Problem 18 - Discrete Mathematics with Applications

Question

Solve each equation, where \(n \geq 0\). $$C(n, 1)=10$$

Step-by-Step Solution

Answer

The given equation is \(C(n, 1) = 10\). Using the combination formula and simplifying, we have: \[\frac{n!}{(n-1)!} = 10\] Solving for n, we find that \(n = 10\).

Step 1: Substitute r = 1 in the combination formula

Since r is given as 1 in the exercise, substitute r = 1 into the combination formula: \[C(n, 1) = \frac{n!}{1!(n-1)!}\]

Step 2: Simplify the combination formula

Since the factorial of 1 is just 1, we can simplify the expression: \[C(n, 1) = \frac{n!}{(n-1)!}\]

Step 3: Set up the given equation

We are given \(C(n ,1) = 10\), so substitute this information into the simplified combination formula: \[\frac{n!}{(n-1)!} = 10\]

Step 4: Solve for n

To solve for n, multiply both sides by \((n-1)!\) to cancel out the denominator on the left side of the equation: \[n! = 10(n-1)!\] Now, we know that \(n! = n \times (n-1)!\). Therefore, our equation becomes: \[n \times (n-1)! = 10(n-1)!\] Divide both sides by \((n - 1)!\) to get: \[n = 10\] Now that we have solved for n, the answer is n = 10.

Key Concepts

Combination Formula

The combination formula is an essential part of combinatorics. It helps us determine how many ways we can select items from a larger group without considering the order. The combination formula is represented as:

• \( C(n, r) = \frac{n!}{r!(n-r)!} \)

Here, \(n\) is the total number of items to choose from, \(r\) is the number of items selected, and \(!\) denotes factorial.

Using the combination formula, you can easily find the number of combinations possible. For example, if you have 5 different books and want to select 2 to read, the formula will help calculate the various combinations of 2 books. This formula only considers the number of ways to pick items, not the order in which they are chosen. This makes it a powerful tool in probability, statistics, and various fields requiring selection without replacement.

Factorials

Factorials are a big part of the concept of permutations and combinations. The factorial of a number, denoted as \( n! \) (read as "n factorial"), is the product of all positive integers up to that number.

• The factorial function grows rapidly with larger values of \(n\).
• \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \)
• For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Factorials are used in the combination formula and are handy when simplifying equations involving permutations or combinations. Understanding how factorials cancel out in equations, like in the combination formula, is key to solving many combinatorial problems.

Remember, the value of \(0!\) is 1 by definition, which is crucial in calculations involving combinations or permutations where all items are chosen or none.

Equation Solving Techniques

Solving equations, especially when they involve factorial expressions, requires familiarity with algebraic manipulation.

In our exercise, the goal was to find \(n\) for the given equation \( C(n, 1) = 10 \). Here's an overview of the steps taken:

• Substitute Known Values: Substitute any given values or constants into the equation. Here, we set \(r = 1\) in the combination formula.
• Simplify the Equation: Reduce the equation using known mathematical properties, such as \(1! = 1\), to make it easier to manage.
• Eliminate Fractions: Multiply through by the denominator to eliminate fractions, as seen when dealing with \(\frac{n!}{(n-1)!} = 10\) by multiplying by \((n-1)!\).
• Isolate the Variable: Manipulate the equation to isolate \(n\) on one side, leading to \(n = 10\).

By systematically following these steps, solving factorial equations becomes more manageable. These techniques can be applied to a wide range of problems, simplifying complex expressions by breaking them into smaller, more workable parts.