Problem 22 - College Algebra with Corequisite Support

Question

For the following exercises, compute the value of the expression. $$ C(26,3) $$

Step-by-Step Solution

Answer

2600.

Step 1: Understanding the Combination Formula

The combination formula is given by \( C(n, r) = \frac{n!}{r!(n-r)!} \). This formula is used to determine the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection.

Step 2: Identifying the Values of n and r

In this problem, \( n = 26 \) and \( r = 3 \). These values are used to compute the combination \( C(26, 3) \).

Step 3: Calculating the Factorials

Compute the factorial values needed for this problem: \( 26! \), \( 3! \), and \( (26-3)! = 23! \). Factorials are the product of all positive integers up to a given number (e.g., \( 3! = 3 \times 2 \times 1 = 6 \)).

Step 4: Applying the Combination Formula

Substitute the computed factorials into the combination formula: \[C(26, 3) = \frac{26!}{3! \times 23!}\] Since \( 26! = 26 \times 25 \times 24 \times 23! \), the \( 23! \) in the numerator and denominator cancel out, simplifying the expression:\[= \frac{26 \times 25 \times 24}{3 \times 2 \times 1}\]

Step 5: Simplifying the Expression

Continue simplifying by calculating the products in the numerator and the denominator:\( 26 \times 25 \times 24 = 15600 \) and \( 3 \times 2 \times 1 = 6 \). Then divide: \[= \frac{15600}{6} = 2600\]

Step 6: Final Result

Thus, the computed value of the expression \( C(26, 3) \) is 2600.

Key Concepts

Combination Formula

The combination formula is a fundamental concept in combinatorics. It calculates how many ways you can select a subset of items from a larger set. In combinations, unlike permutations, the order in which you choose the items does not matter.
For combinations, we use the formula:

• \( C(n, r) = \frac{n!}{r!(n-r)!} \)

Here:

• \( n \) is the total number of items.
• \( r \) is the number of items to be selected.

This formula helps us find the number of possible combinations when selecting \( r \) items from \( n \) items without regard for order. It's especially useful in fields like probability and statistics.

Factorials

Factorials are essential when working with combinations and permutations. They are denoted as \( n! \) and involve multiplying a series of descending numbers.

• For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
• Factorials are used in the combination formula to determine how many ways items can be arranged.

Understanding factorials helps simplify computations in combination and permutation problems. Remember that by definition, \( 0! = 1 \).
This identity is crucial for dealing with certain combinatorial problems.

Permutations

Permutations differ from combinations because the order matters in permutations. If you want to know the number of ways to arrange \( r \) items from a set of \( n \), you look at permutations. The formula for permutations is:

• \( P(n, r) = \frac{n!}{(n-r)!} \)

In permutations:

• Every different order is considered a new arrangement.
• This leads to more arrangements than combinations with the same items and numbers.

Consider permutations when you care about the sequence of events, like assigning prizes to winners.

Binomial Coefficient

The binomial coefficient, often written as \( \binom{n}{r} \), is another way to express combinations. It's a notation used frequently in binomial expansions and probability problems.

• The binomial coefficient is equivalent to the combination formula: \( \binom{n}{r} = C(n, r) = \frac{n!}{r!(n-r)!} \).
• This notation is useful in simplifying expressions, especially when expanding binomials.

In the context of algebra and probability, the binomial coefficient helps calculate the likelihood of events in binomial distributions.