/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 In how many ways can 5 players b... [FREE SOLUTION] | 91Ó°ÊÓ

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In how many ways can 5 players be assigned to the 5 positions on a basketball team, assuming that any player can play any position? In how many ways can 10 players be assigned to the 5 positions?

Short Answer

Expert verified
120 for 5 players and 5 positions. 30,240 for 10 players and 5 positions.

Step by step solution

01

Understanding the Problem

Recognize that the problem involves assigning a number of players to a specific number of positions on a basketball team with unique roles.
02

Identify Total Permutations for 5 Players and 5 Positions

Calculate the number of ways to assign 5 players to 5 positions using permutations: \( P(5, 5) = 5! \)
03

Compute the Factorial for 5 Players

Calculate \(5!\) which is the product of all positive integers up to 5: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
04

Identify Total Permutations for 10 Players and 5 Positions

When you have more players than positions, calculate the permutations for 10 players and 5 positions: \( P(10, 5) = 10! / (10-5)! \)
05

Compute the Factorial for 10 Players and 5 Positions

Calculate the quotient of factorials where first compute \(10!\) and \( (10-5)!\): \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800\) and \(5! = 120\) Therefore, \(P(10, 5) = 3,628,800 / 120 = 30,240\)

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Key Concepts

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

Factorial
In mathematics, the factorial of a non-negative integer is a key concept, especially in permutations and combinatorial problems. The factorial of a number \( n \) is represented as \( n! \) and is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow very rapidly with increasing values of \( n \), which is why they are often used in problems involving large counts of arrangements or combinations. When you encounter a problem in permutations involving factorial, it usually indicates that every item or player is being used in every possible position or role.
Combinatorics
Combinatorics is a branch of mathematics focusing on counting, arrangement, and combination of elements within sets. It's widely used in problems where you need to determine the number of ways items can be selected or arranged. In the context of the problem, combinatorics helps in understanding how to allocate players to positions. For example, if we have 5 players and 5 positions, combinatorics principles guide us to use factorial to count the arrangements. Likewise, if we have more players than positions, such as 10 players for 5 positions, combinatorial rules help us calculate permutations by deciding how to pick and order a subset from a larger set, leading us to use the formula: \( P(n, r) = \dfrac{n!}{(n-r)!} \).
Computation of Permutations
Permutations involve arranging all members of a set in some sequence or order. When dealing with permutations, we usually ask 'in how many ways?' For the given exercise, to assign 5 players to 5 positions, you compute permutations by calculating \( P(5, 5) = 5! \). Thus, there are 120 ways. For a scenario with excess players, like 10 players for 5 positions, the permutation formula \( P(n, r) \) applies. Here, \( P(10, 5) = \dfrac{10!}{(10-5)!} \). You compute each factorial: \( 10! = 3,628,800 \) and \( 5! = 120 \). Dividing these gives you \( P(10, 5) = 30,240 \). Each calculation step involves detailed factorial computation, confirming that permutations help us figure the number of ways to assign players in given positions.

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