Question

Find the number of combinations of 3-lettered numbers that can be created by 3, 4, 5, and 7 with repeating. The three digits cannot be equal to zero.

Hint:

When repetition is allowed, the number of combinations of \( r \) items from \( n \) items is calculated using the formula \( n^r \).

Answer 1

We are tasked with finding the number of combinations of 3-lettered numbers using the digits 3, 4, 5, and 7, with repetition allowed. Additionally, the digits cannot be equal to zero. The number of possible choices for each digit is 4 (since we are choosing from the set {3, 4, 5, 7}). Step 1: Apply the formula for combinations with repetition.
When repetition is allowed, the number of combinations for selecting \( r \) items from \( n \) items is given by the formula: \[ n^r \] where \( n \) is the number of available choices for each item, and \( r \) is the number of items to be selected. Here, \( n = 4 \) (since we have 4 available digits: 3, 4, 5, 7) and \( r = 3 \) (since we are selecting 3 digits for each number).
Step 2: Calculate the total number of combinations.
Using the formula \( n^r \), the total number of combinations is: \[ 4^3 = 4 \times 4 \times 4 = 64 \] Thus, the number of 3-lettered numbers that can be created is: \[ \boxed{64} \]