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Chapter 1: Problem 40

A basketball team has 12 players. However, only five players play at any given time during a game. (a) In how may ways may the coach choose the five players? (b) To be more realistic, the five players playing a game normally consist of two guards, two forwards, and one center. If there are five guards, four forwards, and three centers on the team, in how many ways can the coach choose two guards, two forwards, and one center? (h) (c) What if one of the centers is equally skilled at playing forward? (h)

Short Answer

Expert verified
792 ways; 180 ways; 340 ways.

Step by step solution

01

Calculate total ways to choose 5 players

To determine the number of ways to choose 5 players from 12, use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \( n \) is the total number of players, and \( r \) is the number of players to choose. Here, \( n = 12 \) and \( r = 5 \). So, \( \binom{12}{5} = \frac{12!}{5!(12-5)!} \)
02

Simplify the combination formula

Simplify the factorials in the combination formula for \( \binom{12}{5} \): \[ \frac{12!}{5! \times 7!} \] Calculate as: \( 12 \times 11 \times 10 \times 9 \times 8 \times \frac{7!}{7!} \times \frac{1}{5!} = 3696 \times \frac{1}{120} = 792 \)
03

Calculate ways to choose 2 guards, 2 forwards, and 1 center

We need to choose 2 guards from 5, 2 forwards from 4, and 1 center from 3. Use the combination formula for each selection. For guards: \( \binom{5}{2}\) For forwards: \( \binom{4}{2} \) For centers: \( \binom{3}{1} \) Thus, the total number of ways is \( \binom{5}{2} \times \binom{4}{2} \times \binom{3}{1} \)
04

Calculate combinations for guards, forwards, and centers

Calculate the combinations for each: \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \) \( \binom{4}{2} = \frac{4!}{2!(4-2)!} = 6 \) \( \binom{3}{1} = \frac{3!}{1!(3-1)!} = 3 \) Then multiply these results: \( 10 \times 6 \times 3 = 180 \)
05

Calculate ways with flexible center

If one of the centers can also play as a forward, consider two scenarios: 1. The player plays as a forward: Choose 2 guards from 5, 3 forwards from 4 + the flexible player, 0 centers from 2 remaining. \( \binom{5}{2} \times \binom{5}{3} \times \binom{2}{0} = 10 \times 10 \times 1 = 100 \) 2. The player plays as a center: Choose 2 guards from 5, 2 forwards from 4, 1 center from 3 + 1 flexible. \( \binom{5}{2} \times \binom{4}{2} \times \binom{4}{1} = 10 \times 6 \times 4 = 240 \) Add results: 100 + 240 = 340

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Key Concepts

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

combinations
In mathematics, combinations refer to the selection of items from a larger set, where the order of selection does not matter. This concept is essential in various fields, including sports team selection.

The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where:\
  • n is the total number of items.
  • r is the number of items to choose.
  • ! denotes a factorial, which is the product of all positive integers up to a certain number.

In our basketball team example, the coach needs to choose 5 players out of 12. Using the combination formula, we calculate the number of ways to do this:\ \[ \binom{12}{5} = \frac{12!}{5!(12-5)!} = 792 \] Thus, the coach has 792 ways to choose 5 players from the 12.
combinatorial mathematics
Combinatorial mathematics, or combinatorics, is the branch of mathematics that deals with counting, arranging, and finding patterns in sets of elements. It is a vital tool for solving problems in various fields, including sports.

In our basketball example, we apply combinatorial mathematics to determine the number of ways to choose players for different positions. This involves using combinations and the rules of counting (like the multiplication principle).
Applying these principles, we find different ways to form teams based on specific player positions, such as guards, forwards, and centers.

This helps coaches understand the possibilities available for team selection and create strategic advantages.
sports team strategy
In sports, strategy is crucial for achieving success. Coaches must understand players' strengths and weaknesses and how to strategically place them on the team.

The basketball team selection problem is a perfect example. Coaches need to choose not just any five players but the best combination for guards, forwards, and centers.
For instance:
  • The coach may want two quick and agile guards to handle the ball and defend against opposing guards.
  • Two forwards might be needed to shoot from mid-range or secure rebounds.
  • One strong center to dominate near the basket is essential.
Balancing these positions is key to forming an effective on-court team. By using mathematical combinations, coaches can explore various formations and choose the best strategy.
permutations and combinations
Permutations and combinations are two fundamental concepts in combinatorial mathematics often used in sports team selection.

Permutations involve arranging items in a specific order. This is different from combinations, where the order of selection does not matter. For team selection:
  • Combinations are used to choose which players will form the team.
  • Permutations could be used to decide in what order they play during a game.
In our example, we used combinations to choose how many ways the coach can select certain players. Using permutations might involve deciding the order of players for a play or sequence.
applied mathematics in sports
Mathematics plays a significant role in optimizing strategies in sports. From creating team formations to analyzing games, mathematics provides valuable insights.

Applied mathematics can help answer complex questions such as:
  • How many ways can a basketball coach select the most effective team?
  • What are the odds of winning a game based on current player statistics?
  • How can data be used to improve individual and team performance?
In our basketball team selection problem, applied mathematics helps coaches understand all possible team compositions. This allows for making informed decisions to improve the team's chances of success.

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Most popular questions from this chapter

Show that if \(S\) and \(T\) are finite sets of the same size, then a function \(f\) from \(S\) to \(T\) is one-to-one if and only if it is onto. (h)

There is a generalized pigeonhole principle which says that if we partition a set with more than \(k n\) elements into \(n\) blocks, then at least one block has at least \(k+1\) elements. Prove the generalized pigeonhole principle. (h)

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