Back when I was in University this was known as the "Montmort Letter Problem". The classical name for the derangement problem. Look it up. I was able to obtain the recursion relation in pretty much the same way you did. I was then able to develop the closed form or general nth term formula using Generating Functions and Infinite Power Series. You can also derive this nth term formula, as you said, directly by using inclusion/exclusion - along with a little bit of inductive reasoning. This is how I did it on my first go. However you obtain the general nth term formula, it opens up a whole new way of understanding the problem. Very cool. Anyway, thanks for re-introducing this famous classical problem of derangements to a younger and wider RUclips audience. You just gotta love this stuff.
Small detail you brushed over: in case 2, the second set of people and gifts is !(N-1) because people cannot be matched to their own gifts, except for P2, who could technically get P1's gift, but because case 2 specifically excludes this case, it can be treated as if it was their own gift they cannot give to themselves (as you said, you could relabel everything) and youd get the !(N-1) you were looking for. Great video as always!!
*Factorial = Derangement x e hence directly proportional* ! N is the nearest integer close to N ! / e (and keeps getting closer as N increases to infinity or even greater than N=10). You need to round up or down based on odd vs. even N.
Back when I was in University this was known as the "Montmort Letter Problem". The classical name for the derangement problem. Look it up. I was able to obtain the recursion relation in pretty much the same way you did. I was then able to develop the closed form or general nth term formula using Generating Functions and Infinite Power Series.
You can also derive this nth term formula, as you said, directly by using inclusion/exclusion - along with a little bit of inductive reasoning. This is how I did it on my first go. However you obtain the general nth term formula, it opens up a whole new way of understanding the problem. Very cool.
Anyway, thanks for re-introducing this famous classical problem of derangements to a younger and wider RUclips audience. You just gotta love this stuff.
Thank you so much, that is really cool 😍
Small detail you brushed over: in case 2, the second set of people and gifts is !(N-1) because people cannot be matched to their own gifts, except for P2, who could technically get P1's gift, but because case 2 specifically excludes this case, it can be treated as if it was their own gift they cannot give to themselves (as you said, you could relabel everything) and youd get the !(N-1) you were looking for. Great video as always!!
Thanks so much!! I was actually wondering why this was true haha thanks for clarifying
I was going to ask about this. Thanks for explaining.
*Factorial = Derangement x e hence directly proportional*
! N is the nearest integer close to N ! / e (and keeps getting closer as N increases to infinity or even greater than N=10).
You need to round up or down based on odd vs. even N.
Loved this
It reminds me of the Airplane seats probability problem.
There MUST be an application to (or in) Group Theory!
Like your dashiki
7!/e =1854.112… coincidence?
Nope, it is true in general that !n ~n!/e.
Yes. It’s in fact equal to the nearest integer to n!/e
@@bigdog41407is there a basic explanation of that?
That’s quite interesting!
@@Happy_Abe it has to do with the fact that the series 1-1/1!+1/2!-1/3!... Converges to 1/e.
@@bigdog41407 that I know, but doesn’t fully explain it
Spanish do call them factorials tho /s
Thank you for making fun videos πm
No, their factorial is ¡
Aah I see. Thanks for correcting!
@@ShlokPatel_2310 De nada amigo¡