Changes in Mathematics Research Methods

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The following math problem appeared in a math forum that Ifollow:You have a deck of 56 cards, labeled 1 through 56. All thecards are drawn randomly one at a time. What is the probability that forexactly one card, the face value of the card is one less than the face value ofthe card before it.The reason for considering the number 56 was not specified,but that doesn't matter because it is clear that to solve the problem in anykind of satisfying way we have to solve it in a more general case, where 56 isreplaced by n. In this case, the probability is P(n)/n!, where P(n) is thenumber of permutations p of (1, … ,n) for which p(k – 1) – p(k) = 1 for exactlyone k. The problem reduces to finding P(n). The answer is rather neat, and (SPOILER ALERT) I give it below.However what I find most interesting is the way my finding the solution to thisproblem depended so much on technology. To solve it I did not need to know muchmathematics at all. If I had had to solve the problem 30 years ago, it wouldhave taken much more knowledge and/or much more time.The first thing I did was to generate some numericalevidence. By writing out the acceptable permutations and counting them, I foundthat P(2) = 1, P(3) = 2, P(4) = 9, and P(5) = 43. I sent the problem to afriend of mine, Ken Cutter, who is a MATLAB guru, and he wrote a short programin MATLAB, using perms(1 … n) to list all permutations and the diff function tocompute the differences p(k) – p(k-1). He discovered that I had missed oneacceptable permutation of (1 … 5) so that actually P(5) = 44. He also gave methe list of a few more values of P, including P(6) = 265.I now knew I was looking for a sequence 1 , 2, 9, 44, 265, …, so I went to the Online Encyclopedia of Integer Sequences (OEIS, https://oeis.org/)and entered 1 , 2, 9, 44, 265 into the search field, and out came two knownsequences. Only the first of these two sequences seemed to relate topermutations, and sure enough, one the descriptions of this sequence identifiedit as what I was looking for.The answer turns out to be that P(n) number of derangementsof P(n), that is, the number of permutations with no fixed points. The formulafor P(n) is given on OEIS as P(n) = n!*Sum((-1)^k/k!, k=0..n). On the Wikipedia page for derangements,it is noted that a common notation for P(n) is !n, so the desired probabilityis !n/n! = Sum((-1)^k/k!,k=0..n) which, asWikipedia notes, converges rapidly to 1/e. In fact, it is the first n terms ofthe Taylor expansion for exp(x) about 0, evaluated at x = -1.  For the original problem with n = 56, theanswer for all practical purposes is 1/e.Lookingback, the most crucial tool in my finding the solution was The OnlineEncyclopedia of Integer Sequences, started by Neil J. A. Sloane, and awonderful resource for anyone involved in mathematical work. MATLAB was alsovery helpful. Without it, I might have still gotten the answer if I had goneover my work and discovered my mistake which made me get P(5) off by one. If Ihad entered the correct first 4 terms, Online Encyclopedia of Integer Sequenceswould have returned 17 sequences, but it would still have been pretty easy tofind the correct sequence from amongst these 17. But MATLAB (and Ken Cutter) saveda lot of time. Lastly, Wikipedia was a help in explaining the result.When I wasin school, none of these resources were available. If I knew more aboutcombinatorial mathematics, I might have known the answer immediately, or atleast recognized it once I generated the numerical examples. Otherwise, I wouldhave had to look at my lists of acceptable permutations more carefully and perhapsderived a recursion relation or induction step that would enable me to find thesolution. I would be interested in hearing from anyone would like to send me asolution to this problem from scratch, imagining that they do not know theanswer already.Thisexperiences brings home to me how much the methods of mathematics research havechanged since I was in school. I wonder what changes we should be making inK-12 math education that take into account these changes.