for examble 0123
2006-07-05 14:42:21 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Interestingly, everyone is giving you the number of permutations which are possible--but you specifically asked about 4-digit combinations.
If you are asking how many 4-digit numbers are possible (and allowing that the number could begin with 0), then there are 10000 possibilities if digits can be repeated (10x10x10x10) and 5040 possibilities if numbers cannot be repeated (10x9x8x7).
However, when we are talking about combinations, then the combination 0123 is the same as the combination 3102. In this case, the number of combinations (without repeated digits) is 10C4 which is on 210. (Calculating combinations with repeated digits is really tricky--I'll come back to that if I remember.)
Anyway, your question did not ask about permutations. It asked about 4-digit combinations. The answer is 210.
2006-07-05 15:25:33 · answer #1 · answered by tdw 4 · 0⤊ 0⤋
10000
For each digit you have the choice of 10 numbers (0-9)
Since there are 4 digits, you have the choice of 10^4=10,000 numbers.
Another way to see it is you can have any number between (inclusively) 0000 and 9999, so 9999+1=10,000 numbers.
If you want all the digits to be different numbers, you will obviously have less choices.
For the first digit, you have 10 choices, then you can't use the number you used for the first digit, so you only have 9 choices for the second digit, you only have 8 choices for the 3rd digit, and 7 choices for the 4th digit. Therefore you have 10*9*8*7=5040 choices.
2006-07-05 21:45:22 · answer #2 · answered by Eulercrosser 4 · 0⤊ 0⤋
Can any of the number repeat? If so, there would be 10000 combinations.
Otherwise you have 10 * 9 * 8 * 7 = 5040
2006-07-05 21:47:58 · answer #3 · answered by Poncho Rio 4 · 0⤊ 0⤋
if repeating numbers are allowed then the answer is 10000, but if not, then
nPr = (n!)/((n - r)!)
0 - 9 make up 10 numbers
10P4 = (10!)/((10 - 4)!)
10P4 = (10!)/(6!)
10P4 = 5040
There are 5040 different ways to have 4 digit combinations using 0 through 9.
2006-07-06 00:06:48 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
well if the numbers can repeat, there are 10000 possibilities.
if they can't repeat, then the answer is 10P4 = 5040 = 10x9x8x7
the first digit has 10 different possibilities (0-9) and then once 1 of them is gone, the 2nd digit has 9 possibilities and 3rd digit has 8 and last digit has 7.
so to get the total number of possibilities, you multiply them. itz hard to explain but i hope that helps. laterz
2006-07-05 21:53:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋
10*10*10*10
you have 10 digits to choose from for the first, second, third, and fourth numbers
10000
if you cant have the same number twice, 10*9*8*7=90*56=5040
2006-07-06 16:56:39 · answer #6 · answered by Anonymous · 0⤊ 0⤋
It depends if you can repeat digits. If you're allowed to repeat digits (such as 3347), then you have ten possible choices for the first digit, ten choices for the second, ten choices for the third, and ten choices for the fourth digit.
Total choices = 10 * 10 * 10 * 10 = 10,000
If you're not allowed to repeat digits, then you have ten choices for the first digit, nine possible choices for the second (since it can't be the same as the first digit), eight choices for the third digit and seven choices for the fourth.
Total choices = 10 * 9 * 8 *7 = 5040
2006-07-05 22:36:44 · answer #7 · answered by jackalanhyde 6 · 0⤊ 0⤋
10000 combinations by definition when you include 0000
1 digit = 10 combinations (0-9)
2 digits = 100 combinations (00-99)
3 digits = 1000 combinations (000-999)
4 digits = 10000 combinations (0000-9999)
2006-07-05 21:48:50 · answer #8 · answered by famousblue11 2 · 0⤊ 0⤋
10*10*10*10 = 10000
2006-07-06 00:40:29 · answer #9 · answered by boter_99 3 · 0⤊ 0⤋
9 to the power of 4 or 6561 combinations..btw they can be repetitive numbers like 9999 or 0000
2006-07-05 21:46:19 · answer #10 · answered by pablo_dmc 3 · 0⤊ 0⤋