问题
For a set of size n
, the size of its power set is 2^n
. Generate all permutations for each element of the power set. The power set for set {a, b}
is {{}, {a}, {b}, {a,b}}
. Generate all permutations on each set, we can get {(),(a),(b),(a,b),(b,a)}
. So the number of all subset permutation for a power set generated from a 2-element set is 5. And such a number for a 3-item set is 16. Is there a formula for this number defined in terms of n
?
回答1:
First of all, consider the power set. The number of sets of size k
(for some 0 <= k <= n
) in the power set is
n choose k = n! / (k! * (n - k)!)
Indeed, if we sum the number of sets for all k
, we get 2^n
, see Wolfram Alpha.
How many permutations does a set of size k
have? Well, k!
. So, if we plug that in, we loose the k!
from the denominator and sum n! / (n-k)!
for all k
, which is
n! * Sum(1/k!, 0 <= k <= n)
Again, see the result by Wolfram Alpha.
来源:https://stackoverflow.com/questions/19079038/what-is-the-number-of-all-set-permutations-in-a-power-set