Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no

Q: Wait, why can't we put dividers at the ends? Wouldn't that just mean the first or last box is empty?

Exactly! You've hit the nail on the head. The constraint is that 'no box is empty'. If you place a divider before the first ball, the first box is empty. If you place a divider after the last ball, the last box is empty. By restricting our choices to the $9$ internal gaps, we guarantee that every box contains at least one ball.

Q: What if the balls were distinct? Would the answer still be $^9C_3$?

Great question! No, it would not. If the balls were distinct, the order in which they are placed into the boxes would matter, and we would need to use a different approach, likely involving multinomial coefficients or inclusion-exclusion. The 'Stars and Bars' method specifically relies on the balls being identical.

Q: Why is Statement-2 the correct explanation for Statement-1?

Statement-2 provides the mathematical tool used to solve the physical problem in Statement-1. The physical constraint of 'no empty boxes' forces the distribution problem to map perfectly onto the mathematical problem of 'choosing $3$ gaps out of $9$'. Thus, the logic of Statement-2 is the foundation for the result in Statement-1.

Question

Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is $^9C_3$. Statement-2: The number of ways of choosing any 3 places from 9 different places is $^9C_3$.

Options:

(A) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B) Statement-1 is true, Statement-2 is false.

(C) Statement-1 is false, Statement-2 is true.

(D) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a row of $10$ identical blue balls. Your mission is to partition these balls into $4$ distinct boxes, with one ironclad rule: no box can be empty.

At first glance, this might feel like a daunting task of trial and error. But in the world of combinatorics, we don't guess; we visualize. This is where the **Divider Method**—or as mathematicians affectionately call it, **Stars and Bars**—becomes your most powerful weapon.

### The Geometry of Gaps

To divide our $10$ balls into $4$ groups, we don't need $4$ dividers. Think about a simple piece of string: to cut it into $4$ pieces, you only need $3$ cuts.

Similarly, to create $4$ distinct boxes, we only need $3$ dividers. Now, here is the secret: where do these dividers go? If we place them anywhere, we might accidentally create an empty box.

To satisfy the **no empty box** constraint, we must place our dividers only in the gaps between the balls. With $10$ balls in a row, there are exactly $9$ internal gaps.

By choosing $3$ of these $9$ gaps to place our dividers, we ensure that every box receives at least one ball. It is a beautiful, one-to-one mapping where every unique selection of $3$ gaps corresponds to exactly one unique distribution of balls.

### The Mathematical Elegance

The problem has now transformed from a complex distribution puzzle into a simple selection problem. We have $9$ available gaps, and we need to choose $3$ of them.

The number of ways to do this is given by the combination formula, denoted as $^9C_3$. This is the core of Statement-1.

$$^9C_3 = \frac{9!}{3!(9-3)!} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$$

When we look at Statement-2, it simply defines the fundamental principle of combinations: choosing $r$ items from $n$ distinct items is $^nC_r$.

Since our problem is exactly that—choosing $3$ gaps from $9$—Statement-2 is not just a random fact; it is the very engine that drives the logic of Statement-1.

**Both statements are true**, and Statement-2 provides the essential mathematical justification for the result in Statement-1. You have just mastered a fundamental technique that will serve you throughout your JEE journey.

Animated Solution for Mathematics - Permutations and Combinations: Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is $^{9} C_{3}$ . Statement-2: The number of ways of choosing any 3 places from 9 different places is $^{9} C_{3}$ .

Select Answer:

Visualized Solution

The Setup: Balls and Boxes

The Divider Method for $r = 4$

Identifying the Gaps

The No Empty Box Constraint

Placing the Dividers

Calculating the Ways: $^{9} C_{3}$

Analyzing Statement-2: $^{9} C_{3}$

The Final Verdict

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Geometry of Gaps

The Mathematical Elegance

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