Well, darling, if you're using 4 digits and each one can only be used once, you can make a total of 9,720 different numbers. That's because you have 10 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Multiply all those together and voilà, you've got yourself a whole lot of numbers to play with.
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Because of the question's ambiguity, it is hard to tell exactly what is meant. However, to cover all possibilities:
If there are four numbers and each is used once, there are 4! = 4 x 3 x 2 x 1 = 24 different combinations.
If there are the 10 digits used to make a four-digit number, using each once, we have 10!/(10-4)! = 10!/6! = 10 x 9 x 8 x 7 = 5040 different combinations.
Or does he mean this?
Using the binary numbering system.
Binary Decimal
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 10
1011 11
1100 12
1101 13
1110 14
1111 15
Oh, what a happy little question! You can make 9,720 different numbers with 4 digits, each used only once. Isn't that just wonderful? Just think of all the beautiful combinations you can create!
256
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
234, 243, 324, 342, 423, and 432.
9
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.