Problem 55 - Precalculus with Limits

Question

In Exercises 51 - 56, evaluate \( _nC_r \) using the formula from this section. \( _{25}C_0 \)

Step-by-Step Solution

Answer

The value of \( _{25}C_0 \) is 1.

Step 1: Identify the values

We are given the following to evaluate: \( _{25}C_0 \). So in this case, \(n = 25\) and \(r = 0\).

Step 2: Apply the Combination Formula

The formula for the combination is \(_nC_r = \frac{n!}{r!(n-r)!}\). When you put the values in the formula, we get \(_{25}C_0 = \frac{25!}{0!(25-0)!}\).

Step 3: Simplify the expression

0! is always equal to 1. Replacing 0! with 1 in our formula gives us \(_{25}C_0 = \frac{25!}{1 * 25!} = 1\) as both 25! in the numerator and the denominator will cancel out each other.

Key Concepts

Factorial

A factorial, denoted by the exclamation mark (!), is a basic mathematical function used to multiply a series of descending natural numbers. For any non-negative integer \( n \), the factorial is defined as follows:

• \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \)

This means that you start at the number \( n\) and multiply it by each natural number before it until you reach 1.

For example:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 1! = 1 \)
• By convention, \( 0! = 1 \)

The factorial function is essential in combinations, permutations, and other areas of mathematics because it simplifies the calculation of large numbers. The concept of a factorial helps reduce complex multiplication into simple, manageable steps, making calculations quicker and more efficient.

Binomial Coefficient

The binomial coefficient, often represented as \( \binom{n}{r} \) or \( _nC_r \), is a fundamental part of combinatorics that describes the number of ways to choose \( r \) items from \( n \) total items without regard to the order of selection.

The formula for the binomial coefficient is:

• \( _nC_r = \frac{n!}{r!(n-r)!} \)

This formula can be broken down into the following parts:

• The numerator, \( n! \), accounts for arranging all \( n \) items.
• The denominator, \( r! \), adjusts for the permutations of the \( r \) selected items.
• The additional factor \( (n-r)! \) ensures that we do not overcount the permutations of the \( n-r \) remaining items.

Let's consider the solved example from the original exercise: we are asked to calculate \( _{25}C_0 \).

Using the above formula, we get:

• \( _{25}C_0 = \frac{25!}{0!(25-0)!} = \frac{25!}{1 \times 25!} = 1 \)

Here, the factorial terms cancel each other out, simplifying the calculation significantly. The result is exactly as expected: when choosing zero items from a set, there is only one possible way, which is to choose nothing at all.

Permutation and Combination

Permutations and combinations, though related, describe different ways to arrange or select objects.

• Permutation: Permutations consider the order of arrangement. If the sequence or positioning of items matters, you're dealing with permutations. For example, arranging books on a shelf in a specific sequence is a permutation.
• Mathematically, it is expressed as \( _nP_r = \frac{n!}{(n-r)!} \) where you arrange \( r \) items picked from \( n \) distinct items.
• Combinations, on the other hand, ignore the order of items. Only the selection of items matters, not how they are arranged. This concept is captured in the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) which we've discussed in detail previously.

An easy way to remember this is:

• If order doesn't matter, use combinations.
• If order does matter, use permutations.

For example, in a lottery draw, the order in which the numbers are selected doesn't matter, hence a combination is used. However, if you're organizing a race, the order in which people finish matters, so you would use permutations.
Understanding permutations and combinations is crucial in problem-solving because they provide a framework to tackle a wide range of real-world scenarios, from probability calculations to decision-making processes.