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1–6 Evaluate the expression. $$P(9,2)$$

Short Answer

Expert verified
\(P(9, 2) = 72\).

Step by step solution

01

Understanding "P(n, r)"

The notation \(P(n, r)\) stands for permutations and represents the number of ways to arrange \(r\) items out of \(n\) total items. The formula for permutations is \(P(n, r) = \frac{n!}{(n-r)!}\).
02

Identifying the Values

For this exercise, the expression is \(P(9, 2)\), meaning \(n = 9\) and \(r = 2\). Now we will substitute these values into the permutations formula.
03

Applying the Formula

Substitute \(n = 9\) and \(r = 2\) into the permutations formula: \[P(9, 2) = \frac{9!}{(9-2)!}\]. This simplifies to \[P(9, 2) = \frac{9!}{7!}\].
04

Simplifying the Factorials

Calculate \(9!\) and \(7!\) to simplify the fraction. \(9! = 9 \times 8 \times 7!\), so \(\frac{9!}{7!} = \frac{9 \times 8 \times 7!}{7!}\). The \(7!\) terms cancel out, resulting in \(9 \times 8\).
05

Calculating the Result

Multiply the remaining numbers: \(9 \times 8 = 72\). Thus, \(P(9, 2) = 72\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
Have you ever wondered how many different ways you can arrange items? The factorial, represented by an exclamation mark (!), is a key concept in solving such problems. Factorials are a simple yet powerful tool to calculate permutations or arrangements.
Consider the example of arranging 9 items. The factorial of 9, written as \(9!\), is the product of all positive integers up to 9.
So, \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\). That's a lot of multiplication!
Factorials grow quickly as numbers increase, which is why they are so useful in counting the number of ways to arrange large sets of items.
  • For smaller numbers, you can compute the factorial manually by multiplying.
  • The factorial for 0 is the special case where \(0! = 1\).
Understanding factorials helps in simplifying complex permutation problems, just like in our exercise.
Combinatorics
Combinatorics is the branch of mathematics that deals with counting and arranging objects. It's like solving a giant puzzle where you figure out different possible combinations or orders of things.
One of the main goals in combinatorics is to determine the number of possible arrangements (permutations) or selections (combinations) of a set of items.
  • When order matters, it's a permutation.
  • When order does not matter, it's a combination.
In our exercise, we focused on permutations, evident through the expression \(P(9, 2)\).
We calculated how many different ways we can arrange 2 items out of 9, which is an essential concept in combinatorics.By understanding this, we can solve practical problems, like scheduling or organizing groups, efficiently.
Mathematical Notation
Mathematical notation is a way of writing formulas and equations, making problems easier to understand and solve.It uses symbols and abbreviations, allowing complex ideas to be expressed succinctly.
In the context of permutations, we use the notation \(P(n, r)\) to represent the number of ways to arrange \(r\) items from \(n\) items.
  • \(P(n, r)\) signifies a permutation.
  • \(n!\) shows the factorial of \(n\).
  • The fraction \( \frac{n!}{(n-r)!}\) represents computing the permutation.
Understanding these symbols is crucial to solving problems efficiently.In our example, using the mathematical notation helped us compute \(P(9, 2)\) efficiently, providing a clear path to the solution.With practice, these symbols and notations become second nature, allowing you to tackle more advanced mathematical problems with ease.

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Most popular questions from this chapter

Reliability of a Machine A machine used in a manufacturing process has 4 separate components, each of which has a 0.01 probability of failing on any given day. If any component fails, the entire machine breaks down. Find the probability that on a given day the indicated event occurs. (a) The machine breaks down. (b) The machine does not break down. (c) Only one component does not fail.

55–75 Solve the problem using the appropriate counting principle(s). Choosing a Committee A class has 20 students, of which 12 are females and 8 are males. In how many ways can a committee of five students be picked from this class under each condition? (a) No restriction is placed on the number of males or females on the committee. (b) No males are to be included on the committee. (c) The committee must have three females and two males.

55–75 Solve the problem using the appropriate counting principle(s). Arranging a Class Picture In how many ways can ten students be arranged in a row for a class picture if John and Jane want to stand next to each other, and Mike and Molly also insist on standing next to each other?

55–75 Solve the problem using the appropriate counting principle(s). Casting a Play A group of 22 aspiring thespians contains ten men and twelve women. For the next play the director wants to choose a leading man, a leading lady, a supporting male role, a supporting female role, and eight extras—three women and five men. In how many ways can the cast be chosen?

19–32 These problems involve permutations. Class Officers In how many ways can a president, vice president, secretary, and treasurer be chosen from a class of 30 students?

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