Recent content by Gavran

I agree with all that you said in post #15, but is it okay to use the term “permutation” here? The expression $$ \binom{n-2}{l-1} $$ indicates that we speak about combinations here.

Now we can use the above to calculate how many events with a positive outcome there are. For the general case when there are n passengers, we can define sets ## A=\{1\} ## and ## B=\{2,3,...,n-2,n-1\} ##. Every set ## D=A\cup C ## where ## C\subseteq B ## defines one event with a positive...

People accept mathematical concepts much better if they can see their practical applications and relevance.

I suppose that by the l-th element in this scenario, you are referring to the 100th passenger who is taking seat #1. Any cycle that contains the passenger sitting in seat #100 must end with the 100th passenger sitting in seat #1.

I have found the problem from the initial post online and there is also a solution to the problem. To me, this solution seems to be incomplete, making it difficult to understand. This is why I offer the solution in the initial post, which I believe is easy to understand.

By following the patterns provided by the original poster, it seems to me that the original poster has used the next calculation for getting the sequence 0, 3, 8, 15, 24, … For the general term an there is a pattern 1, 2, …, n-1, n, [n+1], n, n-1, …, 2, 1 and there is the calculation...

0, 3, 8, 15, 24, ... is a known sequence with a general term n2-1 where n=1, 2, 3, ..., but I do not know if it has a particular name. By the way 1, 2, 3, 4, 5, 4, 3, 2, 1 gives 15.

[P. Winkler] One hundred people line up to board an airplane. Each has a boarding pass with an assigned seat. However, the first person to board has lost his boarding pass and takes a random seat. After that, each person takes the assigned seat if it is unoccupied, and one of the unoccupied...

The limit is irrational, and the primary objective now is to provide a rigorous proof of its irrationality.

We can define the function ## f(x)=1+\text{NOT}x ## and prove that ## f^{-1}(x)=f(x) ##. ## \begin{align} f(x)=1+\text{NOT}x&\implies f(x)-1=\text{NOT}x\nonumber\\ &\implies \text{NOT}(f(x)-1)=x\nonumber\\ &\implies f^{-1}(x)=\text{NOT}(x-1)\nonumber\\ \end{align} ## For $$...

A little boy refused to run anymore. When his mother asked him why, he replied, "I heard that the faster you go, the shorter you become."

Mermin, N. David. Space and Time in Special Relativity Chapter 10 (pages 87-90) “Why He Says My Clocks Go Slowly Although I Know that His Go Slowly” https://dokumen.pub/qdownload/space-and-time-in-special-relativity-1stnbsped.html

It seems to me that you did not show that ## \sum_{n\ge0}(a_n-a)\frac{x^n}{n!}\sim0 ##, which is crucial to the proof.