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

[allphotoscloud]

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.

[bstrouble]

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..

[TheAssembler-dryo]

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 😁😁

[anthonychapov]

This is beautiful. I am a CS student and would rather pay my tuition to you! Thank you a lot!

[Christian-envb]

Thank you for the amazing lectures. You are such a great teacher. I hope you can keep uploading math courses. Greetings from Mexico.

[quiksilver]

As a cat in Brighton I can confirm the pigeonhole principle helps the pigeons successfully get away from us

[jh_esports]

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 ❤

[minhnhat]

love the way you explain things, definitely gonna subscribe

[abbasborji]

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)

[geraldfrancis]

U jst made me pass my paper yesterday... My lecturer doesn't know how to teach...

[Yumytumy]

your explanations are very clear, thank you very much, greetings as well from france

[arnesh]

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 < 18.29 => 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.

[nazrie]

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

[swxstik]

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?

[jmsaucedo]

Thank you, these videos are very useful

[prashantnaik]

Thank you for the video, but i didn't got the 10 red balls question, isn't N=20 since we add 10 red and 10 yellow balls?

[helo]

Thanks for uploading these videos. They really do help !

[karumotoart]

for the last computer question, could we argue using a cycle graph?

[gaopinghu]

Excuse me, is there a demonstration of this principle? I'd like to know it.

[karlmoreno]

In the more practice section in problem a. Why do you subtract N-1?