# Number of combinations, repeating and not repeating [closed]

I'm struggling on a question similar to the following:

Your ID can only use 3 digits. Allowed digits are 0-8. How many allowed combinations are there if repeating digits is allowed? What about if they're not allowed?

I feel like it has something to do with nPr and nCr but I really am not sure. Any help would be very much appreciated.

• You are supposed to use the Multiplication Principle. If digits may be repeated, you have nine choices for each digit. If digits may not be repeated, you have nine choices for the first digit, but you only have eight choices for the second digit since it cannot be the same as the first digit. How many choices do you have for the third digit? – N. F. Taussig Nov 22 '19 at 16:40

When repeating digits are not allowed this is a permutation because order matters. You have to choose 3 without repeating and the order matters.

$$_9P_3 = \frac{9!}{(9-3)!} = 504$$

When repeating digits is allowed you have 9 choices for each digit and it doesn't matter what the previous digits were so you can just use the multiplication principle.

$$9^3 = 729$$

• I understand how this works now, thank you for explaining. – melonhead Nov 22 '19 at 16:48

In the first case, the first digit can be anything from $$1$$ to $$9$$. The same applies for the other two digits. Hence there are $$10$$ digits to choose form in each place. So the total number of cases will be $$9{\times}9{\times}9=729$$

Now when digits cannot be repeated, the first digit can be anything from $$1$$ to $$9$$ ($$9$$ options). But the second digit can be anything except the first digit which gives use $$8$$ options. Similarly, we have $$7$$ options for the third digit. Total number of cases $$9{\times}8{\times}7=504$$

• Thank you this helps as well – melonhead Nov 22 '19 at 16:52