It is finite, but there would be many possible combinations.
There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}There are 8 possible combinations:{},{2},{6},{8},{2,6},{2,8},{6,8} and{2,6,8}
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...
The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
86,450
Yes
It is finite, but there would be many possible combinations.
140 possible combinations
you can have 5 24 objects to 120
Combinations.
By making a number tree that could have as many as 1,000,000 combos.
If order doesn't matter, 15 combinations and if order does matter, 360 combinations are possible.
To find the number of combinations possible for a set of objects, we need to use factorials (a shorthand way of writing n x n-1 x n-2 x ... x 1 e.g. 4! = 4 x 3 x 2 x 1). If you have a set of objects and you want to know how many different ways they can be lined up, simply find n!, the factorial of n where n is the number of objects. If there is a limit to the number of objects used, then find n!/(n-a)!, where n is the number of objects and n-a is n minus the number of objects you can use. For example, we have 10 objects but can only use 4 of them; in formula this looks like 10!/(10-4)! = 10!/6!. 10! is 10 x 9 x 8 x ... x 1 and 6! is 6 x 5 x ... x 1. This means that if we were to write out the factorials in full we would see that the 6! is cancelled out by part of the 10!, leaving just 10 x 9 x 8 x 7, which equals 5040 i.e. the number of combinations possible using only 4 objects from a set of 10.
There are millions of possible combinations.