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In how many ways can a 10 -person club select a president and a secretary- treasurer from among its members?

Short Answer

Expert verified
There are 90 ways to select a president and secretary-treasurer from the club.

Step by step solution

01

Understanding the Roles to be Filled

First, we need to recognize that there are two distinct positions to be filled in the club: one president and one secretary-treasurer. These positions are separate, and the same person cannot hold both roles.
02

Selecting the President

Out of the 10 members, any one member can be chosen to be the president. Therefore, there are 10 possible choices for this position.
03

Selecting the Secretary-Treasurer

After selecting the president, there are only 9 members left. Out of these 9 remaining members, we need to select one to be the secretary-treasurer. Therefore, there are 9 possible choices for the secretary-treasurer position.
04

Multiply the Choices Together

Since each choice for president can be combined with each choice for secretary-treasurer, multiply the number of choices for president by the number of choices for secretary-treasurer. This gives us:\[ 10 \times 9 = 90 \]
05

Conclusion: Total Ways to Select the Officers

There are 90 different ways to select a president and a secretary-treasurer from the 10 members of the club.

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Key Concepts

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

Permutations
Permutations are a fundamental concept in combinatorics, which is the branch of mathematics devoted to counting, arranging, and structuring elements. When we talk about permutations, we're interested in the number of ways to arrange a set of elements where the order matters. For instance, if we need to arrange three books on a shelf, the order in which we place them will result in different permutations. In mathematical terms, the number of permutations of a set of elements is determined by factorial notation, where a set of size \( n \) is denoted as \( n! \). This is the product of all positive integers up to \( n \).

In the context of selecting club officers, the concept of permutations helps us understand that selecting individuals for specific roles is an arrangement where their order (i.e., specific roles like president and secretary-treasurer) counts. Thus, choosing a leader and an officer from a group is related to computing permutations because each role must be filled distinctly, highlighting the importance of order in these selections.
Counting principles
Counting principles are rules in mathematics that allow us to determine the number of ways a set of events can occur. One of the most crucial of these is the multiplication principle. It states that if one event can occur in \( n \) ways and a second independent event can occur in \( m \) ways, then there are \( n \times m \) ways for both events to happen.

This principle is perfectly illustrated in our exercise scenario: selecting club officers requires choosing a president and then separately choosing a secretary-treasurer. For the initial choice of a president, there are 10 possible ways (since there are 10 candidates). Once the president is chosen, 9 candidates remain, one of whom will be the secretary-treasurer. According to the multiplication principle, to find the total number of configurations, you multiply these reasons: \( 10 \times 9 = 90 \).

This handy principle simplifies the process of counting complex situations by breaking them down into a series of manageable steps.
Problem solving in mathematics
Problem solving in mathematics involves a systematic approach to finding solutions. It's about breaking down complex problems into simpler parts, applying mathematical principles, and strategically thinking through potential solutions. For our current exercise, problem-solving starts with clearly defining the task: choosing two distinct officers from a group.

By identifying that there are two separate roles, the exercise naturally resolves into smaller tasks. Each role must be filled independently, suggesting an order of operations and an application of counting principles. By addressing sub-tasks sequentially, such as selecting a president first and a secretary-treasurer next, we can systematically arrive at the solution. Mathematicians use problem solving to strategically work through the geometry of the situation, ensuring that each step logically leads to the next while optimizing for ease and efficiency.

Overall, problem solving is not about jumping to conclusions but rather about understanding and meticulously piecing a solution together using solid mathematical logic.

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Most popular questions from this chapter

Partially ordered set, totally ordered set, and well ordered set form a hierarchy, in that each category is more restrictive than the ones that come before it. For each of the following decide if the set with the relation is not partially ordered, partially ordered but not totally ordered, totally ordered but not well ordered, or well ordered. a. "is ancestor of" on the set of people, where a person is considered to be an ancestor of him- or herself b. "is parent of" on the set of people c. "is divisible by" on the set of positive integers d. \(\leq\) on the set of numbers that can be expressed as a positive integer divided by 1,2 , or 3 e. \(\leq\) on the rational numbers \(r\) with \(1 \leq r \leq 2\) f. \(\leq\) on any finite subset of the rational numbers

Use multisets to determine the number of ways to pass out \(k\) identical apples to \(n\) children. Assume that a child may get more than one apple.

Use binomial coefficients to determine the number of ways in which you can line up three identical red apples and two identical golden apples. Use equivalence class counting (in particular, the quotient principle) to determine the same number.

The formula for the number of multisets is \((n+k-1) !\) divided by a product of two other factorials. We want to use the quotient principle to explain why this formula counts multisets. The formula for the number of multisets is also a binomial coefficient, so it should have an interpretation that involves choosing \(k\) items from \(n+k-1\) items. The parts of the problem that follow lead us to these explanations. a. In how many ways can you place \(k\) red checkers and \(n-1\) black checkers in a row? b. How can you relate the number of ways of placing \(k\) red checkers and \(n-1\) black checkers in a row to the number of \(k\)-element multisets of an \(n\)-element set (the set \(\\{1,2, \ldots, n\\}\) to be specific)? c. How can you relate the choice of \(k\) items out of \(n+k-1\) items to the placement of red and black checkers, as in parts a and \(b\) ? Think about how this relates to placing \(k\) identical books and \(n-k\) identical blocks of wood in a row.

What formula do you get for the binomial theorem if, instead of analyzing the number of ways to choose \(k\) distinct \(y\) 's, you analyze the number of ways to choose \(k\) distinct \(x\) 's?

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