Question #261054

State and prove Pascal’s identity using the formula for (nk){n \choose k}


1
Expert's answer
2021-11-08T15:36:40-0500

Pascal's identity states that (nk)=(n1k1)+(n1k){n \choose k} = {n-1 \choose k-1}+{n-1 \choose k}, where (nk)=n!(nk)!k!{n \choose k}={\frac{n!} {(n-k)!*k!}}

(n1k1)+(n1k)=(n1)!(n1k+1)!(k1)!+(n1)!(n1k)!k!=(n1)!(nk)!(k1)!+(n1)!(nk1)!k!=k(n1)!+(nk)(n1)!(nk)!k!=k(n1)!+n(n1)!k(n1)!(nk)!k!=n!(nk)!k!{n-1 \choose k-1}+{n-1 \choose k}= {\frac{(n-1)!} {(n-1-k+1)!*(k-1)!}}+{\frac{(n-1)!} {(n-1-k)!*k!}}={\frac{(n-1)!} {(n-k)!*(k-1)!}}+{\frac{(n-1)!} {(n-k-1)!*k!}}={\frac{k*(n-1)!+(n-k)*(n-1)!} {(n-k)!*k!}}={\frac{k*(n-1)!+n*(n-1)!-k*(n-1)!} {(n-k)!*k!}}={\frac{n!} {(n-k)!*k!}}

The statement has been proven


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