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A major league baseball team roster consists of 40 players of which 25 are considered active. (a) How many ways are there for a manager to select 25 active players from a major league roster? (b) How many ways can a manager select a nine-player batting lineup from the active roster for opening day?

Short Answer

Expert verified
There are \( C(40, 25) = \frac{40!}{25!*(40-25)!} \) ways to select the active players from the roster and \( C(25, 9) = \frac{25!}{9!*(25-9)!} \) ways to form the starting lineup.

Step by step solution

01

Calculate combinations for scenario (a)

Calculate the number of ways to select the active players. We need to choose 25 players from a group of 40. In combinatorics, this is a combination problem, which can be solved using the formula \[C(n, k) = \frac{n!}{k!*(n-k)!}\] where \(C\) stands for combinations, \(n\) is the total number of items, and \(k\) is the number of items to choose. Thus, for scenario (a), \(C(40, 25) = \frac{40!}{25!*(40-25)!}\]
02

Calculate combinations for scenario (b)

We need to choose 9 players from the group of 25 active players. Using the combinations formula, we get \[C(25, 9) = \frac{25!}{9!*(25-9)!}\]
03

Calculate the numerical values

Calculate the values from step 1 and step 2 using the factorials.

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Key Concepts

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

Combination formula
In combinatorial mathematics, the combination formula is fundamental when determining how many ways you can select items from a larger group without regard to the order. This is crucial when solving problems like selecting active players for a baseball team roster. The formula is given as:
  • \( C(n, k) = \frac{n!}{k!(n-k)!} \)
Here, \( n \) represents the total number of items, and \( k \) signifies the number of items to be chosen.

When applying this formula, the denominator \( k!(n-k)! \) accounts for removing the order considerations. It ensures that the solutions focus purely on combinations, not permutations.

For instance, if we need to select 25 players from a roster of 40, we set \( n = 40 \) and \( k = 25 \). The computation becomes:
  • \( C(40, 25) = \frac{40!}{25! \times 15!} \)
This formula helps us systematically find the number of possible combinations.
Factorial
A factorial, represented by an exclamation point (!!), is a key concept in mathematics often seen in combinatorics, especially with the combination formula. The factorial of a number \( n \), written as \( n! \), is the product of all positive integers up to and including \( n \). It is given by:
  • \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \)
For example, \( 5! \) would be:
  • \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

In combinatorics, factorials simplify calculations by efficiently expressing repeated multiplication. They are particularly useful for problems involving combinations and permutations.

Factorials grow very quickly and are a critical component in the combination formula. Without understanding factorials, calculating combinations would be very laborious.
Combinatorial mathematics
Combinatorial mathematics is a fascinating field that deals with counts and arrangements of elements within a set. It provides tools to solve problems related to selecting and organizing items. It's widely used in probability, computer science, and optimization.

One of the essential aspects of combinatorial mathematics is understanding the difference between permutations and combinations. Combinations focus on selecting items without regard to order, whereas permutations consider the order.

Real-life applications include determining lineups in sports, scheduling, and resource allocation problems. The foundational formulas, like the combination and permutation formulas, help solve these problems by breaking down complex arrangements into manageable calculations.
In our baseball team scenario, combinatorial mathematics enables us to calculate how many ways a manager can select players, ensuring the team is organized optimally without being overwhelmed by all potential configurations.

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Most popular questions from this chapter

Consider the sample space \(S=\left\\{o_{1}, o_{2}, o_{3}, o_{4}\right\\} .\) Suppose that \(\operatorname{Pr}\left(o_{1}\right)+\operatorname{Pr}\left(o_{2}\right)=\operatorname{Pr}\left(o_{3}\right)+\operatorname{Pr}\left(o_{4}\right)\) and that \(\operatorname{Pr}\left(o_{1}\right)=0.15\) (a) Find the probability assignment for the probability space when \(o_{2}\) and \(o_{3}\) have the same probability. (b) Find the probability assignment for the probability space when \(\operatorname{Pr}\left(o_{3}\right)=0.22\)

Yahtzee. Yahtzee is a dice game in which five standard dice are rolled at one time. (a) What is the probability of scoring "Yahtzee" with one roll of the dice? (You score Yahtzee when all five dice match.) (b) What is the probability of a four of a kind with one roll of the dice? (A four of a kind is rolled when four of the five dice match.) (c) What is the probability of rolling a large straight in one roll of the dice? (A large straight consists of five numbers in succession on the dice.)

Jack packs two pairs of shoes, one pair of boots, three pairs of jeans, four pairs of dress pants, and three dress shirts for a trip. (a) How many different outfits can Jack make with these items? (b) If Jack were also to bring along two jackets so that he could wear either a dress shirt or a dress shirt plus a jacket, how many outfits could Jack make?

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Find the odds of each of the following events. (a) an event \(E\) with \(\operatorname{Pr}(E)=3 / 11\) (b) an event \(E\) with \(\operatorname{Pr}(E)=0.375\)

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