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. 120
B. 240
C. 164
D. 82

Explanation

To solve the problem of finding the number of ways to draw balls from 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 labels of the balls drawn from the boxes are in strictly increasing order.

### Step 2: Choose Distinct Labels
Since we want \( n_1 < n_2 < n_3 \), we need to select three distinct labels from the set {1, 2, ..., 10}. The order of selection matters because we want them to be in increasing order.

### Step 3: Calculate the Combinations
The number of ways to choose 3 distinct labels from 10 can be calculated using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. Here, \( n = 10 \) and \( r = 3 \).

\[ \text{Number of ways} = \binom{10}{3} \]

### Step 4: Compute the Combination
Now, we compute \( \binom{10}{3} \):

\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 \]

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

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