Prove that Sum(n choose r) = 2^n

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In this video we prove that Sum(n choose r) = 2^n. This proof uses the binomial theorem.

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  1. The Math Sorcerer
  2. In this video we prove that Sum(n choose r) = 2^n. This proof uses the binomial theorem.
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Комментарии
Автор

Worth mentioning the story proof. 2^n is the number of subsets of a set with n elements. n choose r is ur number of subsets of size r. This equation tells us that the total number of subsets of a set with n elements is the number of subsets with 0 elements plus the number of subsets with 1 element plus the number of subsets with 2 elements, and so on.

Singularitarian
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I mean since u choose binomial theorem as a mean of proofing instead of induction over n u should probably proof the theorem to begin with before using it, but very nice video professor.

greyshield
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Shouldnt you be doing these by induction over n?

pablote
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Now I am wondering why I was not able to figure this out!.... thank you so much ...fully understood

seriously-
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would'nt n=2, 3, 4, 5... and then r=n-1 so that the sum is always 1

neilravindran
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Can you please prove the same expression using the induction method please?

feryfez
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MS I have somethings clarify..
How could I contact you ?

hussainfawzer
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Can I represent 3^n using a sum like this or does it only work with two

gyattrizzV
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how would the inductive step work with induction ?

sportmaster
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I thought he'd do induction but I guess this works as well.

__Hmmmmmmmm__
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This is also valid for any pair of integers "" x "" and "" y "" : ( x + y ) = N, where "" x "" and "" y "" are integers, and then "" N "" is also an integer . Then if "" x "" = q, "" y "" = ( N -- q ), then ( x + y ) = N, and so ( x + y )^ n = ( N )^ n, where N = 2, 3, 4, 5, 6, 7, Am I wrong ???

rayvianasampaio
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Hola, podrías explicarme porque 1^n-r * 1^r se cancelan

jhonatanstivendpenaobando
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i don't even study math anymore, why'd i watch this, pretty cool tho

MegaDevilsdad
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That binomial coefficients appear in the binomial theorem is highly related to the number of subsets of cardinality r, so while a correct proof, it would be more pedagogically sound to prove this identity combinatorially.

andrewzhang