The word would be needed for a specific response, but here's a general one:
Take the factorial of the number of letters in the word, and divide that by the product of the factorials of the counts of all the unique letters in the word.
An example:
For the word "levitation", there are 10 letters, with one L, one E, one V, two Is, two Ts, one A, one O, and one N. So the solution is:
10!/(1!1!1!2!2!1!1!1!) = 10!/4 = 907,200 permutations
10! permutations of the word "Arithmetic" may be made.
There are 7 factorial, or 5040 permutations of the letters in the word NUMBERS.
There are 195 3-letter permutations.
There are 5040.
239,500,800 12!/2! * * * * * Actually, as the word "permutation" [not permutations] has 11 letters, the answer is 11!/2! = 19,958,400
39916800 permutations are possible for the word INFORMATION.
10! permutations of the word "Arithmetic" may be made.
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 7 factorial, or 5040 permutations of the letters in the word NUMBERS.
There are 195 3-letter permutations.
There are 30 of them.
2
There are 12.
24.
There are 420.