cleanup of the formatting. Upon examining the list, I saw that 53 have the P (uppercase) moved to the 4th position, and symmetrically, 53 have the p (lowercase) moved to the 1st position. Note that 53 is GREATER THAN !5 (44). By inspection, it appears 53 results from !5+!4. Among those previous two cases, there are 9 cases (!4) where the P is in the 4th position simultaneously with the p being in the first position. Since the overcounted amount was being subtracted from !6, correction to the overcount must be ADDED, just as Ethan Cheung suggested above. Thus it looks like the formula could instead be expressed as: !6-(!5+!4)-(!5+!4)+!4. But note that is equivalent to the !6-!5-!5-!4 that blackpenredpen showed, because two of the !4 terms cancel out. In contrast to how I've just counted, Blackpenredpen's method avoids overcounting in the first place. His two !5 terms count only cases where a single P is in a disallowed position, not both Ps. Since in his approach the !4=9 cases with the p and P has swapped positions weren't yet counted at all, he is correct to subtract them. And to finish, we still need to divide by 2! at the end for the reasons blackpenredpen already explained.

[@maurocruz1824]

@@blackpenredpen Yes you're right. Does that have any "analytical" application just like the ordinary subfactorial has?

[@Kwekinator117]

@@blackpenredpen Another explanation is that, by symmetry, in 1/5 of the derangements, p1 will be in the first place. Similiarly, in 1/5 of derangements, p2 will be in the 4th place. Therefore we can subtract 2*!6/5 from !6 but by doing so, we've double counted the cases where p1 is in the 4th place AND p2 is in the 1st place. Therefore, we must add the number of derangements where p1 is in the 4th place and p2 is in the 1st place which is simply !4. Then we divide by 2! as you did above. Therefore, the final answer can be written as : ( !6-2*(!6/5)+!4)/2!

[@BigDBrian]

@@maurocruz1824 gift exchange but two or more people brought the same gift

[@rksmehul]

Now do derangement of "Mississippi"

[@BabaFroga]

Yes, Mississippi please! 😁

[@davidrheault7896]

Very interesting I am going to count the number of ways it can be done by an improper integral....and another method using rooks polynomial. Notice that Mississippi has the same number of letters as dérangement. So here we have 4 s, 4 i and 2 p. 11 letters in total hence 1 single letter. The answer should be 648 distinct. 746 496 arrangements and then you divide by 4! Twice and 2!

[@Electroneer]

Why am I watching this at 3 am? Math is too interesting sometimes.

[@ErickSousa-lx3qr]

Yeah haha

[@parasjain6404]

Great Video . The only video in the internet with "derangement with repetition"

[@ujjwalchauhan8896]

Thanks ,that is really helpful for jee.

[@blackpenredpen]

Ujjwal Chauhan thanks

[@snatcc]

I can’t wait for t”the color purple” movie”, so balck

[@shubham1999]

Are these topics related to Permutations and Combinations? Thanks I was eagerly waiting for the videos from this topic.

[@ErickSousa-lx3qr]

Thank you so much !!!

[@ethancheung1676]

I have encountered a question if you could solve. Given a bezier curve in parametric form B(t), t0 and L, find t1 such that arc length from B(t0) to B(t1) equals L.

[@naveenrajan537]

I love purple!

[@davidrheault7896]

Oohh I want that shirt

[@protessional6227]

Hi, I'm a high school student and I've stumbled across an interesting limit that I've tested and it seems to converge to a complex number, but I cannot see what value it approaches. I will explain the limit now: Suppose f(n, m) := ln(ln(ln...(ln(m))...) where there are n natural logs (f(0, m)=m. f(1, m)= ln(m)). the limit is then lim n -> ∞ f(n, n).Could you explain this? Edit: Researched this today and worked out that it goes to -W(-1) or if you write -W(-1) as a+bi another fixed point is a-bi.

[@angelmendez-rivera351]

Ooh. Nice. Can you solve the equation Sqrt(x^2 + y^2)*e^(1 - y) = 1 in terms of y? You are allowed to use the Lambert-W function.

[@thehighgupta]

solve......integral limit( minus one to zero) ((x)^(4)-2x^(2))^(1/3)+√(1-√(x^(3)+1))

[@leonardobarrera2816]

Here was the time that Bprp doesn't had Ego

[@aswinibanerjee6261]

So in general derangement of n aliments in which m aliments are same the formula is #=[!n-m×!(n-1)-!(n-m)]/m!

[@mugglerowansquarepantssays7168]

PurplePenGreenPen

[@nicolasgoubin]

Do the same for "pineapple" ! I love pineapples =)

[@Gold161803]

Pen pineapple purple pen

[@arandompersononyoutube3739]

Do arrangements and dearrangements of blackpenredpen.

[@MrQwefty]

PURPLE PEN!?

[@PigZombiePacifico]

if n is the number of letters and m is number of letters that repeat, got that the number of derangements is going to be: (!n-(Σ(mCk)*!(n-k)))/m! Where the sum goes from k=1 to m. Is that right?

[@KuZiMeiChuan]

Very exciting

[@rajatkhandelwal7276]

Sir you have never make video on vectors and 3d like topics plz make interesting videos too on these topics

[@yrcmurthy8323]

Yep

[@aspiringcloudexpert5127]

You should check out 3Blue1Brown's KZbin series called Essence of Linear Algebra. It might help.

[@sword7163]

what if you want to rearrange the letters aabbcc so that same letters never come together ?

[@waynemv]

@blackpenredpen, would you please follow up with a counting of the derangements of the letters in "derangement"?

[@andrescamilohernandezruiz6578]

Can you solve this integral? Integrate (1+x^4)^(1/2) dx

[@Kitulous]

I love purple color!.... and also yellow. But yellow pen wouldn't be too visible on a white board. Can I congratulate you with using a new pen color?

[@samokoribanic9184]

Can you try to solve this question. We have table where sides are a and b long. Question is how many combinations we have to color full table with 4 colors such that no 2 same colored squares will be neighbours... I cant figure that out..

[@ErickSousa-lx3qr]

So the formula of derrangements when there are 2 repeated elements is [!n - 2 × !(n-1) - !(n-2) ] / 2! , right?

[@iole2804]

Pai?

[@davisilvaromao6224]

@@iole2804 Mãe?

[@ErickSousa-lx3qr]

@@iole2804 Mãe?

[@Koisheep]

Purple pen green pen yay!

[@BurningShipFractal]

I was thinking of saying the exact same thing 💜💚💜💚💜💚💜💚

[@calyodelphi124]

Quick homework! Calculate how many arrangements and derangements there are for the words "arrangements" and "derangements"! :P

[@meganweber5057]

How can I get that shirt?

[@alephomega9610]

Can you do i! !i or i^^i with i^2=-1 ( i factorial and subfactorial and i tetrated to i) That would be amazing!

[@angelmendez-rivera351]

Budy i^^i is not defined since tetration is not defined for complex numbers.

[@wojtek9395]

@@angelmendez-rivera351 so lets define it!

[@angelmendez-rivera351]

wo997 It is very difficult to accomplish this. Let me explain why. Suppose we have the function f(x) = a^x, which maps from C to C. We know that f(1) = a = a^^1. Now, f(f(x)) = a^(a^x), hence f(f(1)) = (f^2)(1) = a^(a^1) = a^a = a^^2. Furthermore, f(f(f(x))) = (f^3)(x) = a^(a^(a^x)), such that (f^3)(1) = a^(a^a) = a^^3. This pattern is intuitive, and it is trivial to use induction to prove that, for any n in the set of natural numbers, (f^n)(1) = a^^n. Note that a could be any complex number in principle, as long as the expression presented is well-defined (i.e, no 0^i or any of the sort). However, note that (f^-1)(f(x)) = x as long as f is invertible, a property which is satisfied when a meets a semi-trivial set of conditions. This defines (f^0)(x) = x, so (f^0)(1) = 1, and this defines a^^0 = 1 for all a which satisfy aforementioned conditions. We also know that (f^-1)(x) = Ln x/Ln a is a well-defined function whenever Ln a is well-defined and Ln a = 0 is false, excluding x = 0. Then (f^-1)(1) = a^^(-1) = Ln 1/Ln a = 0, where Ln stands for the natural logarithm, in base e, mapping from the set of complex numbers to the set of complex numbers, in the principal branch. a^^(-2) = Ln 0/Ln a, which is undefined. Therefore, a^^n is well defined for all n > -2 in the set of integers. We can do better. A well-studied analytical function U is known as the half-exponential, with the property that U(U(x)) = e^x. In other words, if a = e in f, then (f^1/2)(x) = U(x). This allows us to recursively define a^^n for ever half-integer n with the property that n > -2. This is the best that we have done. Following this general principle, the problem of extending tetration to complex number heights can be reduced to simply solving the problem of defining (f^c)(x), where c is an arbitrary complex number for any f in [C, C] (this denotes the Cartesian product of C with itself). Once this problem is since, we let f(x) = a^x for any complex a, and we let (f^b)(1) = a^^b. So I’ve stated what is necessary and sufficient to solve the problem. However, defining f^c is much more difficult than it sounds. In fact, it may or may not be impossible. Mathematicians have worked on possible solutions, and we have promising results, but nothing too convincing yet.

[@yaleng4597]

i^i^i=i^(i^i)=i^(e^(-π/2))

[@omrimg]

You can just prove that !n=[n!/e] (which is not that complex) and it's much easier from there :D

[@jasertio]

Why did you delete the area video?

[@blackpenredpen]

The second picture was wrong. It should have been below the x-axis.

[@jasertio]

@@blackpenredpen oh ok. For a second I thought it was because the axes weren't labeled, 😋

[@msolec2000]

So... for derangements of "error" (should be 0), it should be... start with !5 which is 44, then we subtract... 3! * !(eror)... And then... that's too hard...

[@LucasFerreira-hy4sn]

now how do i do this for the general case when there are n repeating letters and m sets of repeating letters in a word?I’ve been trying but its really hard

[@chandankar5032]

This is 4th time i am saying you to integrate e^{-x}cosnx/a^2+x^2

[@ketopp5233]

Please do about nonhomogeneous first order differential equation. Odd enough there is nothing on the internet about this . Helppppppp

[@purushotamgarg8453]

Isn't there any other nice way to calculate subfactorial other than using the recursive formula? If I was to compute say !12, would I have to compute all subfactorials till 12???

[@nathanisbored]

!n = n!/e (rounded to nearest integer)

[@Lordoftheflies234]

General term for !n is is !n = n!*sum (-1)^k / k!

[@jakeandrews8393]

Greenpenpurplepen yaaayy

[@harshsrivastava9570]

How many pens do you have? WTH?!

[@waynemv]

One can never have too many pens or too many pen colors.

[@not_vinkami]

Black red blue green purple And the mixing of any 2 And the mixing of any 3 And the mixing of any 4 And mix up all of them And use different amount of power to write with a pen, or 2-5 Sum up the numbers then you can get all the different colours that he can made with these pens The number of colours that bprp can write has to be approaching infinity

[@Gold161803]

well well

[@dcyxd]

Thumbnail says "arrangments"

[@monjurhussain8022]

Could you help me with variation of calculus?

[@cousinbenjy8905]

When I made a program to test these sorts of things out, it looked like x! / !x approaches e as x approaches infinity (although I don't have a proof for this). Can anyone find a proof for this or show some sort of link?

[@EMorgensztern]

Can you prove the integral formula for the derangement pls

[@blackpenredpen]

Eliott Morgensztern it's already on my to do list. Maybe I will get it done by next weekend.

[@DarkMage2k]

3D geometry please please please please please please

[@sinom]

Yeah use green. Purple is a not green green anyways so it's the same thing

[@Fablle516]

hey blackpenredpen, can you take a look at the very bottom of this wikipedia page? it connects the subfactorial concept with the regular factorial and e, where: !n = [n!/e], when the brackets [x] represents the closest integer to x pt.wikipedia.org/wiki/Desarranjo

[@helloitsme7553]

Purplepengreenpen

[@karthikrambhatla7465]

Hey BlackpenRedPen. I'm not able to find the video area under polar curve. Am I the only one wondering

[@blackpenredpen]

I made it private since I made a mistake on the second picture. I will remake it soon.

[@karthikrambhatla7465]

@@blackpenredpen oh ok Thank you. Waiting for that eagerly. I'm really curious

[@Lordoftheflies234]

Only wanted to see you laughing with a purple pen. Purple pen, purple pen.

[@yrcmurthy8323]

I'm not able to solve this. Can you help me please ? Looks simple but complex one a²+b²=a+ib Actually wanted to find a formula for a²+b² as we have a formula for a²-b². I'm not getting a real formula. So I tried to get it imaginarily. I tried to plot also. I'm not able to solve and get a formula. Please help Harith

[@Rokkc]

no real formula exists

[@ssdd9911]

why is !n~n!/e?

[@CatboyChemicalSociety]

derangements ?? damn you must be derranged!!

[@gouravmadhwal5548]

Plz integrate ln(ln(ln(lnx))) (No limits!)

[@Theraot]

If this vidieo faces demonetization, it is because of the third word of the title.

[@blackpenredpen]

Hmmm, is it that bak? You aren't the first person mentioning that to me. I guess I will just change it just in case.

[@Theraot]

​@@blackpenredpen if you care about my opinion, I'd say keep it. They should not demonetize this video... if they haven't learned that, they have to. I am aware that it is risking money on a protests. So... up to you, it is your revenue, not mine. However, why not contact Brady? He would give better advice than me, and if you get this demonetized, he could help, at least in getting KZbin attention (and having tons of youtube channels with millions of subscribers does not hinder that). I suppose other has already mentioned, the antecedent is this video: kzbin.info/www/bejne/ppO7mGh7fpqnasU Addendum: To be clear, Brady got that one fixed. Brady has faced problems worse than this, for instance there was the case of videos being demonetized because of nudity (see if porn here: kzbin.info/www/bejne/f6LCgJ1spt-Yf7c)... it is unclear if there is a human checking or if it is all automated.

[@blackpenredpen]

Thank you. I will look into it. I also have been wondering what else does the word "derangement" mean that make it so bad.

[@Theraot]

​@@blackpenredpen For what I recall it refers to mental disorders, I suppose it is used pejoratively, or at least that I came to understand with all this mess. There is also the chance that the suffix "phile" did not help Brady's video. He talked about it on Hello Internet - and thankfully that podscast is also uploaded to youtube so I can search the transcriptions - you can listen here: kzbin.info/www/bejne/eIrPkqJ7rrqdaKc - Oh, I did recall incorrectly, it was the extra footage one the one that was facing demonetization, just linten there from Brady himself.

[@Jeetu0301]

I didn't understood anyting.