Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, compute the value of the expression. $$ C(12,4) $$

Step-by-Step Solution

Answer

495

Step 1: Understand the Notation

In this exercise, we're asked to find \(C(12,4)\), which refers to the number of combinations of 12 items taken 4 at a time. Combinations are used to determine how many ways we can choose items where order does not matter.

Step 2: Use the Combinations Formula

The formula for combinations is \(C(n, r) = \frac{n!}{r!(n-r)!}\). Here, \(n = 12\) and \(r = 4\). Thus, we need to calculate \(C(12, 4) = \frac{12!}{4!(12-4)!}\).

Step 3: Simplify the Denominator

First, simplify \((12-4)!\) in the denominator: \((12-4)! = 8!\). So the expression becomes \(\frac{12!}{4!8!}\).

Step 4: Calculate the Factorials

Calculate the factorials: - \(12! = 12 \times 11 \times 10 \times 9 \times 8!\),- \(4! = 4 \times 3 \times 2 \times 1 = 24\),- \(8!\) will cancel with the corresponding part in the numerator.

Step 5: Perform the Cancellation

Cancel \(8!\) from numerator and denominator, transforming the expression into \(12 \times 11 \times 10 \times 9\) divided by \(4!\).

Step 6: Compute the Result

Calculate the remaining expression: \(\frac{12 \times 11 \times 10 \times 9}{24} = \frac{11880}{24} = 495\). Thus, \(C(12, 4) = 495\).

Key Concepts

Factorials

Factorials are fundamental in many areas of mathematics, especially in permutations and combinations. The factorial of a number \(n\), denoted as \(n!\), is the product of all positive integers up to \(n\). For example, the factorial of 4 (written as \(4!\)) is calculated as follows:

• \(4! = 4 \times 3 \times 2 \times 1 = 24\)

Factorials grow very quickly, which is why it's helpful to understand their behavior. If you think about it, when calculating large factorials like \(12!\), a reliable strategy is to start multiplying from 1 up to 12, though in many cases we can simplify our work by canceling out terms when divided.
In combinatorics, you'll often compute expressions like \(12!\) but only up to a necessary point, such as multiplying only necessary terms up to \(9\) and knowing that \(8!\) will cancel out, which simplifies computation drastically.
In our example with \(C(12, 4)\), we simplified \(12!\) by calculating up to \((12 \times 11 \times 10 \times 9)\) before using division—this is important as these simplifications help in making calculations manageable.

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arranging, and grouping objects. One of the most important concepts in combinatorics is combinations. When dealing with combinations, order does not matter. This is key in distinguishing between combinations and permutations.
For combinations, we're interested in how many different ways we can choose a subset of items from a larger set. The formula for combinations is:

• \(C(n, r) = \frac{n!}{r!(n-r)!}\)

Here, \(n\) represents the total number of items to consider, and \(r\) is the number of items we are selecting. For example, if we want to select 4 items from a set of 12, we use \(C(12, 4)\).
Using the formula, we see that:

• \(C(12, 4) = \frac{12!}{4! \times 8!}\)
• Calculating simplifies to \(\frac{12 \times 11 \times 10 \times 9}{24}\)

This gives us the result of 495, indicating there are 495 different ways to choose 4 items from 12 when order does not matter. This concept is particularly useful in probability, game theory, and statistics.

Permutations

Permutations are another critical concept in combinatorics, similar to combinations, but with one key difference: the order of selection matters. If you're permuting a group, swapping positions will create a new permutation. For example, for a set of 3 items \(A, B, C\), the permutations are ABC, ACB, BAC, BCA, CAB, CBA.
The formula for permutations \(P(n, r)\) is:

• \(P(n, r) = \frac{n!}{(n-r)!}\)

Much like combinations, \(n\) is the total number of items, and \(r\) is the items we want to arrange. Unlike combinations, the absence of division by \(r!\) reflects the importance of order.
Understanding when to use permutations versus combinations is crucial. When order matters, permutations provide the correct count. For instance, flashing a pin code sequence is a permutation since 1234 is different from 4321. Understanding permutations helps in cryptography, computer algorithms, and organizational tasks, highlighting their importance across various fields.