A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 India

Q: Wait, why can't I just use one big formula to solve this? Why do I have to break it into cases?

That is a fantastic question! In combinatorics, we use a single formula only when the selection process is uniform. Here, the number of foreigners depends directly on the number of Indians. Because the number of Indians can vary (it could be 2, 3, or 4), the 'size' of the committee changes. These are mutually exclusive scenarios, so we must calculate each scenario separately and then add them up. Think of it as building different types of committees; you can't build them all at once!

Q: What happens if I calculate for $I=5$? The math gives me a number, but is it valid?

This is the classic JEE trap! If you plug $I=5$ into the constraint $F=2I$, you get $F=10$. But look at your pool: you only have $8$ foreigners. You cannot select $10$ people from a group of $8$. Mathematically, $^8C_{10}$ is $0$, but physically, it's an impossible state. Always, always check your constraints against the available pool size before you start crunching numbers.

Q: Why do we add the results of the cases instead of multiplying them?

Great intuition check! We multiply when we need to perform two tasks *together* (e.g., choosing Indians AND choosing foreigners). We add when we have *alternative* scenarios (e.g., Case 1 OR Case 2 OR Case 3). Since you are forming *one* committee, you are either in the scenario where $I=2$ OR the scenario where $I=3$, and so on. Therefore, we sum the possibilities.

Question

A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

Options:

(A) 560

(B) 1050

(C) 1625

(D) 575

Sigma Percentile · Accepted Answer

### The Art of Constrained Selection: A Journey into Combinatorics

Welcome, future engineer. Today, we are not just solving a problem; we are stepping into the shoes of a committee architect.

In the world of JEE Advanced, combinatorics is rarely about memorizing formulas. It is about the art of counting without counting—the ability to visualize a system, identify its boundaries, and systematically dismantle its complexity.

### Phase 1: The Detective Work

Imagine you are standing in front of a room. On the left, you have $6$ Indians. On the right, you have $8$ foreigners.

Your task is to form a committee under two strict laws:
1. You must have at least $2$ Indians ($I \ge 2$).
2. The number of foreigners must be exactly double the number of Indians ($F = 2I$).

This is where most students rush. They see the numbers and start calculating. Don't.

First, be a detective. What are the possible configurations?

If $I=2$, then $F=2(2)=4$. This is valid because $4 \le 8$.

If $I=3$, then $F=2(3)=6$. This is also valid because $6 \le 8$.

If $I=4$, then $F=2(4)=8$. This is valid because $8 \le 8$.

What about $I=5$? Then $F=10$. Stop! You only have $8$ foreigners. You cannot pick $10$. The universe of this problem collapses at $I=5$. By identifying these boundaries, you have already won half the battle.

### Phase 2: The Systematic Construction

Now, let us build these committees one by one. We use the combination formula $^nC_r$, which represents the number of ways to choose $r$ items from a pool of $n$.

**Case 1: The $I=2, F=4$ Committee**

We need to choose $2$ Indians from $6$ and $4$ foreigners from $8$. The math is elegant:

$$^6C_2 	imes ^8C_4$$

Calculating $^6C_2$:

$$^6C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$$

Calculating $^8C_4$:

$$^8C_4 = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$$

Multiplying these gives us $15 	imes 70 = 1050$ ways. This is our first foundation.

**Case 2: The $I=3, F=6$ Committee**

Now we scale up. We need $3$ Indians from $6$ and $6$ foreigners from $8$. The math:

$$^6C_3 	imes ^8C_6$$

Calculating $^6C_3$:

$$^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$

Here is a pro-tip: $^8C_6$ is the same as $^8C_2$ (the symmetry property).

$$^8C_2 = \frac{8 	imes 7}{2 	imes 1} = 28$$

Multiplying these gives us $20 	imes 28 = 560$ ways.

**Case 3: The $I=4, F=8$ Committee**

Finally, the limit. We need $4$ Indians from $6$ and $8$ foreigners from $8$. The math:

$$^6C_4 	imes ^8C_8$$

Calculating $^6C_4$:

$$^6C_4 = ^6C_2 = 15$$

Calculating $^8C_8$:

$$^8C_8 = 1$$

Multiplying these gives us $15 	imes 1 = 15$ ways.

### Phase 3: The Grand Synthesis

We have three distinct, mutually exclusive worlds. In the world of combinatorics, when you have 'OR' scenarios, you add.

We are not forming three committees; we are forming one committee that could belong to any of these three configurations. Therefore, we sum them up:

$$	ext{Total Ways} = 1050 + 560 + 15 = 1625$$

### The Takeaway

Look at that number: **$1625$**. It represents every possible valid committee you could form under these constraints.

The beauty of this problem isn't the final number; it's the discipline of the process. You checked the constraints, you partitioned the problem into manageable cases, and you executed the arithmetic with precision.

This, my friend, is the JEE mindset. Never rush the setup, and the calculation will always follow. Keep this clarity, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Permutations and Combinations: A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

Select Answer:

Visualized Solution

Analyzing the Selection Pool

Understanding the Constraints

Identifying Valid Cases

Case 1: $I = 2, F = 4$ Setup

Calculating Case 1

Case 2: $I = 3, F = 6$ Setup

Calculating Case 2

Case 3: $I = 4, F = 8$ Setup

Calculating Case 3

Summing All Possible Ways

Final Result

The Sigma Insight: Combinations and Selection

The Art of Constrained Selection

Phase 1

Phase 2

Phase 3

The Takeaway

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