The mathematical (and other) thoughts of a (now retired) math teacher,
Saturday, 13 March 2010
Neighboring Binomial Equalities
Quick problem, not easy so plan some time.... Can you find a pair of neighboring combinations so that (hey, you can't use comb(2,0)... Good Luck, and if you find one quickly (tell me how, I would be amazed) find the next.
Unless I misread the question... In a couple of minutes with Excel I learned that comb(15,5)=comb(14,6)=3003. I created a row of integers increasing from zero and a column of integers increasing from zero to make a tabular Pascal's triangle. If you align the left numbers in Pascal's triangle, then the equation comb(a,b)=comb(a-1,b+1) refers to a number and its upper-right neighbor. This is very quick, but I have no way of knowing how common it would be beyond this first number that I noticed.
Using the combination factorial definition, we find that this equality would only occur when [a*(b+1)]/[(a-b)(a-b-1)]=1. Solving that equation for integer solutions would reveal other pairs if there are any. Wolfram Alpha gave me only -1 and -3 but possibly would respond to some finesse.
Nate, Nice job, you found the "little" one... I thought it was "hard" because it is hard to narrow the case of when you get integer solutions. Nice use of technology to make Excel do the work for you, but not sure it will get the next one (at least I think it is the next one) it comb(a,b) > 10^28..
Okay Pat, so this has been on my mind for awhile. The next one that I can find is comb(104,39)=comb(103,40). This pair I found by looking at the simplification of the factors that were in the first example. The expression on the left side of [a*(b+1)]/[(a-b)(a-b-1)]=1 was (for the first pair mentioned) 15*6/(10*9), so the factors of two consecutive integers in the denominator were rearranged as the factors of the numbers in the numerator, whose difference is the smaller of the two denominator factors. I suspected that perhaps the 9 being a square was vital in this arrangement, so after a few tries I found that 64 yielded a result, but I can hardly be certain that I didn't miss a smaller pair.
Holy cow, that is awesome clever detecting, and it is the next one, you didn't miss any. I can tell you that I have never noticed (or known anyone who noticed) what you did... I just checked it with the next solution (I think) which is comb(715,272)= comb(713,273) (I'm totally impressed, great work)
Footnote, I tried to download and unzip the Geogebra sketch you had of type I and type II error and it didn't work (just covered this and drew that kind of diagram 20 times this week, poorly)... can you send me a copy unzipped.(patdotballewat eu.dodea.edu).. would love to have it for the future...
After the previous comment I figured out that you can find solutions by looking for integer solutions of a certain quadratic, which is pretty easy with Excel, though the limited precision could give me extra solutions.
(hey, you can't use comb(2,0)... Good Luck, and if you find one quickly (tell me how, I would be amazed) find the next.
5 comments:
Unless I misread the question... In a couple of minutes with Excel I learned that comb(15,5)=comb(14,6)=3003. I created a row of integers increasing from zero and a column of integers increasing from zero to make a tabular Pascal's triangle. If you align the left numbers in Pascal's triangle, then the equation comb(a,b)=comb(a-1,b+1) refers to a number and its upper-right neighbor. This is very quick, but I have no way of knowing how common it would be beyond this first number that I noticed.
Using the combination factorial definition, we find that this equality would only occur when [a*(b+1)]/[(a-b)(a-b-1)]=1. Solving that equation for integer solutions would reveal other pairs if there are any. Wolfram Alpha gave me only -1 and -3 but possibly would respond to some finesse.
Nate, Nice job, you found the "little" one... I thought it was "hard" because it is hard to narrow the case of when you get integer solutions. Nice use of technology to make Excel do the work for you, but not sure it will get the next one (at least I think it is the next one) it comb(a,b) > 10^28..
Thanks for the message..Pat
Okay Pat, so this has been on my mind for awhile. The next one that I can find is comb(104,39)=comb(103,40). This pair I found by looking at the simplification of the factors that were in the first example. The expression on the left side of [a*(b+1)]/[(a-b)(a-b-1)]=1 was (for the first pair mentioned) 15*6/(10*9), so the factors of two consecutive integers in the denominator were rearranged as the factors of the numbers in the numerator, whose difference is the smaller of the two denominator factors. I suspected that perhaps the 9 being a square was vital in this arrangement, so after a few tries I found that 64 yielded a result, but I can hardly be certain that I didn't miss a smaller pair.
Holy cow, that is awesome clever detecting, and it is the next one, you didn't miss any. I can tell you that I have never noticed (or known anyone who noticed) what you did... I just checked it with the next solution (I think) which is comb(715,272)= comb(713,273) (I'm totally impressed, great work)
Footnote, I tried to download and unzip the Geogebra sketch you had of type I and type II error and it didn't work (just covered this and drew that kind of diagram 20 times this week, poorly)... can you send me a copy unzipped.(patdotballewat eu.dodea.edu).. would love to have it for the future...
I think the next solutions are:
comb(714,272)= comb(713,273)
comb(4895,1869)=comb(4894,1870)
comb(33552,12815)=comb(33551,12816)
comb(229970,87840)=comb(229969,87841)
After the previous comment I figured out that you can find solutions by looking for integer solutions of a certain quadratic, which is pretty easy with Excel, though the limited precision could give me extra solutions.
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