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To be clear, just accepting the pigeonhole principle as an obvious fact is rigorous, because it's so patently obvious The point is connected to (at least) three other points if any of these points are connected to each other, then we have found a triangle of three mutual friends. You want a formal proof, but formality and rigor are distinct notions.
Pigeonhole principle question there is a row of 35 chairs By the pigeonhole principle, the point is either connected to at least three other points or not connected to at least three other points Find the minimum number of chairs that must be occupied such that there is a consecutive set of 4 or more occupied chairs
Another pigeonhole principle question a course has seven elective topics, and students must complete exactly three of them in order to pass the course.
From my book discrete mathematics by rosen, i can't understand the conclusion of the proof If n objects are placed into k boxes, then there is at least one box However, much fewer are aware of the $\bf {probabilistic}$ pigeonhole principle, which answers the question, 'how many do we usually need?' those familiar with the famous birthday problem might have good intuition for this. I have recently heard and read the term dyadic pigeonhole principle (e.g
See these posts by terry tao) However, is dyadic pigeonholing just a special case of classical pigeonholing where there are two (hence the word dyadic) holes Apologies in advance if this post is an overly trivial question. If there are 10,000 people, how many people must have the same birthday (ignoring year)
This is the way i went about this problem
10000 people / 365 days in a year = 27.397 people per day $\\ So one of the exercises we're doing involves the pigeonhole principle I'm asked to prove with the help of the pigeonhole principle that in any set of seven natural numbers, there is always a pair of numbers whose sum or difference is a multiple of 10 I used java to generate a set of random numbers
A = {53, 44, 34, 111, 134, 564, 1}. By, pigeonhole principle the $52^ {nd}$ number is forced to share the same partition set with another of the earlier selected $51$ numbers Thus, the sum of these two numbers is divisible by $100$.
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