Suppose you have n objects and within those, there arem1 objects of kind 1
m2 objects of kind 2
and so on.
Then the number of permutations of the n objects is n!/[m1!* m2!...]
For example, permutations of the word "banana"
n = 6
there are 3 "a"s so m1 = 3
there are 2 "n"s so m2 = 2
therefore, the number of permutations = 6!/(3!*2!) = 720/(3*2) = 120.
10 x 9 x 8 = 720 different "permutations"
3x2x1=6 permutations.
There are 120 permutations and only 1 combination.
10! permutations of the word "Arithmetic" may be made.
There are 195 3-letter permutations.
10 x 9 x 8 = 720 different "permutations"
3x2x1=6 permutations.
There are 8! = 40320 permutations.
There are 6! = 720 permutations.
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
There are 120 permutations and only 1 combination.
39916800 permutations are possible for the word INFORMATION.
10! permutations of the word "Arithmetic" may be made.
There are 195 3-letter permutations.
A freezer is a freezer: it has a fixed shape and a fixed purpose so there are really no permutations.
If there are n objects and you have to choose r objects then the number of permutations is (n!)/((n-r)!). For circular permutations if you have n objects then the number of circular permutations is (n-1)!
There are 7893600 permutations.