Problem 24 - College Algebra with Corequisite Support

Question

For the following exercises, compute the value of the expression. $$ C(10,3) $$

Step-by-Step Solution

Answer

The value of \( C(10,3) \) is 120.

Step 1: Understanding the Problem

We need to compute the value of the combination expression \( C(10,3) \), which represents the number of ways to choose 3 elements from a set of 10 elements without regard to the order of selection.

Step 2: Use the Combination Formula

The formula for combinations is given by \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of elements, and \( k \) is the number of elements to choose. In this problem, \( n = 10 \) and \( k = 3 \).

Step 3: Calculate Factorials

Calculate the factorials needed for the formula:- \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)- \( 3! = 3 \times 2 \times 1 \)- \( (10-3)! = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)

Step 4: Plug Factorials into Formula

Substitute the factorials into the combination formula: \[ C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!} \] Notice that the \(7!\) terms cancel each other out.

Step 5: Simplify the Expression

Now we have: \[ C(10,3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \] Simplify step by step:1. Calculate \( 10 \times 9 \times 8 = 720 \)2. Calculate \( 3 \times 2 \times 1 = 6 \)3. Divide the results: \( \frac{720}{6} = 120 \)

Step 6: Verify the Result

Re-check the division and multiplication steps to ensure the computations are correct. The final calculation confirms that \( C(10,3) \) equals 120.

Key Concepts

Factorials

A factorial, denoted with an exclamation point as "!", is a fundamental concept often used in combinatorics, calculus, and algebra. Calculating the factorial of a number means multiplying all positive integers less than or equal to that number.
For instance, when we see "10!", it means 10 multiplied by all the numbers below it, down to 1:

• 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800

Factorials grow very quickly with increasing values, but they are instrumental in calculating permutations and combinations.
They form the backbone of the formulas used to explore how elements can be arranged or selected in various orders.

Permutations

When we talk about permutations, we're dealing with arrangements of items where the order matters. Imagine having a set of items, and you're interested in understanding how many ways these items can be ordered.
The basic formula for permutations involves factorials and is written as:

• P(n, k) = \( \frac{n!}{(n-k)!} \)

In this expression, "n" represents the total number of items you're choosing from, and "k" is the number of items to arrange.
For example, if you have 3 books and you want to arrange them on a shelf, you would use the permutation formula to find out that you have 6 possible orders.
Permutations are distinct from combinations, where the order doesn't matter, and we'll delve into that next.

Binomial Coefficients

Binomial coefficients, often denoted as \( C(n, k) \), or "n choose k," define combinations. They express how many ways you can choose "k" elements from a total of "n" elements without considering the order.
The combination formula is:

• C(n, k) = \( \frac{n!}{k!(n-k)!} \)

In our specific problem, \( C(10, 3) \) represents the count of ways you can pick 3 items from a group of 10. Let's break down the problem using this formula:

• Calculate the large factorial: \( 10! \).
• Compute \( 3! \) and \( 7! \) (since \( 10-3 = 7 \)).
• Apply these factorials in the formula: \( C(10,3) = \frac{10!}{3! \, 7!} \).

In our step-by-step solution, this simplified to \( \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \), confirming that the number of combinations equals 120.
These coefficients are extensively used in binomial expansions and are key components in statistics and probability.