How many 4 digit numbers can be formed using 2 3 4 and 5 as many times as possible?

Answer 1

If you are asking to use each of those 4 numbers, than we would simply say there are 4 choices for the first digit, 3 for the second etc. and there would be 4! possible numbers, or 24 of them. Here they are Think of each one of the entries as a number. This was done for you with the program Mathematica. {{2, 3, 4, 5}, {3, 2, 4, 5}, {4, 2, 3, 5}, {2, 4, 3, 5}, {3, 4, 2, 5}, {4, 3, 2, 5}, {5, 3, 2, 4}, {3, 5, 2, 4}, {2, 5, 3, 4}, {5, 2, 3, 4}, {3, 2, 5, 4}, {2, 3, 5, 4}, {2, 4, 5, 3}, {4, 2, 5, 3}, {5, 2, 4, 3}, {2, 5, 4, 3}, {4, 5, 2, 3}, {5, 4, 2, 3}, {5, 4, 3, 2}, {4, 5, 3, 2}, {3, 5, 4, 2}, {5, 3, 4, 2}, {4, 3, 5, 2}, {3, 4, 5, 2}} If we can use a number more than once or not at all than that changes it. For example, you do not see 2222 above. Look at each number, let's start with 2, it is either in the first digit or not, in the second or not, in the third digit or not,and in the fourth digit or not.-. there are 2 choices for the number 2, same for 3,4,and 5. Then there are 2^4 choices for each (2^4)^4 or 16^4 possible numbers. That is much bigger than 24 as you can see. It is 65536. BUT that includes numbers such as 2 or 22 where we include 2 twice but not the other numbers, and we need to subtract those if we are talking about 4 digit numbers. That makes it a little harder! Do you seen why? Let me know if you need more help.