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Pascal Triangle

In Probability,

(nk)=(n1k1)+(n1k){n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}

Proof

(n1k1)+(n1k){n-1 \choose k-1} + {n-1 \choose k}
=(n1)!(k1)! (nk)!+(n1)!(n1k)! k!= {(n-1)! \over {(k-1)!~(n-k)!}} + {(n-1)! \over {(n-1-k)!~k!}}
=(kk(n1)!(k1)! (nk)!)+((n1)!(n1k)! k!(nk)(nk))= ({k \over k} {(n-1)! \over {(k-1)!~(n-k)!}}) + ({(n-1)! \over {(n-1-k)!~k!}} {(n-k) \over (n-k)})
=(n1)!(k+nk)k! (nk)!=n!k! (nk)!= {{{(n-1)!} (k + n - k)} \over {k!~(n-k)!}} = {n! \over {k!~(n-k)!}}
=(nk)          = {n \choose k} ~~~~~~~~~~ \blacksquare

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