Solving Cricket Team Selection with Captain Constraint

Let's break down the problem of selecting a cricket team of 11 players from a batch of 15 players, with the specific condition that the captain must always be included in the team.

Understanding the Cricket Team Selection Problem

We need to form a team of 11 players. We have 15 players available in total. A key condition is that one specific player, the captain, is automatically selected for the team.

Step-by-Step Team Selection Process

Since the captain is already included in the team of 11, we effectively need to choose the remaining players for the team from the rest of the available batch.

• Total players in the batch = 15
• Size of the team to be selected = 11
• Player who is always included = 1 (the captain)

Because the captain is already chosen, we need to select fewer players from a smaller pool of players.

• Number of players already selected (the captain) = 1
• Number of players remaining to be selected for the team = 11 - 1 = 10
• Number of players remaining in the batch after the captain is set aside = 15 - 1 = 14

So, the problem simplifies to choosing 10 players from the remaining 14 players.

Calculating the Number of Ways

The number of ways to choose 10 players from 14 players is a combination problem, as the order in which the players are selected does not matter. We use the combination formula:

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

Where:

• \(n\) is the total number of items to choose from (remaining players in the batch) = 14
• \(k\) is the number of items to choose (remaining players for the team) = 10

Substituting the values into the formula:

\(\binom{14}{10} = \frac{14!}{10!(14-10)!}\)

\(\binom{14}{10} = \frac{14!}{10!4!}\)

Let's expand the factorials and simplify:

\(\frac{14 \times 13 \times 12 \times 11 \times 10!}{10! \times 4 \times 3 \times 2 \times 1}\)

Cancel out \(10!\) from the numerator and denominator:

\(\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}\)

Simplify the denominator: \(4 \times 3 \times 2 \times 1 = 24\)

\(\frac{14 \times 13 \times 12 \times 11}{24}\)

Simplify by dividing 12 by 24:

\(\frac{14 \times 13 \times (12/12) \times 11}{(24/12)} = \frac{14 \times 13 \times 1 \times 11}{2}\)

Now, divide 14 by 2:

\(\frac{(14/2) \times 13 \times 11}{1} = 7 \times 13 \times 11\)

Calculate the product:

\(7 \times 13 = 91\)

\(91 \times 11 = 1001\)

The number of ways to choose the remaining 10 players from 14 is 1001.

Since the captain is always included, every combination of these 10 players with the captain forms a unique valid team of 11. Therefore, the total number of ways to choose the team with the captain always included is 1001.

Summary of Calculation

Revision Table: Team Selection Concepts

Additional Information: Combinatorics in Sports

Combinatorics is a branch of mathematics concerned with counting, arrangement, and combination. Problems like selecting teams, forming batting orders, or scheduling tournaments in sports often involve combinatorics.

• Team Selection: Choosing a group of players from a larger set (combination).
• Batting Order: Arranging a set of players in a specific sequence (permutation).
• Tournament Scheduling: Determining the pairings and sequence of matches (more complex combinatorial problems).

In this specific problem, because the captain's position is fixed and we are just selecting the remaining members to fill the team spots without considering their specific roles within the team (other than being a player), it's a straightforward combination problem.