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Question
Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
n · n − 1Cr − 1 = (n − r + 1) nCr − 1
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Solution
\[LHS = n . {}^{n - 1} C_{r - 1} \]
\[ = \frac{n \left( n - 1 \right)!}{\left( r - 1 \right)! \left( n - 1 - r + 1 \right)!} \]
\[ = \frac{n!}{\left( r - 1 \right)! \left( n - r \right)!}\]
\[RHS = \left( n - r + 1 \right) {}^n C_r \]
\[ = \left( n - r + 1 \right) \frac{n!}{\left( r - 1 \right)! \left( n - r + 1 \right)!} \]
\[ = \left( n - r + 1 \right)\frac{n!}{\left( r - 1 \right)! \left( n - r + 1 \right)\left( n - r \right)!} \]
\[ = \frac{n!}{\left( r - 1 \right)! \left( n - r \right)!}\]
∴ LHS = RHS
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