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Fungrim entry: 0d92f6

(2nn)=(2n)!(n!)2{2 n \choose n} = \frac{\left(2 n\right)!}{{\left(n !\right)}^{2}}
Assumptions:nZ0n \in \mathbb{Z}_{\ge 0}
{2 n \choose n} = \frac{\left(2 n\right)!}{{\left(n !\right)}^{2}}

n \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
Binomial(nk){n \choose k} Binomial coefficient
Factorialn!n ! Factorial
Powab{a}^{b} Power
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(Equal(Binomial(Mul(2, n), n), Div(Factorial(Mul(2, n)), Pow(Factorial(n), 2)))),
    Assumptions(Element(n, ZZGreaterEqual(0))))

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2021-03-15 19:12:00.328586 UTC