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Chapter 9: Problem 92

Prove the identity. $$_{n} C_{n}=_{n} C_{0}$$

Short Answer

Expert verified
\( _{n} C_{n}=_{n} C_{0} = 1 \)

Step by step solution

01

Understand the combinatorics notation

The symbol \( _{n} C_{r} \) is read as the number of ways to choose r elements from a set of n elements. It's also known as 'n Choose r'. This quantity is computed using the formula: \( _{n} C_{r}= \frac{n!}{r!(n-r)!} \) where '!' denotes the factorial of a number.
02

Compute the binomial coefficients in the identity

Substitute the values of r in the formula. For \( _{n} C_{n} \), r is equal to n: \( _{n} C_{n}= \frac{n!}{n!(n-n)!}= \frac{n!}{n!0!} \) where 0! is conventionally taken as 1. Hence, \( _{n} C_{n} = 1 \). For \( _{n} C_{0} \), r equals zero: \( _{n} C_{0} = \frac{n!}{0!(n-0)!}= \frac{n!}{0!n!} \). Again using the convention that 0! is 1, then \( _{n} C_{0} = 1 \).
03

Prove the identity

Comparing the results of \( _{n} C_{n} \) and \( _{n} C_{0} \), it's evident that both are equal to 1. Hence, the identity \( _{n} C_{n}=_{n} C_{0} \) is proven.

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Most popular questions from this chapter

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