Question 570976
For this problem we will assume that *[tex \LARGE 1 \le m \le n].


The best way is probably to try small cases (for m). Let P(m) be the probability that m is the largest number drawn. If m = 1, then every number drawn must be a 1, and


*[tex \LARGE P(1) = (\frac{1}{n})^k]


If m = 2, every number must be a 1 or a 2, and we would try to claim something like


*[tex \LARGE P(2) = (\frac{2}{n})^k]


However, this is incorrect, because this includes the case where every number is a 1 (which violates the constraint that at least one number must be a 2). To fix this, we subtract P(1):


*[tex \LARGE P(2) = (\frac{2}{n})^k - P(1) = (\frac{2}{n})^k - (\frac{1}{n})^k]


Finding P(3) is similar to finding P(2), except that we subtract P(2):


*[tex \LARGE P(3) = (\frac{3}{n})^k - P(2) = (\frac{3}{n})^k - (\frac{2}{n})^k +(\frac{1}{n})^k]


This generalizes to


*[tex \LARGE P(m) =  (\frac{m}{n})^k - P(m-1)], or


*[tex \LARGE P(m) = (\frac{m}{n})^k - (\frac{m-1}{n})^k + (\frac{m-2}{n})^k  ... \pm (\frac{1}{n})^k]


where the plus/minus of (1/n)^k depends on the parity of m (i.e. if m is odd, +; if m is even, -).