Problem 62 - College Algebra with Modeling and Visualization

Question

Evaluate the expression. \(C(10,2)\)

Step-by-Step Solution

Answer

The value of \(C(10,2)\) is 45.

Step 1: Understand the Combination Formula

The expression \(C(10,2)\) refers to the number of combinations of 10 items taken 2 at a time. This is calculated using the formula for combinations: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] where \(n\) is the total number of items, and \(r\) is the number of items to choose.

Step 2: Substitute Values into the Formula

Substitute \(n = 10\) and \(r = 2\) into the combination formula:\[ C(10, 2) = \frac{10!}{2!(10-2)!} \] This simplifies to \[ C(10, 2) = \frac{10!}{2! \cdot 8!} \]

Step 3: Simplify the Factorial Expressions

Calculate \(10!\), \(2!\), and \(8!\):- \(10! = 10 \times 9 \times 8!\) - \(2! = 2 \times 1 = 2\) - \(8!\) remains unchanged because it will cancel out. Replace these in the expression:\[ C(10, 2) = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} \]

Step 4: Cancel Out the Common Terms

The \(8!\) in the numerator and the denominator cancel each other out, simplifying the expression to: \[ C(10, 2) = \frac{10 \times 9}{2 \times 1} \]

Step 5: Calculate the Remaining Expression

Continue with the arithmetic:\[ C(10, 2) = \frac{90}{2} = 45 \]

Key Concepts

Factorial

The factorial, denoted by an exclamation point as in "10!", is a mathematical operation used in various calculations, including permutations and combinations. Here's how it works: you multiply a series of descending natural numbers, starting from the number itself down to 1.
For example, the factorial of 4, written as "4!", is calculated as:

• 4! = 4 × 3 × 2 × 1 = 24

Factorials grow very quickly as numbers increase. This property is useful in solving problems involving large sets of items where you need to calculate the number of ways to arrange or choose these items. Factorials are the backbone of the binomial coefficient, which is used to find combinations.

Binomial Coefficient

The binomial coefficient, commonly expressed as \( C(n, r) \) or \( \binom{n}{r} \), is a fundamental concept in combinatorics. It represents the number of ways to choose \( r \) items from a total of \( n \) items without considering the order.
To calculate it, you use the formula:

• \( C(n, r) = \frac{n!}{r!(n-r)!} \)

This formula is just saying that the total number of ways to select \( r \) objects from \( n \) objects hinges on dividing the total number of arrangements \( n! \) by the arrangements of those \( r \) objects \( r! \) and the remaining unchosen objects \((n-r)!\).
In our exercise with \( C(10, 2) \), by substituting \( n = 10 \) and \( r = 2 \), we divide \( 10! \) by \( 2! \cdot 8! \) to find the number of combinations, which turns out to be 45 after simplification and calculation.

Permutation and Combination Concepts

Permutations and combinations are two key concepts in probability and combinatorics. Both deal with ways to arrange or select items from a collection.
A permutation considers all possible arrangements of a set, where the order matters. For example, the permutation of choosing 3 items out of 4 is different if chosen in a different order.
Combinations, on the other hand, do not consider the order of selection. When calculating combinations, as illustrated by the previous exercise with \( C(10, 2) \), all that's crucial is which items are selected, not the sequence of selection.
Here’s how combinations differ from permutations:

• Permutations: More concerned with arranging. Formula: \( P(n, r) = \frac{n!}{(n-r)!} \)
• Combinations: More concerned with selecting. Formula: \( C(n, r) = \frac{n!}{r!(n-r)!} \)

Understanding these concepts can help solve a wide range of mathematical problems related to counting and probability.