Question

Using the digits 2 through 9, find the number of different 5-digit numbers such that (a) digits can be used more than once, (b) digits cannot be repeated but can come in any order, and (c) digits cannot be repeated, and the order doesn't matter.

Answer 1

For (a), the number of possible outcomes is 9 to the power of 5. For (b), there are 9 x 8 x 7 x 6 x 5 different 5-digit numbers. For (c), the number of different 5-digit numbers is 9 choose 5.

Step-by-step explanation:

a. The number of possible outcomes (microstates) of any repeated independent situation is equal to the number of possibilities in one iteration to the power of the repetitions. So in this case, the answer is 9 to the power of 5.

b. In this case, since digits cannot be repeated, any 5-digit number can be formed using any 5 of the given digits. So the number of different 5-digit numbers is 9 x 8 x 7 x 6 x 5.

c. In this case, since both repetition and order do not matter, we can use combinations. The number of different 5-digit numbers is 9 choose 5 which is equal to 9! / (5! x (9-5)!).