Videos

The Multinomial Theorem

https://www.youtube.com/watch?v=2UHlMA-Qcg0 The multinomial theorem generalizies the binomial theorem by replacing the power of the sum of two variables with the power of the sum of an arbitrary finite number of variables $$(x_1+x_2+cdots+x_m)^n.$$ This is the result that gives rise to the concept of multinomial coefficients.

The Binomial Theorem

https://www.youtube.com/watch?v=5n8qXRTVUps In the binomial theorem, the goal is to expand the polynomial expression $(x+y)^n$ and find the coefficients after like terms have been collected. This is a very useful result in mathematics as a whole, and it is particularly useful in combinatorics for proving combinatorial identities.

Vandermonde’s Identity

https://www.youtube.com/watch?v=-yLJ8G_i_vM If we have $m$ cats and $n$ dogs, in how many ways can we choose $k$ of the animals? By iterating through the possible combinations of dogs and cats, we can derive a combinatorial identity, which is called Vandermonde’s identity. We derive this identity here using the described combinatorial or committee-forming proof.

Hockey Stick Identity

https://www.youtube.com/watch?v=hk98kUxk-V0 The hockey stick identity in combinatorics tells us that if we take the sum of the entries of a diagonal in Pascal’s triangle, then the answer will be another entry in Pascal’s triangle that forms a hockey stick shape with the diagonal. Although proofs by induction or Pascal’s identity are possible, we show a …

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Sum of Squares of a Row of Pascal’s Triangle

https://www.youtube.com/watch?v=jAcbcWV9CyE Similar to the question of summing a row of Pascal’s triangle, we can consider summing the squares of the entires of a row of Pascal’s triangle. By a combinatorial argument involving cats and dogs, we show that $$binom{n}{0}^2+binom{n}{1}^2+cdots+binom{n}{n}^2=binom{2n}{n}.$$

Sum of a Row of Pascal’s Triangle

https://www.youtube.com/watch?v=1Ovnz74AasA If one takes the sum of a row of entries in Pascal’s triangle, one finds that $$binom{n}{0}+binom{n}{1}+cdots+binom{n}{n}=2^n.$$ In this video, we prove this remarkable combinatorial identity in two ways: by forming a committee and substituting numbers into the binomial theorem.