Question: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. Solution 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..