Problem 25 - College Algebra

Question

Evaluate each expression. $$ \frac{_{7} C_{3}}{_{5} C_{4}}-\frac{98 !}{96 !} $$

Step-by-Step Solution

Answer

-9499

Step 1: Calculate the combinations

The first part of the problem asks to evaluate \(\frac{_{7} C_{3}}{_{5} C_{4}}\), which can be done using the combination formula, \(\frac{n!}{ r!(n - r)!} \). Hence, calculate \(_{7} C_{3}\) and \(_{5} C_{4}\) separately. \(_{7} C_{3} = \frac{7!}{3!(7 - 3)!} = \frac{7*6*5}{3*2*1} = 35\) and \(_{5} C_{4} = \frac{5!}{4!(5 - 4)!} = 5\). And their ratio is \(\frac{35}{5} = 7\).

Step 2: Compute the factorial

We next find the ratio \(\frac{98 !}{96 !}\). Since \(n! = n*(n-1)*(n-2)*...*3*2*1\), we can simplify \(\frac{98 !}{96 !}\) to \(\frac{98*97}{1} = 9506\).

Step 3: Combine the results

Finally, subtract the value obtained in step 2 from the value obtained in step 1. So the solution will be \(7 - 9506 = -9499\).

Key Concepts

Factorials

Factorials play a crucial role in combinatorics and are essential for solving problems that involve arrangements or selections. A factorial is denoted by an exclamation mark after a number, such as \(n!\), and it represents the product of all positive integers less than or equal to \(n\). This means that to compute \(n!\), you multiply this series of decreasing integers until you reach 1. For example:

• \(3! = 3 \times 2 \times 1 = 6\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

The concept of factorial is a foundation for other combinatorial methods, making it indispensable for calculating permutations and combinations.
When evaluating expressions with factorials, it's often possible to simplify them by canceling out common terms as shown in the problem. For example, in the expression \(\frac{98!}{96!}\), the \(98!\) term can be expanded to \(98 \times 97 \times 96!\), allowing the \(96!\) parts to cancel out, simplifying the operation to \(98 \times 97\). This approach is valuable for reducing computational complexity in larger problems.

Combinations Formula

Combinations are used when the order of selection does not matter, and they are calculated using the combinations formula. The formula \[_{n} C_{r} = \frac{n!}{r!(n - r)!}\]helps determine how many ways \(r\) items can be chosen from \(n\) items. The problem involved finding \(_{7}C_{3}\) and \(_{5}C_{4}\).

• For \(_{7}C_{3}\), substitute the values into the formula: \(\frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).
• And for \(_{5}C_{4}\), it simplifies to \(\frac{5!}{4!(5-4)!} = 5\).

With combinations, you can find the number of possible groupings when the sequence is irrelevant, vital in probabilities and various planning scenarios.

Arithmetic Operations

Performing arithmetic operations correctly is fundamental in solving expressions and equations. The problem given involved operations of division and subtraction, each performed step-by-step:

• First, computing the quotient of two combinations, \(\frac{35}{5} = 7\), is a division where you distribute or group values evenly.
• Next, a division of factorials \(98 \times 97 = 9506\) shows the simplification process to reduce laborious calculations.
• Finally, combining results from these arithmetic operations with subtraction brings it all together: \(7 - 9506\), leading to \(-9499\).

Understanding how to systematically perform these operations ensures accuracy and efficiency, especially in larger, more complex expressions.

Problem Solving

In combinatorics, problem solving often involves breaking down complex tasks into smaller, more manageable pieces. This problem showed a full example. The key steps were to:

• Identify what operations are needed (combinations, division, subtraction).
• Apply the combinations formula to evaluate each part separately.
• Simplify expressions by recognizing patterns or canceling unnecessary terms, particularly in factorials which can quickly grow large.
• Combine results with arithmetic operations ensuring each step connects correctly to the next.

Systematic and structured approaches to problem solving not only help in arriving at the correct solution but also build deeper understanding of combinatorial concepts and enhance critical thinking skills in other mathematical or real-world problems.