The problem asks to determine how many 3-digit numbers can be formed from the digits 2, 3, 4, 5, 7, and 9 that are less than 400. We assume repetition of digits is allowed.

1. Problem Description

The problem asks to determine how many 3-digit numbers can be formed from the digits 2, 3, 4, 5, 7, and 9 that are less than

4

0

0. We assume repetition of digits is allowed.

2. Solution Steps

A 3-digit number is less than 400 if its first digit is less than

4. From the digits 2, 3, 4, 5, 7 and 9, only 2 and 3 are less than

4. Therefore, the first digit can be either 2 or

3. So, we have 2 choices for the first digit.

Since repetition is allowed, we have 6 choices for the second digit (2, 3, 4, 5, 7, or 9) and 6 choices for the third digit (2, 3, 4, 5, 7, or 9).

The number of possible 3-digit numbers less than 400 is the product of the number of choices for each digit.

Therefore, the total number of such 3-digit numbers is $2 \times 6 \times 6$.

$2 \times 6 \times 6 = 2 \times 36 = 72$

3. Final Answer

72