How do you simplify #(2n!) / (n!)#?

1 Answer
Cesareo R.
Jul 2, 2016

#(2n!) / (n!) = prod_{k=n+1} ^{k=2n}k#

Explanation:

Suppose that we need

#(8!)/(4!) = (1 cdot 2 cdot 3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8)/(1 cdot 2 cdot 3 cdot 4) = 5 cdot 6 cdot 7 cdot 8#

then

#(2n!) / (n!) = prod_{k=n+1} ^{k=2n}k#

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