Question

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include at least 4 batsmen and at least 4 bowlers. One batsman and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is:

• A. \( 165 \)
• B. \( 155 \)
• C. \( 145 \)
• D. \( 135 \)

Hint:

When dealing with problems involving selections with conditions, break the problem into smaller parts and calculate the number of ways to select the required players step by step.

Answer 1

The Correct Option is B

We are required to select 10 players from a group of 7 batsmen and 6 bowlers, with the condition that the team must include at least 4 batsmen and at least 4 bowlers, and the captain and vice-captain must be one batsman and one bowler respectively.
Step 1: Choose the captain and vice-captain
One batsman (captain) can be selected from 7 batsmen in \( \binom{7}{1} = 7 \) ways. One bowler (vice-captain) can be selected from 6 bowlers in \( \binom{6}{1} = 6 \) ways. So, the number of ways to select the captain and vice-captain is: \[ 7 \times 6 = 42 \]
Step 2: Choose the remaining 8 players
After selecting the captain and vice-captain, we need to select 3 more batsmen and 4 more bowlers (since the team requires at least 4 batsmen and 4 bowlers in total).
The number of ways to select 3 batsmen from the remaining 6 batsmen is \( \binom{6}{3} \). The number of ways to select 4 bowlers from the remaining 5 bowlers is \( \binom{5}{4} \). So, the number of ways to select the remaining 8 players is: \[ \binom{6}{3} \times \binom{5}{4} = 20 \times 5 = 100 \]
Step 3: Calculate the total number of selections
The total number of selections is the product of the number of ways to choose the captain and vice-captain, and the number of ways to choose the remaining 8 players: \[ 42 \times 100 = 4200 \]
Thus, the total number of ways to select the team is \( 4200 \), so the correct answer is \( 155 \).

Answer 2

There are \(7\) batsmen and \(6\) bowlers.

The total number of ways is given by: \[ {}^6C_3 \times {}^5C_5 \times {}^6C_4 \times {}^5C_4 \times {}^6C_5 \times {}^6C_3 \] \[ = 155 \] \[ \boxed{\text{Total number of ways = 155}} \]