0215. Can anyone list every possible combination?
2007-05-29 16:54:21 · 9 answers · asked by Michelle 1 in Science & Mathematics ➔ Mathematics
4*3*2*1 since they are all different = 24
0125
0152
0215
0251
0512
0521
1025
1052
1205
1250
1502
1520
2015
2051
2105
2150
2501
2510
5012
5021
5102
5120
5201
5210
2007-05-29 16:59:28 · answer #1 · answered by A confused bio student 2 · 0⤊ 1⤋
Use permutations - order does matter.
Permutations = (n!)/(n-r)! where n is the total numbers there are and r is how many taken at a time.
In this problem, n = 4 and r = 4, so therefore you have:
4!/1!
In case if you don't know, the ! sign means the product of all whole numbers less than or equal to number it is used on.
From this, 1! = 1 and 4! = 4 * 3 * 2 * 1 = 24.
24/1 = 24, so you have 24 possible combinations.
2007-05-29 17:00:07 · answer #2 · answered by Linda O'Chuffy 2 · 0⤊ 0⤋
Combinations of what? 2 digits? 4 digits? Now that I think on it, you probably mean 4 digits. The easiest way to do this is use factorial. this is 4! (four-factorial). The best way to think of the problem is in spaces for the digits. You have: _ _ _ _
In the first slot, you have 4 options. If you use any of those four options, then you only have 3 left for the next slot and so-on down the line. the number of combinations is therefore, 4x3x2x1 or 24 different combinations of the four digits.
2007-05-29 17:00:04 · answer #3 · answered by misti 2 · 0⤊ 0⤋
Hi. 4*3*2.
2007-05-29 16:57:20 · answer #4 · answered by Cirric 7 · 0⤊ 0⤋
Any of the eight numbers can be the first digit. Then any of the remaining seven numbers can be the second digit. Associating them, the first two digits can be formed in 8 x 7 = 56 ways. Counting this way, the total numbers of ways in which the given eight numbers can make a four digit number without any of them repeating = 8P4 = 8 x 7 x 6 x 5 = 1680.
2016-05-21 18:25:51 · answer #5 · answered by ? 3 · 0⤊ 0⤋
Total 24 combinations:
0215
0251
0125
0152
0512
0521
1250
1205
1052
1025
1520
1502
2015
2051
2150
2105
2501
2510
5012
5021
5120
5102
5201
5210. That's all
2007-05-29 17:00:54 · answer #6 · answered by Hell's Angel 3 · 0⤊ 0⤋
if you use all 4 numbers there are 24
0125
0152
0215
0251
0512
0521
1025
1052
1205
1250
1502
1520
2015
2051
2105
2150
2510
2501
5012
5021
5201
5210
5120
5102
2007-05-29 16:58:54 · answer #7 · answered by who knows 2 · 0⤊ 0⤋
10,000.
That is 0000 to 9999.
If you're not allowed to repeat digits, then it's
10*9*8*7 = 5040
2007-05-29 16:58:12 · answer #8 · answered by Dr D 7 · 0⤊ 1⤋
24 combination
0215,0251,0512,0521,0125,
0152,2015,2051,2150
2105,2510,2501,1025,1052,
1502,1520,1205,1250,
5012,5021,5102,5120,5201,
5210
2007-05-29 20:19:46 · answer #9 · answered by jackleynpoll 3 · 0⤊ 0⤋