There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketke

Q: Wait, why do we add the cases instead of multiplying them?

Great question! We add them because these cases are mutually exclusive. You are either forming a team with $2$ wicketkeepers OR a team with $6$ batsmen OR a team with $5$ bowlers. You cannot have all these scenarios happening at the same time for a single team. In combinatorics, when you have 'OR' conditions, you add the possibilities.

Q: Why can't I just select the minimums first and then pick the remaining $1$ player from the remaining $5$ players?

That is a classic trap! If you pick the minimums first, you are essentially treating the players as 'already selected' and then picking the last one. This approach fails because it doesn't account for the different ways the groups (bowlers, batsmen, wicketkeepers) can be formed. For example, picking a bowler as the 'extra' player is a completely different set of combinations than picking a batsman. You must define the exact composition of the team first.

Q: Is there a way to use complementary counting to solve this?

Technically, yes, but it would be much more complex. You would have to subtract the invalid cases (where $B < 4$ or $Bt < 5$ or $W < 1$) from the total possible ways to choose $11$ from $15$, which is $\binom{15}{11}$. Given the constraints, the direct case-by-case method is much faster and less prone to errors.

Question

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

Sigma Percentile · Accepted Answer

### The Art of Selection: Mastering Combinatorics

Imagine you are the chief selector for a national cricket team. You have a pool of $15$ talented players: $6$ bowlers, $7$ batsmen, and $2$ wicketkeepers. Your mission is to assemble a balanced playing eleven under specific constraints.

You need at least $4$ bowlers, $5$ batsmen, and $1$ wicketkeeper. This is the kind of problem that separates the casual observer from the master strategist. Let's break it down.

### Phase 1: The Boundary Conditions

In any combinatorial problem, the first step is to understand the boundaries. We have a total of $15$ players and we need to select $11$. The constraints are:

- Bowlers ($B$): $B \ge 4$
- Batsmen ($Bt$): $Bt \ge 5$
- Wicketkeepers ($W$): $W \ge 1$

If we simply take the minimum requirements, we have $4 + 5 + 1 = 10$ players. Since we need $11$ players, we have exactly one 'extra' spot to fill. This realization transforms a daunting problem into three manageable, distinct cases.

### Phase 2: The Three Scenarios

We can now define our three mutually exclusive scenarios based on who takes that extra spot:

1. **Case 1: The Wicketkeeper Focus.** We take the minimums and add an extra wicketkeeper. The composition becomes $(4, 5, 2)$.

2. **Case 2: The Batsman Focus.** We take the minimums and add an extra batsman. The composition becomes $(4, 6, 1)$.

3. **Case 3: The Bowler Focus.** We take the minimums and add an extra bowler. The composition becomes $(5, 5, 1)$.

### Phase 3: The Calculation

We use the power of combinations, denoted as $\binom{n}{r}$, which calculates the number of ways to choose $r$ items from a set of $n$.

**For Case 1 $(4, 5, 2)$:**
We choose $4$ bowlers from $6$, $5$ batsmen from $7$, and $2$ wicketkeepers from $2$.

$$	ext{Ways} = \binom{6}{4} 	imes \binom{7}{5} 	imes \binom{2}{2} = 15 	imes 21 	imes 1 = 315$$

**For Case 2 $(4, 6, 1)$:**
We choose $4$ bowlers from $6$, $6$ batsmen from $7$, and $1$ wicketkeeper from $2$.

$$	ext{Ways} = \binom{6}{4} 	imes \binom{7}{6} 	imes \binom{2}{1} = 15 	imes 7 	imes 2 = 210$$

**For Case 3 $(5, 5, 1)$:**
We choose $5$ bowlers from $6$, $5$ batsmen from $7$, and $1$ wicketkeeper from $2$.

$$	ext{Ways} = \binom{6}{5} 	imes \binom{7}{5} 	imes \binom{2}{1} = 6 	imes 21 	imes 2 = 252$$

### The Grand Total

Because these cases are mutually exclusive, we apply the Addition Principle to find the total number of valid team combinations.

$$	ext{Total Ways} = 315 + 210 + 252 = 777$$

There you have it! By systematically breaking down the constraints and identifying the 'extra' player, we have navigated through the complexity to find that there are exactly **$777$** ways to select your team. Remember, in JEE Advanced, the secret lies in the logical structure you build before you even touch your pen to the paper.

Enter Numerical Value:

Visualized Solution

Problem Overview

Identifying Constraints

Formulating Valid Cases

Case 1: $(4, 5, 2)$ Setup

Case 1: Calculation

Case 2: $(4, 6, 1)$ Setup

Case 2: Calculation

Case 3: $(5, 5, 1)$ Setup

Case 3: Calculation

Total Number of Ways

The Sigma Insight: Combinations and Selection

The Art of Selection

Phase 1

Phase 2

Phase 3

The Grand Total

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