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Chapter 9: Problem 93

Prove the identity. $$_{n} C_{n-1}=_{n} C_{1}$$

Short Answer

Expert verified
The identity \( _nC_{n-1} = _nC_{1} \) is proven by using the combination formula. After simplification, both sides of the equation are equal, thus confirming the identity.

Step by step solution

01

Write the formula for combination

First, write the formula for combinations for each side of the identity separately: \( _nC_{n-1} = \frac{n!}{(n-1)!(n-(n-1))!} = \frac{n!}{(n-1)!1!} \) and \( _nC_{1} = \frac{n!}{1!(n-1)!} \).
02

Simplify

Simplify the equations. This yields \( _nC_{n-1} = \frac{n!}{(n-1)!} \) and \( _nC_{1} = \frac{n!}{(n-1)!} \).
03

Compare results

After simplification, both sides of the identity are equal. Thus, it is proven that \( _nC_{n-1} = _nC_{1} \).

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