0
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
What else can I help you with?
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.
How many different numbers can you make with the digits 1 2 3 and 4 using each digit only once?
If you want 4-digit numbers, there are 24 of them.
How many different 3 digits numbers can you write with the digits 347?
If you use each number once, there are six combinations.
How many different numbers can make using digits 1 3 and 5 only once?
Using the digits 1, 3, and 5 exactly once, you can create different 3-digit numbers by permuting these digits. The number of permutations of 3 distinct digits is calculated as 3! (3 factorial), which equals 6. Therefore, the different numbers you can create are: 135, 153, 315, 351, 513, and 531. Thus, there are 6 different numbers that can be formed.
How many 5-digit numbers can you make with the digits 1 2 3 5 7 8 if each digit can only be used once?
There are six choices, the 5 digits and the one you leave blank. That makes 6! or 720 different numbers.
How many 4 digit numbers can be made by he digits 0123 using the digit once?
2
How many 4-digit numbers can be formed by the digits 1 1 2 2 using the digits more than once?
6
How many 3 digit numbers can you make using the digits 1 3 5 and 7 by using the digits more than once?
Using the combination C(4,3) I got the answer 24. Its either a combination or a permutation.
How many numbers can be formed using digits 1245?
Well honey, you've got 4 digits there, so you can form 4! (4 factorial) which is 24 numbers. That's right, you can make 24 different combinations with those digits. Math can be fun when you've got some sassy numbers to play with!
How many three digit numbers can you make using the digits 1 3 5 7 and 9 if no digit appears more than once in a number?
There are 5*4*3 = 60 such numbers.
How many whole numbers between 1 and 100 use the digits 6 at least once?
19 do.
How many 3 digit numbers can be formed from the digits 0 through 9 in a set of three?
If the digits can be used more than once, then 900. If not, then 648.
How many 4-digit numbers can be formed by the digits 0 1 2 3 4 5 6 7 8 9 using the digits once?
There are 9*9*8*7 = 4536 such numbers.
