Problem 49 - Algebra and Trigonometry

Question

41–54 These problems involve combinations. Violin Recital A violinist has practiced 12 pieces. In how many ways can he choose eight of these pieces for a recital?

Step-by-Step Solution

Answer

495 ways.

Step 1: Understand the Problem

We need to find the number of ways to choose 8 pieces out of 12. This is a typical combinations problem where the order of selection does not matter.

Step 2: Use the Combination Formula

The combination formula is used as follows: \( C(n, r) = \frac{n!}{r!(n-r)!} \) where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose. Here, \(n = 12\) and \(r = 8\).

Step 3: Calculate the Factorials

We need to calculate the factorials involved in the formula: - Calculate \(12!\)- Calculate \(8!\)- Calculate \( (12-8)! = 4!\)

Step 4: Substitute Values into the Formula

Substitute the factorials into the combination formula: \[ C(12, 8) = \frac{12!}{8! \, 4!} \]

Step 5: Simplify the Expression

Now, simplify the factorial expression: - First, cancel \(8!\) in both numerator and denominator:\[ C(12, 8) = \frac{12 \times 11 \times 10 \times 9 \times 8!}{8! \times 4 \times 3 \times 2 \times 1} \]- Simplify further to get: \[ C(12, 8) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \]

Step 6: Solve the Expression

Multiply and divide the numbers to find the result:- Calculate \(12 \times 11 = 132\) - Calculate \(132 \times 10 = 1320\)- Calculate \(1320 \times 9 = 11880\)- Calculate \(4 \times 3 = 12\) and \(12 \times 2 = 24\)- Finally, divide \(11880 \div 24 = 495\)

Step 7: Interpret the Result

The number of combinations or ways the violinist can choose 8 pieces out of 12 is 495.

Key Concepts

Combination Formula

When choosing items from a larger pool without regard to order, combinations are the perfect tool. For this, we use the combination formula, which is also known as binomial coefficient. This formula allows us to find how many different ways we can select a subset of items. It is written as: \[C(n, r) = \frac{n!}{r!(n-r)!}\]Here, \(n\) is the total number of items, and \(r\) is the number of items we want to select. This formula accounts for the fact that in combinations, unlike permutations, the order of selection does not matter. Thus, it avoids over-counting similar selections.For instance, in our exercise about the violinist, we have 12 pieces of music and want to select 8 of them. Plugging into our combination formula: \[ C(12, 8) = \frac{12!}{8! \, 4!} \] shows us there are 495 unique ways to pick 8 pieces.

Factorials

Factorials are essential elements in the calculation of combinations. The factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers up to \( n \). For example:

• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(0! = 1\) by definition.

In the combination formula, factorials help in enumerating all possible arrangements of items, where the division part of the formula cancels out repeated arrangements due to groupings that are identical in composition.During our violin exercise, calculating the factorials for \(12!\), \(8!\), and \(4!\) is crucial. This is how we adjust our problem setup into numerical form, simplifying the coil of possibilities into a clean number.

Permutations vs Combinations

Both permutations and combinations involve selection, but they differ significantly based on the importance of order. Understanding the distinction is essential when tackling problems related to arranging or selecting objects.- **Permutations**: Order matters. For example, if we were arranging the violin pieces to play in a specific performance order, we'd use permutations. The formula is \[ P(n, r) = \frac{n!}{(n-r)!} \] Here every distinct arrangement is counted separately.- **Combinations**: Order does not matter. As seen in our violinist example, when choosing 8 pieces to play, the order of chosen pieces isn't relevant, hence using combinations.By contrasting the two, we notice permutations answer the "How many ways to arrange?" question, while combinations answer "How many ways to choose?" This focus on whether the order is significant directly affects which formula to apply.