On this pagePascal TriangleIn Probability, (nk)=(n−1k−1)+(n−1k){n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}(kn)=(k−1n−1)+(kn−1) Proof (n−1k−1)+(n−1k){n-1 \choose k-1} + {n-1 \choose k}(k−1n−1)+(kn−1) =(n−1)!(k−1)! (n−k)!+(n−1)!(n−1−k)! k!= {(n-1)! \over {(k-1)!~(n-k)!}} + {(n-1)! \over {(n-1-k)!~k!}}=(k−1)! (n−k)!(n−1)!+(n−1−k)! k!(n−1)! =(kk(n−1)!(k−1)! (n−k)!)+((n−1)!(n−1−k)! k!(n−k)(n−k))= ({k \over k} {(n-1)! \over {(k-1)!~(n-k)!}}) + ({(n-1)! \over {(n-1-k)!~k!}} {(n-k) \over (n-k)})=(kk(k−1)! (n−k)!(n−1)!)+((n−1−k)! k!(n−1)!(n−k)(n−k)) =(n−1)!(k+n−k)k! (n−k)!=n!k! (n−k)!= {{{(n-1)!} (k + n - k)} \over {k!~(n-k)!}} = {n! \over {k!~(n-k)!}}=k! (n