From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should includ

Q: Wait, why do we subtract the captain and vice-captain from the pool? Shouldn't we just pick 10 players and then assign the roles?

That is a great question! If the problem said 'choose 10 players and then assign a captain and vice-captain', you would be right. But the problem states that one specific batsman and one specific bowler *are* the captain and vice-captain and *must* be included. This means they are already fixed in the team. We are essentially selecting the remaining 8 players to complete the team of 10.

Q: Why can't we have 6 batsmen and 2 bowlers in the remaining selection?

Remember the constraint: the team must have at least 4 bowlers. We already have 1 bowler fixed as the vice-captain. So, we need at least 3 more bowlers in our selection of 8. If we chose 6 batsmen and 2 bowlers, we would only have $1 + 2 = 3$ bowlers in total, which violates the 'at least 4 bowlers' rule. Always check your total count against the original constraints!

Q: Is there a faster way to solve this without breaking it into cases?

While you could use the Principle of Inclusion-Exclusion, it is often more complex for this type of problem. Breaking it into mutually exclusive cases (like we did: 3B-5Bo, 4B-4Bo, 5B-3Bo) is the most robust and error-proof method for JEE Advanced. It keeps your logic clean and prevents over-counting.

Question

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen 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

Options:

(A) 165

(B) 155

(C) 145

(D) 135

Sigma Percentile · Accepted Answer

### The Art of Selection: Mastering Combinatorics

Welcome, future engineers! Today, we are going to tackle a classic combinatorics problem that often appears in JEE Advanced. It is not just about plugging numbers into a formula; it is about the art of logical partitioning.

Imagine you are the team selector for a prestigious cricket match. You have a pool of talent, strict rules to follow, and a team to build. Let us break this down step-by-step.

### Phase 1: The Fixed Reality

The most common mistake students make is trying to select all ten players at once. But look closely at the problem: "One batsman and one bowler who are captain and vice-captain respectively of the team should be included." This is your anchor.

These two individuals are already on the team. They are not part of the "selection" process; they are the "given."

We started with $7$ batsmen and $6$ bowlers. Since we have fixed $1$ batsman and $1$ bowler, our remaining pool is $6$ batsmen and $5$ bowlers.

We needed a team of $10$, but with $2$ already fixed, we only need to select $8$ more players. This is the first step to clarity: simplify the problem by handling the fixed elements first.

### Phase 2: The Shrinking Pool and New Constraints

Now, let us look at the constraints. The original rule was at least $4$ batsmen and at least $4$ bowlers. Since we have already placed $1$ batsman and $1$ bowler in the team, our new requirement for the remaining $8$ players is:

- We need at least $3$ more batsmen (because $1 + 3 = 4$).
- We need at least $3$ more bowlers (because $1 + 3 = 4$).

This is where the logic gets exciting. We have $8$ slots to fill, and we must satisfy these new, tighter constraints. Let $x$ be the number of additional batsmen and $y$ be the number of additional bowlers. We know $x + y = 8$, with $x \ge 3$ and $y \ge 3$.

### Phase 3: The Case Study

To find the total number of ways, we must partition our problem into mutually exclusive cases. Let us list the possible values for $(x, y)$ that satisfy our conditions:

**Case 1: $3$ Batsmen and $5$ Bowlers**

Here, $x=3$ and $y=5$. The number of ways to choose these is given by:

$$\binom{6}{3} 	imes \binom{5}{5} = 20 	imes 1 = 20 	ext{ ways}$$

**Case 2: $4$ Batsmen and $4$ Bowlers**

Here, $x=4$ and $y=4$. The number of ways is given by:

$$\binom{6}{4} 	imes \binom{5}{4} = 15 	imes 5 = 75 	ext{ ways}$$

**Case 3: $5$ Batsmen and $3$ Bowlers**

Here, $x=5$ and $y=3$. The number of ways is given by:

$$\binom{6}{5} 	imes \binom{5}{3} = 6 	imes 10 = 60 	ext{ ways}$$

Why stop here? Could we have $6$ batsmen and $2$ bowlers? No, because that would violate the requirement of at least $3$ additional bowlers.

Could we have $2$ batsmen and $6$ bowlers? No, that would violate the requirement of at least $3$ additional batsmen. The boundaries are clear, and the logic is sound.

### Phase 4: The Final Summation

Finally, we use the addition principle. Since these cases are mutually exclusive—you cannot have a team that is simultaneously $(3B, 5Bo)$ and $(4B, 4Bo)$—we simply add the results:

$$	ext{Total Ways} = 20 + 75 + 60 = 155$$

And there you have it! The **final answer is $155$**.

It is a beautiful example of how breaking a complex constraint into smaller, manageable cases leads you directly to the solution. Keep practicing this mindset, and you will find that even the most intimidating combinatorics problems become a thrilling puzzle to solve. You have got this!

Select Answer:

Visualized Solution

Initial Setup

Fixed Captains

New Constraints

Possible Cases

Case 1: $3$ Batsmen, $5$ Bowlers

Case 1 Calculation

Case 2: $4$ Batsmen, $4$ Bowlers

Case 2 Calculation

Case 3: $5$ Batsmen, $3$ Bowlers

Case 3 Calculation

Total Combinations

The Sigma Insight: Combinations and Selection

The Art of Selection

Phase 1

Phase 2

Phase 3

Phase 4

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