Discrete Math II - 6.2.1 The Pigeonhole Principle
In this video, we will explore the Pigeonhole Principle, which is a topic we didn't touch on in Discrete Math I. The concept itself it quite simple, stating that if we have more objects than places to put objects, then one of the places will have to have more than one object (paraphrased, of course). We will focus on the direct applications of the principle in this video, with the more complex applications saved for 6.2.2.
Video Chapters:
Intro 0:00
The Pigeonhole Principle Introduced 0:06
Easy Pigeonhole Practice 0:57
Generalized Pigeonhole Principle 2:15
Pigeonhole Practice 8:07
More Practice 10:47
Up Next 14:03
This playlist uses Discrete Mathematics and Its Applications, Rosen 8e Power Point slide decks to accompany the videos can be found here: bellevueuniversity-my.sharepo...
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Commenting on every single video in this playlist because you deserve any revenue that you can pickup from this whole series on youtube as you're the only teacher who has taken time and effort to concentrate on one book topic wise in such a meticulous and proper manner. Every single topic is covered and the consistency is amazing.
You are my saving grace for discrete math! There arent too many discrete math videos here on youtube unlike the common subjects such as algebra, calculus, etc..
2 years later and this series is still a great help to students around the globe, so thankful you uploaded this publicly instead of just for your students 😁😁
You are SO amazing! I don‘t need this for a class but simply like this kind of math and you couldn’t have explained it any better ❤
U jst made me pass my paper yesterday... My lecturer doesn't know how to teach...
@SawFinMath
2 жыл бұрын
Sorry 😞 But glad I could help!
@kedareswarp7402
10 ай бұрын
Which college are you from
Thanks for uploading these videos. They really do help !
This is beautiful. I am a CS student and would rather pay my tuition to you! Thank you a lot!
Thank you for the amazing lectures. You are such a great teacher. I hope you can keep uploading math courses. Greetings from Mexico.
@SawFinMath
2 жыл бұрын
Greetings! I'm working on redoing Discrete Mafh II and Calc I, as well as working on Abstract Algebra.
your explanations are very clear, thank you very much, greetings as well from france
Beautiful. Thanks!
Thank you, these videos are very useful
Thank you for a plain video
Thanks for the hard work you put into this course! The whole list helped me prepare for my final exam in Discrete Mathematics. However, kindly note that there are some tiny topics missing from the list. (ex: Chinese Remainder Theorem / Fermat's Little Theorem in section 4.4)
@SawFinMath
Жыл бұрын
Thanks! My students don't cover section 4.4 so I haven't created videos for that section.
@calvinripley9093
Жыл бұрын
Those damn Chinese
Thank you
As a pigeon who lives at Brighton I can confirm that I use this principle when we try to escape from cats
@SawFinMath
5 ай бұрын
Hahahhahahaha
Hi Kimberly, @12:21 I'm thinking of the 677 different classes as the pigeons. Then k amount of rooms as the pigeonholes. Now, I want to stuff at most 38 different classes into such rooms. ceil(677/k) = 38 => k k = floor(18.29) = k = 18. 18 such rooms. Is the validity of my argument correct? It made sense for me to visualize the pigeonholes as the rooms.
In the more practice section in problem a. Why do you subtract N-1?
for the last computer question, could we argue using a cycle graph?
Thank you for your explanation and you really are amazing. However, about the question that number of students from 50 states, aren't we asked what will be the minimum number of students if each states at least has 100 students? But if we go with your answer it will be " there is at least one state which has 100 members". I don't know if you get me or not but please answer for me
U are the best
@SawFinMath
2 жыл бұрын
Thanks!
hey @ 10:55 idk why there was an n-1 and 100-1 pls reply if u can
thankssss
Can you explain how we arrive at N-1 ?
Hey, sorry I'm late to this but at 9:15 I am confused, how did you get N/2. Could you maybe elaborate a little further if you read this?
@SawFinMath
Жыл бұрын
We are looking at the ceiling function of n/k where k represents the number of boxes. Since there are two colors, essentially that means 2 boxes. Does that make sense?
@swxstik
Жыл бұрын
@@SawFinMath yess, understood it now. Thanks for the response :)
Excuse me, is there a demonstration of this principle? I'd like to know it.
Thank you so much teacher. I sent you an email I hope you saw it. Thanks again!
@SawFinMath
8 ай бұрын
I haven’t seen it. You may have to try to send it again.
@Tubeswax
8 ай бұрын
@@SawFinMath I sent it through your Gmail address 😢
the book is by kenneth rosen right ?? ... i have similar problems in that book
@madhavgupta2002
Жыл бұрын
Yeah It's given in description too
Hey Kim, Can you please explain why K is 5 in the problem of computer networks why can't it be 6 where all computers are connected to all other computers?
@_7__716
Ай бұрын
I think because a computer can't be connected to itself, at most it would connect to the 5 others
Mam please upload abstract algebra lectures ...
@SawFinMath
2 жыл бұрын
They are on the way! Hopefully by the end of summer.
What textbook do you advise using
@SawFinMath
9 ай бұрын
This series uses the Rosen text
@AmCanTech
9 ай бұрын
@@SawFinMath thank you so much! great videos btw... for recurrence problems, we need to do it with complex roots... does your video on the subject happen to have that?
@SawFinMath
9 ай бұрын
@@AmCanTech I believe I have one example with complex roots, yes
@AmCanTech
9 ай бұрын
@@SawFinMath Thank you so much!!
Can anyone help me solve this problem: From 1 to 100 (including 1 and 100), at least how many numbers are drawn randomly to ensure that there are two numbers which have 2 or more common factors?
@hr3nk
9 ай бұрын
you need to count all prime numbers from 1 to 100, let's say it's k. Then whatever next number you pick is not prime and can be factored with primes from 1 to 100. And for any prime from this decomposition you can say that it and number you factored have 2 common factors - 1 and prime number itself. So answer would be k+1, where k is number of primes from 1 to 100.
621 (sorry)
@cara1629
Жыл бұрын
you have been seen and recognized! this is the best place to be talking about this XD
4:43 why cant i have 50 people all born in the same month?
not very clear