How do I find the $n$ $n$ th row of Pascal's triangle?

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Answer 1 · 2015-07-15T11:21:57

The $n$ $n$ th row of Pascal's triangle is:

$((n-1),(0))$ $((n-1),(1))$ $((n-1),(2))$ ... $((n-1), (n-1))$

That is:

$((n-1)!)/(0!(n-1)!)$ $((n-1)!)/(1!(n-2)!)$ ... $((n-1)!)/((n-1)!0!)$

It's generally nicer to deal with the $(n+1)$ th row, which is:

$((n),(0))$ $((n),(1))$ $((n),(2))$ ... $((n),(n))$

or if you prefer:

$(n!)/(0!n!)$ $(n!)/(1!(n-1)!)$ $(n!)/(2!(n-2)!)$ ... $(n!)/(n!0!)$

How do I find the nnnth row of Pascal's triangle?