what is 2n choose n

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what is 2n choose n The symbol is usually read as n choose k because there are ways to choose an unordered subset of k elements from a fixed set of n elements For example there are ways to choose 2 elements from 1 2 3 4 namely 1

Compute answers using Wolfram s breakthrough technology knowledgebase relied on by millions of students professionals For math science nutrition history The closely related Catalan numbers C n are given by C n 1 n 1 2 n n 2 n n 2 n n 1 for all n 0 displaystyle C n frac 1 n 1 2n choose n 2n choose n 2n choose n 1 text for all n geq 0

what is 2n choose n

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what is 2n choose n
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Combinatorial Proof C 2n 2 2 C n 2 N 2 Combinatorial Proofs 3
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On the right side you re choosing n objects from 2n objects On the left side it s equal to n k n n k So divide the 2n objects into 2 groups both of n size Then the total number of way of choosing n objects is partitioning over how many elements you choose from one group and the remaining n k elements from the other group In full generality the binomial theorem tells us what this expansion looks like small begin align a b n C 0a n C 1a n 1 b C 2a n 2 b 2 C nb n end align a b n C 0an C 1an 1b C 2an 2b2 C nbn where C k C k C k is the number of all possible combinations of k

The theorem defined in binomial coefficient as 2n choose n frac 2n n 2 for n geq 0 and it approaches frac 4 n sqrt pi n asymptotically Given Stirling s formula we have Submit your answer In the expansion of 2x frac k x 8 where k is a positive constant the term independent of x is 700000 Find k Show that 2 n sum k 0 n n choose k Proof Set x y 1 in the binomial series to get

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Theorem limn 2n n 4n n lim n 2 n n 4 n n Proof From Approximation to x y y x y y limx y x y y 1 2 1 x 1 y 1 y x x 1 x y y lim x y x y y 1 2 1 x 1 y 1 y x x 1 x y y Thus Sources 23 So i was given this question Show by combinatorial argument that 2 n 2 2 n 2 n 2 Here is my solution Given 2 n objects split them into 2 groups of n A and B 2 combinations can either be assembled both from A both from B or one from each There are n 2 from A and n 2 from B

The sequence 4 n 2n n n c eventually decreases if c 1 4 this is straightforward divide the squares of two consecutive terms and subtract 1 and its limit equals 1 Thus for large enough n we get the upper bound 1 n c For c 1 this just works from the very beginning Fedor Petrov Obviously 1 x n 1 x n 1 x 2n On the right the coefficient by x n is 2n choose n On the left it is sum k 0 n n choose k n choose n k In fact with a little change of the language we can prove a more general identity displaystyle sum k ge 0 n choose k m choose k n m choose n

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what is 2n choose n - In full generality the binomial theorem tells us what this expansion looks like small begin align a b n C 0a n C 1a n 1 b C 2a n 2 b 2 C nb n end align a b n C 0an C 1an 1b C 2an 2b2 C nbn where C k C k C k is the number of all possible combinations of k