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 ith box, (I = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that `n_(1) lt n_(2) lt n_(3)` is

Options

A. 82
B. 120
C. 240
D. 164

Explanation

To solve the problem of how many ways we can choose one ball from each of the three boxes such that the labels of the balls satisfy the condition \( n_1 < n_2 < n_3 \), we can follow these steps:

### Step 1: Understand the Problem
We have three boxes, each containing balls labeled from 1 to 10. We need to select one ball from each box such that the label of the ball from the first box is less than the label of the ball from the second box, which in turn is less than the label of the ball from the third box.

### Step 2: Choose Distinct Labels
Since we need \( n_1 < n_2 < n_3 \), we can think of this as choosing 3 distinct labels from the set of labels {1, 2, ..., 10}. The labels must be distinct because they need to satisfy the strict inequality.

### Step 3: Calculate the Number of Ways to Choose Labels
To choose 3 distinct labels from 10, we can use the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

In our case, we want to choose 3 labels from 10:
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

### Step 4: Conclusion
Thus, the total number of ways to choose the balls such that \( n_1 < n_2 < n_3 \) is **120**.