Hi Candace.
Let's call the four numbers a, b, c, and d.
{a, b} is one combination, and {a, b, c, d} is another. We could list them all out, but let's approach this in a more systematic way.
We have four original numbers, so any given combination Q may have from 0 to 4 of these numbers in it.
How many ways are there of choosing zero items from a group of four? Just one.
How many ways are there of choosing one item from a group of four?
Count them out: a, b, c, d  that's four.
How many ways are there of choosing two items from a group of four?
First we select one (four choices), then select another (from the remaining three, so that's three choices). But if we select a and then b, that is the same as selecting b and then a, so we divide by two to eliminate the duplicates. 4x3/2 = 6.
Now let's look back at what we've done. When we selected items for our combinations, we divided the group into two parts: one that is selected and one that is not. For every combination we selected, the remainder is another combination of the original four.
So when we found that there is just one way of selecting zero items, we are also saying that there is just one combination with all four items (that makes sense).
It also means that when we selected one item (four choices), we created four different combinations with three items.
Now we know all of the combinations and can add them up:
1 + 4 + 6 + 1 + 4 = 16 different possibilities.

A related topic to combinations is "permutations". When order matters, you use permutations because combinations consider order to be irrelevant ( a, b is the same as b, a).
If you meant to say "permutations", then you are probably asking the question "how many different ways can I arrange the order of four numbers?"
The answer to this question (which you got right) is 24.
Here's how to observe this:
1. Pick one of the four numbers (there are four choices in this step).
2. Pick one of the remaining three numbers (there are three choices).
3. Pick one of the remaining two numbers (two choices)
4. Stick the last number on the end.
You multiply these choices together to get your result: 4 x 3 x 2 (x 1) = 24.
Combinations and permutations are often confused by students  they are related, but they mean different things and can lead to totally different interpretations of situations and questions. It doesn't help that "combination locks" are really "permutation locks" because order certainly matters!
Hope this helps,
Stephen La Rocque.
