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Pascal Triangle

In Probability,

(nk)=(nโˆ’1kโˆ’1)+(nโˆ’1k){n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}

Proofโ€‹

(nโˆ’1kโˆ’1)+(nโˆ’1k){n-1 \choose k-1} + {n-1 \choose k} =(nโˆ’1)!(kโˆ’1)!ย (nโˆ’k)!+(nโˆ’1)!(nโˆ’1โˆ’k)!ย k!= {(n-1)! \over {(k-1)!~(n-k)!}} + {(n-1)! \over {(n-1-k)!~k!}} =(kk(nโˆ’1)!(kโˆ’1)!ย (nโˆ’k)!)+((nโˆ’1)!(nโˆ’1โˆ’k)!ย k!(nโˆ’k)(nโˆ’k))= ({k \over k} {(n-1)! \over {(k-1)!~(n-k)!}}) + ({(n-1)! \over {(n-1-k)!~k!}} {(n-k) \over (n-k)}) =(nโˆ’1)!(k+nโˆ’k)k!ย (nโˆ’k)!=n!k!ย (nโˆ’k)!= {{{(n-1)!} (k + n - k)} \over {k!~(n-k)!}} = {n! \over {k!~(n-k)!}} =(nk)ย ย ย ย ย ย ย ย ย ย โ– = {n \choose k} ~~~~~~~~~~ \blacksquare

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