How Pascal’s Triangle Teaches Us How To Count

How Pascal’s Triangle Teaches Us How To Count - Sum Of A Row

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10 ай бұрын

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There are lots of patterns within Pascal's triangle. What do these patterns teach us about counting? This video examines the sum of a row of Pascal's triangle, how to prove the pattern, and describes some different abstract approaches to problems like this.
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0:04 Sum of a Row
1:22 A Counting Problem - Solution 1
3:14 A Counting Problem - Solution 2
5:03 Pascal's Triangle Proof
5:57 Comparing Approaches
8:26 Conclusion - Three Perspectives
Intro riff taken from: Nikolai Kapustin - 8 Concert Etudes, Op. 40: III. Toccatina
Music Credit:
Moving Ahead / Megan Wofford
Cottage In Wales / Sam Eber
Elegance Becomes Her / Howard Harper-Barnes
Blues for the Lonesome / Rikard From
courtesy of www.epidemicsound.com
This video was sponsored by Brilliant.

Пікірлер: 13

@zhulimath 9 ай бұрын

To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/zhuli . The first 200 of you will get 20% off Brilliant’s annual premium subscription.

@gametimewitharyan6665 9 ай бұрын

I don't understand why your videos have so less views, there are pure gold

@spiderjerusalem4009 9 ай бұрын

Principles and Techniques in Combinatorics by Chen Chuan Chong and Koh Khee Meng Undergraduate combinatorics by Titu Andreescu (and indeed his 102 combin problems) introductory combinatorics by richald brualdi. These are top recommended combinatoric books for those who may have interest in the subject

@zhulimath 9 ай бұрын

Not only are these resources excellent, but one of these I forgot about and have been searching for for a while now. Thanks for sharing!

@kyay10 9 ай бұрын

Edit: turns out you've addressed this. This row doubling proof is actually really useful because it generalises to seeing patterns in recursive functions. Pascal's triangle is defined recursively, and that's why we can reason about it inductively like that Currently watching the video, so not sure if you've addressed this, but a simple way to show the 2^n relationship is inductively. Base case is 2^0 = 1, and inductively you can show that for row k+1, its sum will be "double counting" every element of row k, (notice that the 1s at the edges always exist, so in a way they're counted twice as well), and hence the sum of row k+1 is 2*2^k = 2^(k+1)

@Alkhazov94 9 ай бұрын

Amazing explanation (as always)!!! If you replace “toppings” with 1, 10, 100, 1000 (in binary). You will get this table: { } {1} {1, 10 } {1, 10, 100 } {1, 10, 100, 1000} {10} {1, 100 } {1, 10, 1000} {100} {1, 1000 } {1, 100, 1000} {1000} {10, 100 } {10, 100, 1000} {10, 1000 } {100, 1000} Meaning that { } = 0 up to {1, 10, 100, 1000} = 1111. I think this gives additional insight into why sums of the rows are all powers of 2.

@kyay10 9 ай бұрын

The power set argument you presented is beautiful, and especially because it leads to the simple fact that any function that goes from a domain A to a codomain B has |B|^|A| possibilities, and so when going from set S to a Boolean, you have 2^|S| possibilities, which is why that's even the notation for a power set