Problem 24 - College Algebra

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: Understand the Expression

The expression \( C(10,3) \) represents a combination. We want to find the number of ways to choose 3 items from a set of 10 without regard to the order of selection.

Step 2: Identify the Combination Formula

The formula for combinations is \( C(n, r) = \frac{n!}{r!(n-r)!} \). Here, \( n = 10 \) and \( r = 3 \).

Step 3: Substitute Values into the Formula

Substitute \( n = 10 \) and \( r = 3 \) into the combination formula to obtain \[ C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}. \]

Step 4: Simplify the Factorials

We calculate the factorials needed: \(10! = 10 \times 9 \times 8 \times 7!\), and \(3! = 3 \times 2 \times 1\). Notice that the \(7!\) in \(10!\) will cancel with the \(7!\) in the denominator.

Step 5: Calculate the Result

After canceling \(7!\), we are left with \[ \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120. \] Therefore, \( C(10,3) = 120 \).

Key Concepts

Combinations

In mathematics, combinations are a way to select items from a larger set, where the order of selection doesn't matter. This is in contrast to permutations, where the order does matter. When solving problems involving combinations, we often use the notation \( C(n, r) \), which represents the number of ways to choose \( r \) items from \( n \) items without regard to the order. This is calculated using the combination formula:

• \( C(n, r) = \frac{n!}{r! (n-r)!} \)

Here, \( n! \) (read as "n factorial") means the product of all positive integers up to \( n \), and "\( r! \)" and "\((n-r)!\)" are calculated similarly.
It's important in combinatorics to remember that:

• In combinations, the arrangement of items doesn't change the outcome. For example, choosing red, blue, and green balls is the same as choosing blue, green, and red.
• The combination formula helps simplify problems by eliminating redundant counting that occurs when order isn't important.

Factorials

Factorials are a mathematical function that are especially crucial when dealing with permutations and combinations. The factorial of a non-negative integer \( n \), written as \( n! \), is the product of all positive integers less than or equal to \( n \).
Here’s how factorials work:

• \( 0! = 1 \) by definition.
• \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \).

For example, \( 5! \) (or "5 factorial") is:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow very quickly with larger numbers, making them useful for counting possible arrangements or selections. In the context of combinations, factorials allow us to remove the ordering of selected items, simplifying complex counting problems into more manageable calculations.

Permutations

Permutations are a foundational concept in combinatorics, related to the arrangement of items where the order does matter. Unlike combinations, here the sequence or order influences the outcome. The permutation formula to calculate the number of ways to arrange \( r \) items out of \( n \) is:

• \( P(n, r) = \frac{n!}{(n-r)!} \)

Using permutations, you can determine in how many different ways a subset of items can be ordered, which is useful for scenarios like scheduling or ordering performances.
When comparing permutations to combinations:

• Permutations focus on order, meaning \( ABC \) is different from \( BCA \).
• Permutations typically result in a larger number since they account for different sequences of the same items.

To apply permutations, remember:

• Start by calculating the total possible arrangements, then reduce based on the number of selections and desired orderings.

Understanding permutations alongside combinations can vastly enhance your problem-solving skills in any counting or arrangement scenarios.