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

The **Answer to the Question**

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**Here's the Solution to this Question**

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

${n-1 \choose k-1}+{n-1 \choose k}= {\frac{(n-1)!} {(n-1-k+1)!*(k-1)!}}+{\frac{(n-1)!} {(n-1-k)!*k!}}={\frac{(n-1)!} {(n-k)!*(k-1)!}}+{\frac{(n-1)!} {(n-k-1)!*k!}}={\frac{k*(n-1)!+(n-k)*(n-1)!} {(n-k)!*k!}}={\frac{k*(n-1)!+n*(n-1)!-k*(n-1)!} {(n-k)!*k!}}={\frac{n!} {(n-k)!*k!}}$

The statement has been proven