How many combinations can you get with four numbers?

Answer 1

Depends on how you use them
4^4 or 4!

Answer 2

How Many Combinations With 4 Numbers

Answer 3

4 numbers

assuming that they DO need to be in a specific order
(1st 2nd 3rd 4th) then there are 24 possible combinations:

1234 1243 1324 1342 1423 1432
2134 2143 2314 2342 2413 2431
3124 3142 3214 3241 3412 3421
4123 4132 4213 4231 4312 4321

This can be figured with factorial notation as "4!" which is the same as 1*2*3*4 (Both =24)

(*FYI: exception of factorial rule, 0!=1 1!=1)

I noticed that someone answered 4^4 (4 to the power of 4)...

The best way to think of that one is this way:
Say you have 4 multiple choice questions on a test, and EACH individual answer has 4 possible answers (a,b,c and d), how many possible ways can ALL 4 questions can be answered?

That is 4^4 or 256 possible solutions

Another way to look at your question:
(Sometimes another view helps)
If you are into horse racing, a trifecta box is a racing bet where you pick three horses and say that these three horses are going to come in 1st, 2nd, and 3rd but you dont care about the order that they do come in just that all three do cross the finish either 1st, 2nd or 3rd. That (as a $1 ticket per possible combination) cost $6.00 (3*2*1 or 3!) because there are 6 possible combinations, but a trifecta box ticket with 4 horses means 3 out of 4 horses are going to cross the finish line as 1st, 2nd, or 3rd and that, as a $1 ticket, cost $24 dollars! Because its three places (6 possible cominations) out of 4 horses! But dont worry thats a different formula :) which is n!/(n-r)! n=4, r=3 in this case so 24/1=24. This is known as a permutation.

Now the question was combinations and everything that was mentioned was assumed that it had to be in a specific order, but what if there was no specified order?

Remember that horse formula of n!/(n-r)! well here is a new one
:
n!/r!(n-r)!

In case you are wondering, I'm not breaking the order of operations [Please Excuse(My Dear)(Aunt Sally)], for the formula is a fraction:

n!
____________

r!(n-r)!


This is the combination formula
This is thought of this way:

A number of objects (in your case 4 numbers) are considered only on how many ways they can be grouped so lets continue using horses to show the combos:

still n=4,r=3
4!/3!(4-3)!
or
(4*3*2*1)/(1*2*3)*[(1)!]
(24)/(6)*(1)
(24)/(6)
4


There are 4 ways to GROUP the 4 horses into groups of three:
123, 234, 134, 124

any other combination will repeat itself:

123: ---132, 213, 231, 312, 321
234: ---243, 324, 342, 423, 432
134 : ---143, 314, 341, 413, 431
124: ---142, 241, 214, 412, 421

see that there ARE 24 possible combinations but note also that the groups of horses after the --- are REPEATS of the first in the row, that is why there are only 4 GROUPS possible cause we are not concerned about order.


So how does work for your question?
n=4,r=4
4!/4!(4-4)!
or
(4*3*2*1)/(1*2*3*4)*[(1)!]
(24)/(24)*(1)
(24)/(24)=1
1 possible way to group 4 numbers in a set of 4
meaning the 4 horses are 1 group


But I'm sure you'll find that the answer to your question is 24. Just wanted to show the other ways as well :)

Hope that helps

Answer 4

If you can repeat the numbers, then number of combinations is infinite. For example, if your numbers are: 1,2,3,4 then here are a few combinations:
1234
123434
1212121212121212

If each of your numbers must be used once and only once, and the number are different (e.g. same 1,2,3,4) then you can chose which number goes first, then second etc - you get:
4*3*2*1 = 24 combinations.

If each number can be used NOT MORE than once, then you can have combinations like:
1
12
342
etc.
The number of combinations is of course:
4 + 4*3 + 4*3*2 + 4*3*2*1 = 4+12+24+24 = 64.

Thus, depending on definition of what is allowed, we got at least three answers: infinity, 24 and 64. You can consider a few more definitions.

Answer 5

4 unique numbers eg 1234 = 4 x 3 x 2 x 1 = 24
1 repeating number eg 1233 = 4 x 3 x 2 x 1 / 2 = 12
2 sets of 2 repeating numbers eg 1122 = 4 x 3 x 2 x 1 / (2 x 2) = 6
3 repeating numbers eg 1222 = 4 x 3 x 2 x 1 / (3 x 2 x 1) = 4

Answer 6

depends if you mean 'how many ways can I arrange these four different numbers' or 'i've got a hefty padlock with 4 numbers on it, how many combinations are there'

first: 4 choices for the first number, 3 for the second, 2 for the third, 1 for the last. So you have 4*3*2*1=24 combinations.

second: assuming 10 numbers on each wheel, then there are 10 possible numbers for each of the 4 wheels, so there are 10*10*10*10 = 10000 possible combinations

Answer 7

If you are being specific, combinations means order does not matter, therefore !@#$ is the same as $#@!.

It depends how many numbers you have to choose from. If you have only 4, then there is only 1 combination.

If you are talking about the 10 digits, that is 0 to 9, then there are
10C4 = 10P4/4! = (10!/6!)/4!= 210.

Answer 8

If your combinations can't contain repeated numbers then 4*3*2*1=24 otherwise it is 4^4 = 256

Answer 9

Case 1 : 4 different numbers : 9*8*7*6

Case 2 : 3 different numbers: 9*8*7*3*3

Case 3: 2 different numbers: 6+8 (counted by hand!)

Case 4: 1 different number: 9

Result = Case 1+ Case 2+ Case 3+ Case 4
(Zero in not included in numbers)

Answer 10

24

Answer 11

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How many combinations can you get with four numbers?