Show that: $nCr = n-1Cr + n-1Cr-1$ ??

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Answer 1 · 2018-03-24T13:12:29

See the proof below

This is the proof of the Pascal's Triangle

$RHS=((n-1),(r))+((n-1),(r-1))$

$=((n-1)!)/((n-r-1)!(r)!)+((n-1)!)/((n-r)!(r-1)!)$

$=(n-1)!(1/((n-r-1)!(r)!)+1/((n-r)!(r-1)!))$

$=(n-1)!((n-r)/((n-r)(n-r-1)!(r)!)+r/((n-r)!r(r-1)!))$

$=(n-1)!((n-r)/((n-r)!(r)!)+r/((n-r)!(r)!))$

$=(n-1)!*n/((n-r)!(r)!)$

$=(n!)/((n-r)!(r)!)$

$=((n),(r))$

$=LHS$

$QED$

Show that: nCr = n-1Cr + n-1Cr-1nCr = n-1Cr + n-1Cr-1 ??