A three digit number is chosen, what is the probability of having digits in increasing order from left to right .

Answer

To find the probability of a randomly chosen three-digit number having its digits in increasing order from left to right, we can follow these steps:

### Step 1: Determine the total number of three-digit numbers.
The range of three-digit numbers is from 100 to 999. Therefore, the total number of three-digit numbers is:
\[ 999 - 100 + 1 = 900 \]

### Step 2: Identify the conditions for digits to be in increasing order.
For a three-digit number to have its digits in increasing order, we need to select three different digits from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. The digit '0' cannot be included because it cannot be the leading digit in a three-digit number.

### Step 3: Calculate the number of ways to choose three digits.
We can choose 3 digits from the 9 available digits (1 to 9). The number of ways to choose 3 digits from 9 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n = 9 \) and \( r = 3 \):
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

### Step 4: Calculate the probability.
The probability \( P \) of choosing a three-digit number with digits in increasing order is given by the ratio of the number of favorable outcomes to the total number of outcomes:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{84}{900} \]

### Step 5: Simplify the probability.
To simplify \( \frac{84}{900} \):
\[ \frac{84 \div 12}{900 \div 12} = \frac{7}{75} \]

### Final Answer:
The probability of having digits in increasing order from left to right in a three-digit number is:
\[ \frac{7}{75} \]