Points to Remember:

  • Combinations: This problem requires understanding combinations (choosing a subset from a larger set), denoted as nCr or ⁿCᵣ = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to choose.
  • Conditions: The problem involves two separate conditions, each requiring a different combinatorial calculation.
  • Addition Principle: For part (b), we'll use the addition principle of counting, summing the possibilities for different scenarios that satisfy the condition "at least one wicket keeper."

Introduction:

This question deals with combinatorial mathematics, specifically calculating the number of ways to form a cricket team under given constraints. We are given a pool of 16 cricketers (5 bowlers, 2 wicket-keepers, and 9 other players). The problem asks us to determine the number of ways to select an 11-member team based on specific conditions regarding the inclusion of bowlers and wicket-keepers. This involves applying the principles of combinations to solve two distinct scenarios.

Body:

(a) Exactly three bowlers and one wicket-keeper:

To form a team with exactly three bowlers and one wicket-keeper, we need to choose:

  • 3 bowlers from 5: ⁵C₃ = 5! / (3!2!) = 10 ways
  • 1 wicket-keeper from 2: ²C₁ = 2! / (1!1!) = 2 ways
  • Remaining players: We need to select 7 more players from the remaining 9 (16 total - 5 bowlers - 2 wicket-keepers = 9). This can be done in ⁹C₇ = ⁹C₂ = 9! / (7!2!) = 36 ways.

Therefore, the total number of ways to form a team with exactly three bowlers and one wicket-keeper is:

10 (ways to choose bowlers) * 2 (ways to choose wicket-keeper) * 36 (ways to choose remaining players) = 720 ways.

(b) At least one wicket-keeper:

This condition can be satisfied in two ways:

  • One wicket-keeper: We can choose one wicket-keeper from two in ²C₁ = 2 ways. Then we need to choose 10 more players from the remaining 14 (16 - 2 wicket-keepers). This can be done in ¹⁴C₁₀ = ¹⁴C₄ = 1001 ways. So, there are 2 * 1001 = 2002 ways to have exactly one wicket-keeper.

  • Two wicket-keepers: We choose both wicket-keepers (²C₂ = 1 way) and then choose 9 more players from the remaining 14 players (¹⁴C₉ = ¹⁴C₅ = 2002 ways). This gives us 1 * 2002 = 2002 ways.

Therefore, the total number of ways to have at least one wicket-keeper is the sum of the ways to have one wicket-keeper and the ways to have two wicket-keepers: 2002 + 2002 = 4004 ways.

Conclusion:

In summary, we have determined the number of ways to form an eleven-member cricket team under two different conditions. For (a), exactly three bowlers and one wicket-keeper, there are 720 possible team combinations. For (b), at least one wicket-keeper, there are 4004 possible team combinations. These calculations demonstrate the application of combinatorial principles in solving real-world problems involving selection and constraints. This approach ensures fairness and allows for a wide range of team compositions, promoting inclusivity within the selection process. Further analysis could explore the optimal team composition based on player statistics and strategic considerations, moving beyond simple combinatorial calculations.