Calculating combinations of four numbers depends on what you're trying to combine from and how many you're choosing at a time, and whether order matters. We need to define these parameters. Let's explore a couple of common scenarios:
Scenario 1: Choosing combinations from four specific numbers (e.g., 1, 2, 3, 4) and choosing a subset of them (order doesn't matter)
This is the most common interpretation. "Combination" here means we're selecting a group where the order doesn't matter. For example, choosing 1 and 2 is the same combination as choosing 2 and 1. We can find all possible combinations of varying sizes:
- Choosing 1 number at a time: {1}, {2}, {3}, {4}. There are 4 combinations.
- Choosing 2 numbers at a time: {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. There are 6 combinations.
- Choosing 3 numbers at a time: {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}. There are 4 combinations.
- Choosing 4 numbers at a time: {1, 2, 3, 4}. There is 1 combination.
- Choosing 0 numbers at a time: {}. There is 1 combination (the empty set).
In total, there are 4 + 6 + 4 + 1 + 1 = 16 combinations. Notice that the original short answer reference stated 15, which is incorrect, as it missed the null set.
A more general formula to calculate the number of combinations of choosing k items from a set of n items (where order doesn't matter) is:
nCk = n! / (k! * (n-k)!)
where "!" denotes the factorial (e.g., 5! = 5 4 3 2 1).
For example, choosing 2 numbers from a set of 4:
4C2 = 4! / (2! 2!) = (4 3 2 1) / ( (2 1) (2 * 1) ) = 24 / 4 = 6
This confirms our listing above.
Scenario 2: Creating a sequence of four numbers from a larger set (repetition allowed, order matters - permutations with repetition)
Suppose you want to create a sequence of four digits, where each digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 possible digits). Here, order does matter (1234 is different from 4321), and repetition is allowed (1111 is valid). In this case, for each of the four positions, you have 10 choices. Therefore, the total number of combinations is:
10 10 10 * 10 = 10,000
Scenario 3: Creating a sequence of four numbers from a larger set (no repetition allowed, order matters - permutations without repetition)
Suppose you want to create a sequence of four digits, where each digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 possible digits). Here, order does matter (1234 is different from 4321), and repetition is not allowed (1111 is invalid). In this case, for the first position you have 10 choices, for the second you have 9 choices, for the third you have 8 choices and for the fourth you have 7 choices. Therefore, the total number of combinations is:
10 9 8 * 7 = 5,040
In Summary: The way to calculate combinations of four numbers depends entirely on the specific problem you are solving. Consider whether the order of selection matters and whether repetition is allowed, and the size of the set you are selecting from.
