Understanding the Team Selection Problem

The problem asks us to find the number of ways to select a team of 5 players from a group of 8 players, with a specific condition: a particular player must not be included in the team selection.

This is a classic problem involving combinations, as the order in which the players are selected does not matter; only the final group of 5 players forms the team.

Applying the Constraint: Excluding a Specific Player

The key constraint here is that one specific player out of the original 8 cannot be part of the team. If this particular player is excluded from the selection pool from the start, we are effectively selecting our 5-player team from a smaller group of players.

The number of players available for selection becomes:

Total players - Excluded player = \(8 - 1 = 7\) players.

So, the problem simplifies to finding the number of ways to choose a team of 5 players from these 7 available players.

Calculating Combinations

The number of ways to choose \(k\) items from a set of \(n\) distinct items, where the order of selection does not matter, is given by the combination formula:

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

In our case:

• \(n\) = Number of players available for selection = 7
• \(k\) = Number of players to select for the team = 5

We need to calculate \(C(7, 5)\).

Step-by-Step Calculation of C(7, 5)

Let's calculate the value using the formula:

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

First, calculate the difference in the denominator:

\((7-5)! = 2!\)

So, the formula becomes:

\(\frac{7!}{5!2!}\)

Now, expand the factorials. Remember that \(n! = n \times (n-1) \times \dots \times 1\). We can expand \(7!\) until \(5!\) to simplify the calculation:

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

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

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

Substitute these values back into the formula:

\(\frac{7 \times 6 \times 5!}{5! \times 2!}\)

Cancel out the \(5!\) terms in the numerator and denominator:

\(\frac{7 \times 6}{2!}\)

Substitute the value of \(2!\):

\(\frac{7 \times 6}{2}\)

Perform the multiplication and division:

\(\frac{42}{2} = 21\)

So, the number of ways to select a team of 5 players from 7 players is 21.

This is the number of ways to select the team of 5 players from the original 8 players, while ensuring that the particular player is not included.

Conclusion

The number of ways to select a team of 5 players from 8 players so as not to include a particular player is 21.

Revision Table: Key Concepts in Team Selection

Additional Information: Understanding Combinations with Constraints

Problems involving combinations often include specific conditions or constraints that affect the selection process. When a problem states that certain items must be excluded, the approach is to first reduce the total number of items available for selection by the number of items to be excluded. Then, you perform the combination calculation on the reduced set and the required number of items to be selected.

For example, if you had to select a committee of 3 from 10 people, but 2 specific people cannot serve together:

1. Calculate total combinations without constraints: \(\binom{10}{3}\).
2. Calculate combinations where the 2 specific people ARE together (select the 2, then 1 more from the remaining 8: \(1 \times \binom{8}{1}\)).
3. Subtract the constrained combinations from the total: \(\binom{10}{3} - \binom{8}{1}\).

In our specific team selection problem, the constraint was simpler: one player is simply unavailable, reducing the pool directly.