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In how many ways can 5 starting positions on a basketball team be filled with 8 men who can play any of the positions?

Short Answer

Expert verified
There are 560 ways for the 5 positions on a basketball team to be filled with 8 men.

Step by step solution

01

Understanding Combinations

In combinatorics, a branch of mathematics, the concept of combination refers to the selection of items without considering the order. It is often represented as \(C(n, k)\) where \(n\) is the total number of items and \(k\) is the number of items to select. In this scenario, there are 8 men (the total number of items) and 5 positions to be filled (the number of selections to make).
02

Applying the Formula

The standard formula for combinations or \(C(n, k)\) is given by \[C(n,k) = \frac{n!}{k!(n-k)!}\]where \(n!\) is the factorial of \(n\), \(k!\) is the factorial of \(k\), and \((n – k)!\) is the factorial of \((n – k)\). Factorial, denoted by an exclamation point (!), is the product of all positive integers up to the given number. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Now, substitute the given values into the formula: \(C(8, 5) = \frac{8!}{5!(8-5)!}\).
03

Solving the Equation

Calculate the factorial for each value. \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\), \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\), and \((8-5)! = 3! = 3 \times 2 \times 1 = 6\). Substitute these values back into the formula. \(C(8, 5) = \frac{40320}{120 \times 6} = 560\).

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Key Concepts

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

Combination Formula
Combinations are a fundamental concept in combinatorics which focus on choosing a subset of items from a larger set without considering the order. The combination formula is written as \( C(n, k) \), showing how many ways you can select \( k \) items from \( n \) items. This is useful in scenarios like team selection, especially when the order of selection does not matter.

The formula used is:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]

This formula helps in calculating the number of possible combinations by dividing the factorial of the total number of items \( n! \) by the product of the factorials of the number of items to choose \( k! \) and the difference \((n-k)!\).
Factorial Calculation
Understanding factorials is crucial for applying the combination formula. Factorials, denoted by \(!\), are products of an integer and all the positive integers below it.

For instance:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

For larger numbers, factorials grow very quickly:
  • \( 3! = 3 \times 2 \times 1 = 6 \)
  • \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)

These calculations are necessary steps in solving combination problems. Ensure that each factorial is calculated accurately to get the correct number of combinations.
Selection Without Order
When selecting items without considering the order, it means rearranging the chosen items does not create a new selection. This is the principle behind combinations. It is key to understanding when order matters (like permutations) and when it does not (like combinations).

In our example, filling 5 positions on a basketball team out of 8 players does not matter in which order you select the players. The focus is purely on which players are chosen, not the sequence. The order of players' names doesn’t affect the distinct grouping of 5 people.

This concept is vital in scenarios requiring arrangements where order doesn't alter the outcome, ensuring clarity in problems where grouping is more important than line-up.

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