Probability basics Arranging books on a shelf, with duplicates Assume that some of the books have one or even several duplicate copies, which can- not be distinguished from each other, as in a bookstore. For instance, the series of 10 books includes two brand-new English-Russian dictionaries. Having these two in shelf positions (a, b) or (b, a) represents the same arrangement. So as not to count this arrangement twice, we should divide the previous result (10!) by two (2!), which is the number of possible permutations for the duplicated dictionaries. If we had three identical books in the series, we should divide the result by six (3!), and SO on. It is clear, then, that the number of ways of arranging n elements containing p indistinguishable elements and n - p distinguishable ones is given by the ratio: (1.8) The above theorem defines the number of possible arrangements "without repetition" (of the indistinguishable copies). Consistently, if the series contains n indistinguishable duplicates, the number of arrangements is simply An = n!/n! = 1, namely, leaving a unique possibility. The next example will make us progress one step more. Assume that we must make a selection from a set of objects. The objects can be all different, all identical, or partly duplicated, which does not matter. We would only like to know the number of possibilities there are to make any random selection of these objects. Fruit-market shopping In a fruit market, the stall displays 100 fruits of various species and origins. We have in mind to pick at random up to five fruits, without preference. There are 100 different possibilities to pick up the first fruit, 99 possibilities to pick up the second, and SO on until the last fifth, which has 96 possibilities left. The total number of possibilities to select five specific fruits out of 100 is therefore Q = 100 X 99 X 98 X 97 X 96. Based on the definition of factorials in Eq. (1.6), we can write this number in the form Q = 100!/95! = 100!/(100 - 5)! But in each selection of five fruits we pu