[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1].
A set of n elements has n! (n factorial) permutations: you have n options to pick the first element, n-1 to pick the second element, and so on.
A combination of a set is one of its subsets, including the set itself and the empty set. The set {1,2,3} has the following subsets:
{}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}.
A set of n elements has exactly 2^n combinations; for each element you either include or exclude it.
With permutations, you get—IIRC, the factorial?—which is much, much worse. Mistaking permutations for combinations isn’t a small error.
More precisely, when taking N objects out of M, the number of permutations is always computed by multiplying N factors, which are either all equal to M when repetitions are allowed (i.e. the power M^N) or they are decreasing by one at each factor when the extracted objects must be unique (i.e. M*(M-1)*(M-2) ...), which gives the factorial in the case of N taken out of N.
With combinations either with or without repetitions you also get a product of N factors, but each factor is much smaller, being a ratio of two integers, instead of the integer that is the numerator. This is usually written in a form that is useless for actual computation, as the ratio between a factorial and the product of other two factorials (which differ between the two kinds of combinations).
The sum of all combinations without repetitions is 2^N (when repetitions are allowed, the sum is infinite).
Thanks for the correction.