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Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

Short Answer

Expert verified
a. 5005, b. 1816214400, c. 2100 ways.

Step by step solution

01

Understand the Context

We have a Little League team with 15 players and need to select a lineup of 9 players. The exercise also involves ordering and specific left-handed/right-handed constraints.
02

Step 2a: Calculate Combinations for Part A

We need to calculate how many ways we can select 9 players from a roster of 15. This is a combination problem, denoted by \( \binom{n}{k} \) where \( n \) is the total number of players and \( k \) is the number of players to choose. Thus, we calculate \( \binom{15}{9} = \frac{15!}{9!(15-9)!} = \frac{15!}{9!6!} \). Calculating, we get 5005 distinct ways.
03

Step 3a: Combination Calculation Details

To expand the factorial terms: \( 15! = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6! \), and the cancelations reduce the calculation to \( 15 \times 14 \times 13 \times 12 \times 11 \times 10 \div (6 \times 5 \times 4 \times 3 \times 2 \times 1) = 5005 \).
04

Step 2b: Calculate Permutations for Part B

After selecting the 9 players, we need to arrange them in a batting order. This requires calculating permutations of the 9 players, as the order matters now. Therefore, we calculate \( 9! \) for the permutations of these players: \( 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \).
05

Step 3b: Calculate Total for Part B

Multiply the combinations from Part A with the permutations from this part: \( 5005 \times 362880 = 1816214400 \). So, there are 1,816,214,400 distinct ways to both select and order the players.
06

Step 2c: Understand Constraints for Part C

We need to select 3 left-handed players as outfielders from the 5 available and the remaining 6 position players must be right-handed. This is two sub-problems: choosing left-handed players and right-handed players separately.
07

Step 3c: Select Left-handed Players

Calculate the number of ways to choose 3 left-handed players from 5: \( \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \).
08

Step 4c: Select Right-handed Players

Calculate how many ways to choose 6 right-handed players from the remaining 10 players (since 15 total players minus 5 left-handed ones): \( \binom{10}{6} = \frac{10!}{6!4!} = 210 \).
09

Step 5c: Calculate Total for Part C

Multiply the combinations from the separate groups to find total combinations: \( 10 \times 210 = 2100 \). So, there are 2100 ways to select the players under these constraints.

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Key Concepts

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

Combinations
In solving problems involving selecting a subset of items from a larger set, we often rely on the concept of combinations. When the order in which we select items does not matter, combinations come into play. For example, in the context of a Little League team, when selecting 9 players from a roster of 15, we use combinations. The formula to calculate combinations is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items to choose from, and \( k \) is the number of items to be chosen. In this case, we plug in \( n = 15 \) and \( k = 9 \) to find \( \binom{15}{9} \). After calculations with factorials, we find there are 5005 distinct ways to select the team. Remember, factorial (\(!\)) stands for a product of all positive integers up to a given number.
  • Combinations prioritize selection, not arrangement.
  • Key formula: \( \binom{n}{k} \)
  • Used when order doesn't matter.
Permutations
When it comes to arranging a set of items in a specific order, the idea of permutations is essential. Permutations are used when the order of arrangement is important, such as arranging players in a batting order. For our Little League example, once 9 players are selected for the starting lineup, permutations are applied to arrange them in order. The formula for permutations of \( n \) items is given by: \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \] In this scenario, we calculate \( 9! \) for arranging the 9 players, resulting in 362,880 permutations. By combining these permutations with the earlier calculated 5005 combinations, you determine there are 1,816,214,400 distinct ways to both select and order the players.
  • Permutations consider arrangement order.
  • Key formula: \( n! \)
  • Used when order matters.
Factorials
Factorials are a foundational element in combinatorics, crucial for calculating both combinations and permutations. The notation for factorial is \( n! \), which represents the multiplication of all positive integers up to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials simplify the process of understanding combinations and permutations as they form the basis for their respective formulas. They help streamline calculations, ensuring the correct number of selections and arrangements are accounted for in a given problem. Understanding factorials can significantly enhance problem-solving in probability and statistics as a whole.
  • Factorial notation denoted as \( n! \).
  • Essential for calculations in combinatorics.
  • Useful in both combinations and permutations.
Probability Problem Solving
Probability problem solving often involves combinations, permutations, and factorials to determine the likelihood of certain outcomes. In the context of the Little League team, these calculations are crucial to solving complex selection and arrangement problems effectively. For example, determining the number of ways to assemble a team using specific player constraints enhances understanding of the probability of various scenarios occurring. Like in the problem scenario, constraints such as choosing specific left-handed or right-handed players involve calculating combinations under different conditions. This holistic approach merges multiple concepts in probability problem solving:
  • Recognize selection and arrangement requirements.
  • Utilize combinations for selection without regard to order.
  • Employ permutations when order is key.
Mastering these elements equips one to approach probability problems more strategically, enhancing decision-making and solution accuracy.

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Most popular questions from this chapter

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