4 variables how many combos of 3 can you make

Question

If you have 4 variables

A, B, C, D

How many combos can you make that use 3 of the variable and are unique (order matters), so I mean A,B,C and B,A,C only counts as 1, combo.

I have these are there any others?

A,B,C
A,B,D
B,C,D
C,D,A

Are these the only possibilities?

Thanks

Answer 1

The order matters. hence the first choice has 4 then 3 then 2 so you will get $4(3)(2) =24$ Here are some A,B,C

A,C,B

A,B,D

A,D,B

A,D,C

A,C,D

If the order is not matter you will divide this by $3(2)(1)$ The number of orders of three elements.

Answer 2

For how many combinations, you have it.

C is combination. n is the number of items. r is the number of items to be chosen

nCr = n!/(r!(n-r)!)
4C3 = 4!/(3!(4-3)!)
    = 24/(6*1)
    = 4

Permutations is 24. P is permutations. n and r are same as above.

nPr = n!/(n-r)!
4P3 = 4!/(4-3)!
    = 24/1
    = 24

Another way to think of permutations in this case is you have 4 items to choose from. When you pick one, you now have 3. When you pick the from 3, you now have two. Or

4*3*2=24

The formula simply takes 4*3*2*1 (4!) divided by the number of spots you didn't fill, which is one.