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# How many combinations are there - 3 digit from a pool of 4 digits, no repeats?

Ok. I need all possible 3 digit combinations from a pool of 4 numbers. No repeats of a digit in a combination (001, 011, 022, etc).

So, how many combinations are there?

### 3 Answers

- ?Lv 58 years agoFavorite Answer
Are you allowing cases like 101? The examples you gave are all consecutive digits. If you are simply not allowing any re-use of digits within the number, then it's simply choosing 3 unique digits from a group of 4. Thus you can look at is as a standard combination problem: You can choose any of 4 digits for the first, then 3 for the second, then 2 for the last. Thus you have 24 total permutations.

Now, if it's permutations (123 =/= 321) then you're done. If you're actually dealing with combinations, then the order doesn't matter, so 123 = 132 = 213 = 231 = 321 = 312. You'll find there's always 6 repeats of each set (due to rearrangements of 3 items), resulting in 4 combinations.

If you allow cases like 101, 232, etc. then it becomes more complicated, but having done the work about it's not too bad. Since each of these cases is the same form (xyx for 2 numbers x and y) you can think of it as simply how many ways you can choose 2 numbers, with the 3rd number being identical to the first. Thus you have 4 choices for the first and 3 for the 2nd resulting in 12 choices under permutations. Now you have 12 options with reuse of digits, and 24 without re-use. So the total would be 36 permutations.

There is likely to be another answer simply using standard combination and permutation formula, which is basically just the short-hand written form of what's above.

- 8 years ago
If the order matters, then:

The first number has 4 options, right?

After you choose 1 option, you only have 3 left because you can't repeat, right?

After you choose another, you only have 2 left because you can't repeat either of the other 2 numbers, right?

So do 2x3x4=24

That's called a permutation because the order matters.

But what if the order doesn't matter?

But there can be multiply combinations of the same number!

So, you do 3x2x1 to find the total number of ways you can arrange each combination. 3x2x1=6

Divide:

24/6=4

So there are 4 combinations. It's called a combination because the order doesn't matter.

Hope this helps!

- wysockiLv 44 years ago
First Digit = 4 2nd Digit = 3 0.33 Digit = a million Fourth Digit = 2 upload 2d and third = 3+a million = 4 >> First Digit Subtract third Digit from 2d Digit = 3-a million= 2 Subtract answer (2) from First Digit= 4-2=2 >> Fourth Digit additionally 0-0-0-0 works in simple terms as 4-3-a million-2 i'm hoping i ought to get 10 factors for that thank you!