Mike has four number cards: 2, 2, 2, and 5. He wants to use each card once to make a four-digit number.

Since there are repeated numbers (three 2s), we need to calculate the number of unique four-digit combinations that can be made from these digits.

Step-by-step solution:

1. The total number of arrangements if all digits were different would be $4!$, which is: $4! = 4 \times 3 \times 2 \times 1 = 24$

2. However, since the digit 2 repeats three times, we must divide by the number of ways to arrange the three 2s, which is $3!$: $3! = 3 \times 2 \times 1 = 6$

3. Therefore, the total number of unique four-digit numbers is: $\frac{4!}{3!} = \frac{24}{6} = 4$

Answer:

Mike can make 4 different four-digit numbers using the cards.

Would you like more details on this explanation or have any further questions?

Here are some related questions you could explore:

1. What if Mike had two 2s and two 5s instead? How many different numbers could he make?
2. What if the digits were all different, like 1, 2, 3, and 4?
3. How many different five-digit numbers can be made from the digits 1, 1, 1, 2, 3?
4. How would the answer change if Mike could repeat some digits?
5. What is the probability of randomly selecting a number where all digits are distinct?

Tip: When calculating arrangements with repeated items, always divide by the factorial of the number of repeated elements to avoid overcounting.