A set of n objects has 2n combinations.
In each combination, each element can either be included or excluded. Two possibilities for each of n objects.
One of these combinations will be the empty set - where none of the objects are selected.
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To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.
It is finite, but there would be many possible combinations.
10 possible numbers on each wheel equals 10x10x10 or 1000 combinations possible.
There are 167960 combinations.
You did not give us the set to work with...