Problem 22 - Algebra and Trigonometry

Question

Evaluate the expression. $$ C(99,3) $$

Step-by-Step Solution

Answer

C(99, 3) = 156849.

Step 1: Understand the Combinatorial Expression

The expression \( C(99,3) \) represents a combination. It is used to determine the number of ways to choose 3 items from a total of 99 items without considering the order of selection. The formula for combinations is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items and \( k \) is the number of items to choose.

Step 2: Substitute Values into the Combination Formula

We substitute \( n = 99 \) and \( k = 3 \) into the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), giving us: \[ C(99, 3) = \frac{99!}{3!(99-3)!} = \frac{99!}{3! \times 96!} \]

Step 3: Simplify the Factorials

Recall that the factorial \( n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1 \). We can replace \( 99! \) with \( 99 \times 98 \times 97 \times 96! \) and cancel the \( 96! \) from both the numerator and the denominator: \[ \frac{99 \times 98 \times 97 \times 96!}{3! \times 96!} = \frac{99 \times 98 \times 97}{3!} \]

Step 4: Simplify the Denominator \(3!\)

Calculate \(3!\), which represents the factorial of 3: \[ 3! = 3 \times 2 \times 1 = 6 \] Now the expression becomes: \[ \frac{99 \times 98 \times 97}{6} \]

Step 5: Complete the Calculation

Now divide the product \( 99 \times 98 \times 97 \) by 6. First, calculate the product of the numerator: \( 99 \times 98 = 9702 \), then \( 9702 \times 97 = 941094 \). Now divide by 6: \[ \frac{941094}{6} = 156849 \]

Step 6: Final Step: Write the Answer

The number of ways to choose 3 items from 99 items is \( C(99, 3) = 156849 \). This is the final answer for the combination of 99 items taken 3 at a time.

Key Concepts

Factorials

Imagine trying to find the number of different ways to arrange a set of items. Factorials make this calculation easy. A factorial, represented by an exclamation point (!), signifies a product of all positive integers up to a given number. For instance:

• 4! = 4 × 3 × 2 × 1 = 24
• 5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials are fundamental in counting and arrange theories because they help in scaling up calculations where multiple outcomes are possible.

When handling combinations, such as finding how many ways you can choose a certain number of items from a set, factorials reduce complex multiplication into manageable terms. In our example, calculating combinations requires you to determine the factorial of both the number of items to choose from and the number of items being chosen. This can simplify the problem dramatically.

Combinatorial Mathematics

Combinatorial mathematics, or combinatorics, involves counting, arranging, and analyzing how sets of objects can differ. It deals with the principles of counting as well as arrangement permutations and combinations.

This branch of mathematics answers questions like how many ways you can select items, the efficiency of arrangements, and even leads to further insights in probabilities. Combinatorics is essential in fields such as computer science, cryptography, and operations research. In our example, choosing 3 items out of 99 requires finding the total number of combinations, focusing on selection rather than arrangement.

• Combinatorics helps identify efficient methods to count groupings
• It aids in optimizing decision-making processes where options are numerous
• Any field requiring optimization of resources employs combinatorial mathematics

Combinatorics Formula

The heart of combinatorics lies in its formulas, which allow us to calculate complex scenarios easily. One key formula is for combinations, expressed as:\[ C(n, k) = \frac{n!}{k!(n-k)!} \]In this formula:

• \( n \): total items available
• \( k \): items to be chosen

The power of this formula is that it calculates the number of ways to choose \( k \) items from \( n \) items without the concern of order. Using this formula, it becomes straightforward to find solutions like our example, where combinations from 99 items down to selecting 3 is needed.
The application of the formula involves substituting numbers directly and simplifying where needed, by managing factorial calculations. This process streamlines what could otherwise be an overwhelming counting task.