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Chapter 7: Problem 378

Find the value of \(\mathrm{C}(\mathrm{n}, 0)\).

Short Answer

Expert verified
The value of \(\mathrm{C}(\mathrm{n}, 0)\) is always 1, regardless of the value of n.

Step by step solution

01

Recall the formula for combinations

The formula for combinations is given by \(\mathrm{C}(\mathrm{n}, \mathrm{k})=\frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}\), where \(\mathrm{n}!\) represents the factorial of n (the product of all the positive integers less than or equal to n).
02

Substitute the value of k

Since we want to find the value of \(\mathrm{C}(\mathrm{n}, 0)\), substitute "0" for k in the formula: \(\mathrm{C}(\mathrm{n}, 0)=\frac{\mathrm{n}!}{0!(\mathrm{n}-0)!}\).
03

Simplify the expression

Recall that 0! = 1, so the formula becomes: \(\mathrm{C}(\mathrm{n}, 0)=\frac{\mathrm{n}!}{1*(\mathrm{n}-0)!}\). As \(\mathrm{n}-0=\mathrm{n}\), the formula can be written as: \(\mathrm{C}(\mathrm{n}, 0)=\frac{\mathrm{n}!}{1*(\mathrm{n})!}\).
04

Evaluate the expression

The expression now simplifies to: \(\mathrm{C}(\mathrm{n}, 0)=\frac{\mathrm{n}!}{\mathrm{n}!}\). Since the factorial of any number divided by the same factorial is 1, the combination expression evaluates to: \(\mathrm{C}(\mathrm{n}, 0)=1\). So, the value of \(\mathrm{C}(\mathrm{n}, 0)\) is always 1, regardless of the value of n.

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