Problem six. Lisa has four wooden digits. She can use them to form the number 2025. How many different numbers greater than 2025 can she form with these digits? So we see we have four digits, two twos, one zero, and one five, and we need to form them into numbers and count how many are greater than 2025. So the numbers, since they have to be greater than 2025, it can't start with zero, so it has to either start with a two or with a five. So we'll look at those different cases. So if we start with a two as the first digit, we'll find that there's six numbers. This is because we have three digits left. They're all different, so we have three choices for which one of them we put in the second position, two choices for which one we put in the third position, and then the last one is determined, so three times two is six, but we can also just list out the numbers. So 2025 is the smallest, then the next largest is 2052, and then 2205, 2250, 2502, and 2520. But out of these, we need only numbers that are strictly greater than 2025. So since 2025 is the smallest, that's the only one that's not strictly greater than 2025, and so only five of them are left. Now if we don't start with the two as the first digit, we have to start with a five. Here there's only three numbers. Basically we pick which of the three remaining slots is containing the zero. So we have a 5022, 5202, 5220. So we have these three numbers, and they're all greater than 2025. So now we had five numbers starting with the two, and three numbers starting with five, and so in total we have eight numbers, and the answer is C.

numbers

combinations

digits

greater than 2025

Lisa