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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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Maxine
LaoBecause 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
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How many 4 digit numbers can be formed by the digits 1 2 3 4 using the digits more than once in each number example 1112 1122?
256
How many three even digit numbers by using 12457 if each digit is used once?
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.
Write all the three digit numbers you can make for 2 3 and 4 using each of the digits once?
234, 243, 324, 342, 423, and 432.
How many five-digit numbers can be created using the digits 0-9?
9
How many even 3-digit numbers can be formed from the digits 1 2 3 and 5 if each digit can be used only once?
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.
