Find the probability that X(1), X(2)... X(n) is a permutation of
1,2... n when X(1), X(2)... X(n) are independent and each is equally
likely to be any of the values 1,2... n.

I assume the meaning of "permutation of 1,2,...n" means that the digits 1 through n can be selected in any order, but all must be present precisely once.

For the first digit you can select any one of the n numbers. For the second digit you can select any one of the remaining (n-1) numbers that have not yet been picked - the probability of doing so is (n-1)/n. For the third number there are (n-2) numbers remaining that haven't been picked yet, so the probability of picking a "good" number is (n-2)/n. And so on.

Can you take it from here?

Thanks ebanies I can across this question in Sheldon Ross A first course in probability book chapter 6 jointly disributed random variable self test problem and exercise the answer with this approach doesn't match is there any other meaning to this question which I am not able to think!