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

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Answer 1 · 2015-07-15T11:21:57

The #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!)#