How Can the Binomial Theorem Be Derived from Discrete Math Concepts?

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Discussion Overview

The discussion revolves around the derivation of the binomial theorem from concepts in discrete mathematics, particularly focusing on combinatorial identities and their implications. Participants explore the relationship between the powerset of a set and the binomial coefficients.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant notes the identity |P(S)|=2^n for a set S with n elements and connects it to the sum of binomial coefficients: C(n,0)+C(n,1)+...+C(n,n)=2^n.
  • Another participant suggests that induction could be used to prove the identity mentioned.
  • A participant questions how to transition from the identity \sum^{n}_{i=0}\left(\stackrel{n}{i}\right)=2^n to the binomial expansion \sum^{n}_{i=0}\left(\stackrel{n}{i}\right)x^{n-i}*y^{i}=(x+y)^{n}, expressing uncertainty about the steps involved.
  • Another reply points out that setting x=y=1 provides an easy way to see the first identity but questions how to derive the second from the first, noting that the second contains "more information."

Areas of Agreement / Disagreement

Participants express varying levels of understanding and uncertainty regarding the derivation process. There is no consensus on the steps needed to connect the two identities or on the correctness of the initial claims.

Contextual Notes

The discussion highlights potential gaps in understanding the transition between combinatorial identities and their applications in the binomial theorem, with participants not fully resolving the steps or assumptions involved.

Enjoicube
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Ok, so after a little discussion with my discrete math teacher today, he sent me on a little "quest". Here is how it happened:

The topic we were covering was set theory, and as I had been studying very basic combinatorics the night before, I noticed something about the powerset, namely:

Assuming a set S with n elements:

|P(S)|=2^n

however, if S has n elements, and the powerset is compose of S's subsets, then:

|P(S)|= C(n,0)+C(n,1)...+C(n,n)

so

C(n,0)+C(n,1)...+C(n,n)=2^n

so

[tex]\sum^{n}_{i=0}(\stackrel{n}{i})=2^n[/tex]

I asked about this after class, and he said the binomial theorem could be derived through this identity, I sort of see how, but I doubt the corectness of these ways. Does anybody know about a way to go about this?
 
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If you mean how to prove this identity, induction would do a fine job.
 
I guess what I really meant was, is there any step you can see that can be taken to get from:

[tex]\sum^{n}_{i=0}\left(\stackrel{n}{i}\right)=2^n[/tex]

to

[tex]\sum^{n}_{i=0}\left(\stackrel{n}{i}\right)x^{n-i}*y^{i}=(x+y)^{n}[/tex]
 
The other way is easy: let x= y= 1. I don't see how, just from the first, you can get to the second: the second contains "more information" than the first.
 

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