Problem 92

Question

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

Step-by-Step Solution

Verified

Answer

$ _{n} C_{n}=_{n} C_{0} = 1 $

1Step 1: 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.

2Step 2: 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 $.

3Step 3: 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.

Problem 91

Problem 92

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