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There are 4 choices for the first digit (1, 2, 3, or 4) since zero cannot be the first digit in a four-digit number. For the second digit, there are 3 remaining choices, for the third digit there are 2 choices, and for the fourth digit, there is only 1 choice left. Therefore, the total number of four-digit numbers that can be formed using the digits 1, 2, 3, and 4 is 4 x 3 x 2 x 1 = 24.

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ProfBot

3w ago
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Wiki User

14y ago

24 possible combinations:

1234,

1243,

1324,

1342,

1423,

1432,

2134,

2143,

2314,

2341,

2413,

2431,

3124,

3142,

3214,

3241,

3412,

3421,

4123,

4132,

4213,

4231,

4312,

4321

Generalizing it, you can only make n! combinations out of n different digits.

n! = 1 * 2 * 3 * ... * n . (Math definition: the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n)

When n = 4 , n! = 1 * 2 * 3 * 4 = 24.

If 16 digits are available, let's say 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F you can make

1 * 2 * 3 * 4 * ... * 16 = 2 ^15 * 3 ^6 * 5 ^3 * 7 ^2 * 11 * 13 = Excuse me, but I am not going to bring the product. Use your calculator (and be prepared for an overflow).

Anyway, it is:

16! = 32768 * 729 * 125 * 49 * 11 * 13 = 32768 * 729 *125 * 7007 = 32768 * 729 * 875875 = whatever... almost 20 trillion combinations of 16 different digits.

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Q: How many four digit numbers can you make using the digits 1234?
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