The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is

Options

A. A) 127
B. B) 128
C. C) 502
D. D) 512

Answer

To find the number of natural numbers of two or more digits in which the digits are in increasing order, we can follow these steps:

### Step 1: Identify the available digits
The available digits for forming natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. We do not include 0 because a natural number cannot start with 0.

### Step 2: Determine the total number of digits
We have a total of 9 digits (1 to 9).

### Step 3: Understand the requirement for increasing order
When digits are in increasing order, each selection of digits will automatically be in increasing order. Therefore, we can simply choose the number of digits we want to use from the available 9 digits.

### Step 4: Calculate combinations for different digit lengths
We will calculate the combinations for numbers with 2 digits up to 9 digits.

- **For 2-digit numbers:** We need to choose 2 digits from 9.
\[ \text{Number of 2-digit combinations} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \]

- **For 3-digit numbers:** We need to choose 3 digits from 9.
\[ \text{Number of 3-digit combinations} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

- **For 4-digit numbers:** We need to choose 4 digits from 9.
\[ \text{Number of 4-digit combinations} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

- **For 5-digit numbers:** We need to choose 5 digits from 9.
\[ \text{Number of 5-digit combinations} = \binom{9}{5} = \binom{9}{4} = 126 \quad (\text{by property of combinations}) \]

- **For 6-digit numbers:** We need to choose 6 digits from 9.
\[ \text{Number of 6-digit combinations} = \binom{9}{6} = \binom{9}{3} = 84 \]

- **For 7-digit numbers:** We need to choose 7 digits from 9.
\[ \text{Number of 7-digit combinations} = \binom{9}{7} = \binom{9}{2} = 36 \]

- **For 8-digit numbers:** We need to choose 8 digits from 9.
\[ \text{Number of 8-digit combinations} = \binom{9}{8} = \binom{9}{1} = 9 \]

- **For 9-digit numbers:** There is only one way to choose all 9 digits.
\[ \text{Number of 9-digit combinations} = \binom{9}{9} = 1 \]

### Step 5: Sum all combinations
Now we sum all the combinations calculated for each digit length:
\[ \text{Total} = 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 502 \]

### Final Answer
The total number of natural numbers of two or more digits in which the digits are in increasing order is **502**.

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