Consider three boxes, each containing 10 balls labelled 1, 2, ..., 10. Suppose one ball is randomly

Q: Wait, why can't we just use $10 \times 9 \times 8$?

That is a fantastic question! If you calculate $10 \times 9 \times 8$, you are calculating the number of permutations—that is, the number of ways to pick 3 balls where the order *matters* and is *distinct*. However, in our problem, the order is not free; it is strictly constrained by $n_1 < n_2 < n_3$. For any set of 3 numbers you pick, there is only one way to arrange them to satisfy this inequality. Multiplying $10 \times 9 \times 8$ would overcount by a factor of $3!$ (which is 6), because it treats the sequence $(2, 5, 7)$ as different from $(7, 5, 2)$, even though only one of those satisfies our condition.

Q: What if the condition was $n_1 \le n_2 \le n_3$?

If the condition were $n_1 \le n_2 \le n_3$, the problem would become significantly more complex! This is known as 'combinations with repetition' or 'stars and bars'. You would no longer be picking 3 distinct numbers; you could pick the same number multiple times. The formula would then change to $\binom{n+r-1}{r}$. But for our problem, the strict inequality $n_1 < n_2 < n_3$ keeps it elegant and simple: we only need to select distinct numbers.

Q: Why do we treat the boxes as distinct but the numbers as a pool?

The boxes are distinct because they represent the positions $n_1, n_2,$ and $n_3$. However, the *labels* on the balls are what we are choosing. Because we need to satisfy $n_1 < n_2 < n_3$, we are essentially choosing a subset of 3 labels from the 10 available. Once you have that subset, the 'distinctness' of the boxes is automatically handled by the fact that there is only one way to place those 3 labels into the boxes to satisfy the inequality.

Question

Consider three boxes, each containing 10 balls labelled $1, 2, ..., 10$. Suppose one ball is randomly drawn from each of the boxes. Denote by $n_i$, the label of the ball drawn from the $i^{th}$ box, $(i = 1, 2, 3)$. Then, the number of ways in which the balls can be chosen such that $n_1 < n_2 < n_3$ is :

Options:

(A) 82

(B) 240

(C) 164

(D) 120

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three boxes, each containing balls labeled $1$ through $10$. We draw one ball from each box, resulting in values $n_1, n_2,$ and $n_3$.

The objective is to determine the number of ways these draws can satisfy the strict inequality $n_1 < n_2 < n_3$.

At first glance, this might appear to be a complex task involving permutations and case-based counting. However, the most elegant solutions in JEE Advanced often arise from observing hidden symmetries rather than brute-force calculation.

### The Insight: Order is an Illusion

Imagine you have drawn three distinct numbers from the boxes, such as $7, 2,$ and $5$.

Consider the condition $n_1 < n_2 < n_3$. Is there any ambiguity in how these numbers must be assigned to the boxes?

No. You must place $2$ in the first box, $5$ in the second, and $7$ in the third to satisfy the inequality. The condition acts as a gravitational force, pulling the numbers into a single, unique, sorted configuration.

Because the order is strictly dictated by the values themselves, we do not need to worry about permutations. If we pick any three distinct numbers, there is exactly one way to arrange them to satisfy the condition.

The problem of "arranging" effectively vanishes, leaving us with a much simpler task: **selection**.

### The Mathematical Execution

Since we only need to select $3$ distinct numbers from the $10$ available, we are looking for the number of combinations of $10$ items taken $3$ at a time. This is represented by the binomial coefficient $\binom{n}{r}$, where $n=10$ and $r=3$.

The general formula is defined as:

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Substituting our specific values into the formula, we obtain:

$$\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! 	imes 7!}$$

To simplify the calculation, we expand the numerator just enough to cancel the largest factorial in the denominator:

$$\frac{10 	imes 9 	imes 8 	imes 7!}{3! 	imes 7!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1}$$

The $7!$ terms cancel out, leaving us with a straightforward arithmetic expression. The numerator is $10 	imes 9 	imes 8 = 720$, and the denominator is $3 	imes 2 	imes 1 = 6$.

Dividing $720$ by $6$ yields the final result.

### Final Calculation

The total number of ways to satisfy the condition $n_1 < n_2 < n_3$ is **$120$**.

This problem serves as a vital lesson in combinatorial strategy. When you encounter a constraint like $n_1 < n_2 < n_3$, always ask: "Does this constraint fix the order?"

If the answer is yes, you have successfully transformed a permutation problem into a combination problem. This perspective is the hallmark of a master student, allowing you to uncover the simplicity hidden beneath complex problem statements.

Select Answer:

Visualized Solution

Visualizing the Setup

The Core Constraint: $n_{1} < n_{2} < n_{3}$

The Logic of Selection

Reducing to Combinations

Applying the Formula

Substituting Values

Simplifying Factorials

Atomic Calculation

Final Result

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Insight

The Mathematical Execution

Final Calculation

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