Three digit numbers are to be formed out of natural numbers 1 to 9 so that each number has digits in increasing order from left to right. Numbers of such numbers is

Options

A. 343
B. 84
C. 126
D. none of these

Answer

To solve the problem of forming three-digit numbers from the natural numbers 1 to 9 such that the digits are in increasing order, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to form three-digit numbers using the digits from 1 to 9, where the digits must be in strictly increasing order. This means that once we choose three digits, there is only one way to arrange them (in increasing order).

2. **Choosing the Digits**:
Since the digits must be in increasing order, we can think of this as a combination problem. We need to choose 3 different digits from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}.

3. **Using Combinations**:
The number of ways to choose 3 digits from 9 can be calculated using 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.

4. **Calculating the Combinations**:
Here, \( n = 9 \) and \( r = 3 \):
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!} \]
Simplifying this:
\[ = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

5. **Final Result**:
Therefore, the total number of three-digit numbers that can be formed from the digits 1 to 9 in increasing order is **84**.

### Summary:
The total number of three-digit numbers that can be formed from the digits 1 to 9, with the digits in increasing order, is **84**.