For the life of me I am having a hard time understanding how to do problems of this nature. As I understand it, were using the multiplication rule here with a twist. a. How many integers from 1 through 100,000 contain the digit 6 exactly once? 5 * 9 * 9 * 9 * 9 = 38805 is what I have. Because the digit 6 can appear in 5 different locations. b. How many integers from 1 through 100,000 contain the digit 6 at least once? c. If an integer is chosen at random from 1 through 100,000, what is the probability that it contains two or more occurrences of the digit 6? I am clueless here. Add up all the possibilities of containing two 6's...three 6's... four 6's... five 6's and then divide by the total possibilities.That's my line of thinking... I am just not sure how to find the amount of possibilities for each partition.