how many 3 digit combinations can you make out of 4 numbers?

Answer 1

1 2 3 4

1 2 3-------1
1 2 4 --------2
1 3 2-------3
1 3 4------4
1 4 2------5
1 4 3------6
2 1 3------------7
2 1 4-------------8
2 3 1-------------9
2 3 4------------10
2 4 1----------11
2 4 3----------12
there is a patter, 6 per # and there is 4 # so, there is 24

Answer 2

The argument in some of the other answers appears to be with whether or not repetition is allowed. While the problem doesn't say, chances are it isn't, and 4 x 3 x 2 = 24 would be the correct answer. If it is, 4 x 4 X 4 = 64 would be right.

"Combination", by the way is also a bad choice of word from whoever made up this problem. From a strict mathematical point of view, technically 123 and 231 are the same "combination" of numbers. What your problem almost certainly refers to is actually permutations, where putting the numbers in different orders changes their value. If in fact you are in a class learning probablilty and "combinations" really is what you want, the answer would be 4 (because you can leave out any one of the 4 numbers).

Answer 3

I believe that the answer above is wrong. If you only had 3 digits the answer would be 27. With 4 digits available it is 64.

Four digits with three possible combinations would be 4^3 or 64.
As 'DMB' points-out below - can digits be repeated? This makes a HUGE difference in the answer.

The long answer (allowing repetition) is below:

1.111
2.112
3.113
4.114
5.121
6.122
7.123
8.124
9.131
10.132
11.133
12.134
13.141
14.142
15.143
16.144
17.211
18.212
19.213
20.214
21.221
22.222
23.223
24.224
25.231
26.232
27.233
28.234
29.241
30.242
31.243
32.244
33.311
34.312
35.313
36.314
37.321
38.322
39.323
40.324
41.331
42.332
43.333
44.334
45.341
46.342
47.343
48.344
49.411
50.412
51.413
52.414
53.421
54.422
55.423
56.424
57.431
58.432
59.433
60.434
61.441
62.442
63.443
64.444

Answer 4

Hey dmb, how did you come up with 4x3x2? It comes out to be 24 which is what I counted on my fingers, so I know it's correct, but where does 4x3x2 come from?