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Monday, 26 October 2015

Grove HSC 3U book EX 10.1 Q15 Pascal Triangle identity

Prove \left({n\atop k}\right)=\left({n-1\atop k-1}\right)+\left({n-1\atop k}\right)


LHS =\left({n\atop k}\right)={n!\over k! (n-k)!}

RHS =\left( {n-1\atop k-1}\right)+\left({n-1 \atop k}\right)

={(n-1)!\over (k-1)! (n-1-(k-1))!}+ {(n-1)!\over k! (n-1-k)!}

={(n-1)!\over (k-1)! (n-k)!}+ {(n-1)!\over k! (n-k-1)!}

={k(n-1)!\over k(k-1)! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)(n-k-1)!}

={k(n-1)!\over k! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)!}

={k(n-1)! + (n-1)!(n-k)\over k! (n-k)!}

={(n-1)!(k + (n-k))\over k! (n-k)!}

={(n-1)!n\over k! (n-k)!}

={n!\over k! (n-k)!}
=LHS

as required.

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