Understanding the Player Selection Problem

This question asks us to find the number of ways to select a team of 5 players from a group of 9 players, with a specific condition: two particular players must not be included in the team. This is a problem involving combinations, as the order in which the players are selected for the team does not matter.

Analyzing the Constraints

Let's break down the given information:

• Total number of players available initially = 9
• Size of the team to be selected = 5
• Constraint: Two specific players must be excluded from the selection.

Adjusting the Pool of Players

Since two particular players are excluded, they are removed from the pool of available players. We need to find the number of players remaining from whom we can select the team.

Number of players available for selection = Total players - Players to be excluded

Number of players available for selection = \(9 - 2 = 7\)

So, we now need to select a team of 5 players from this reduced group of 7 players.

Applying the Combination Formula

We are selecting 5 players from 7 available players, and the order of selection doesn't matter. This is a combination problem, which can be solved using the combination formula:

The number of combinations of selecting \(k\) items from a set of \(n\) items is given by:

\[ C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

In this case, \(n = 7\) (the number of players available for selection) and \(k = 5\) (the number of players to select for the team).

So, the number of ways to select the team is:

\[ \binom{7}{5} = \frac{7!}{5!(7-5)!} \]

\[ \binom{7}{5} = \frac{7!}{5!2!} \]

Calculating the Number of Ways

Let's calculate the factorial values:

\(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)

\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

\(2! = 2 \times 1 = 2\)

Now, substitute these values into the formula:

\[ \binom{7}{5} = \frac{5040}{120 \times 2} = \frac{5040}{240} \]

Alternatively, we can simplify the factorial expression directly:

\[ \binom{7}{5} = \frac{7 \times 6 \times 5!}{5! \times 2 \times 1} = \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21 \]

So, there are 21 ways to select a team of 5 players out of 9, excluding two particular players.

Summary of the Solution

We started with 9 players and needed to exclude 2, leaving 7 players. We then selected a team of 5 from these 7 players using the combination formula \( \binom{7}{5} \). The calculation showed that there are 21 possible ways to form the team under the given conditions.

Revision Table: Key Concepts

Let's quickly review the concepts used in this problem.

Additional Information: Combinations vs. Permutations

It's important to understand the difference between combinations and permutations in counting problems.

• Combinations: Used when the order of selection does not matter. For example, selecting a team of players or choosing a set of items. The formula is \( C(n, k) \).
• Permutations: Used when the order of selection does matter. For example, arranging items in a line or assigning specific roles to selected individuals. The formula is \( P(n, k) = \frac{n!}{(n-k)!} \).

In this problem, selecting players A, B, C, D, E for the team is the same as selecting players E, D, C, B, A. Therefore, it is a combination problem.

The constraint of excluding specific players simplifies the problem by reducing the total pool from which selections are made. We simply subtract the excluded players from the initial total before applying the combination formula for the required team size.