To make a 5-digit number using the digits 8, 9 and 0, you have to consider two cases: with repetition and without repetition.
- With repetition, you can use any of the three digits as many times as you want. For example, 88888, 90909, and 09876 are all valid 5-digit numbers with repetition. To count how many such numbers are possible, you can use the multiplication principle: for each of the 5 positions, you have 3 choices of digits, so the total number of choices is 3 x 3 x 3 x 3 x 3 = 3^5 = 243. Therefore, there are 243 5-digit numbers that can be made using the digits 8, 9 and 0 with repetition.
- Without repetition, you can use each of the three digits only once. For example, 89076, 90876, and 08976 are all valid 5-digit numbers without repetition. To count how many such numbers are possible, you can use the permutation formula: the number of ways to arrange n distinct objects in r positions is n! / (n - r)!. In this case, n = 3 and r = 3, so the number of ways to arrange the three digits in the first three positions is 3! / (3 - 3)! = 3! / 0! = 6. For the last two positions, you have to use the remaining digits that are not 8, 9 or 0. There are 7 such digits: 1, 2, 3, 4, 5, 6 and 7. The number of ways to arrange 7 distinct objects in 2 positions is 7! / (7 - 2)! = 7! / 5! = 42. Therefore, the total number of 5-digit numbers that can be made using the digits 8, 9 and 0 without repetition is 6 x 42 = 252.
- **With repetition:** 243 numbers
- **Without repetition:** 252 numbers
However, the number "09876" is counted in both cases, so you need to subtract it once from the total.
243 + 252 - 1 = 494
Therefore, the correct total is 494.
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