# Seminaire d'Algebre Paul Dubreil et Marie-Paule Malliavin by M.P. Malliavin

By M.P. Malliavin

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Example text

Thus, by (MP), the number of ways to arrange any r elements from N n in a row is given by n(n - l)(n - 2 ) . . (n - r + 1). For convenience, let us call an arrangement of any r elements from N n in a row, an r-permutation of N n , and denote by P™ the number of r-permutations of N n . Thus, we have P r " = n(n - l)(n - 2 ) . . 2. 1) for P™ looks a bit long. We shall make it more concise by introducing the following useful notation. Given a positive integer n, define n! to be the product of the n consecutive integers n, n — 1 , .

4 • 3 • 2 • 1 = 24. " is read "n factorial". By convention, we define 0! = 1. Using the "factorial" notation, we now have P r n = n(n - 1 ) . . (n - r + 1) n{n — 1 ) . . (n — r + l)(n — r)(n — r — 1 ) . . 2 • 1 _ n\ (n — r)(n — r — 1 ) . . 3) When n = 4 and r = 3, we obtain 4! r 3 — (4 - 3)! P4 - 4! 1! 4• 3• 2• 1 = 4 • 3 • 2 = 24, 1 which agrees with what we found before. 3) is valid when 0 < r < n. Consider two extreme cases: when r = 0 and r = n respectively. 3), pn __ n! = ^ = 1 (n - 0)!

1. ,n n choices n-\ choices 1 n-(r-2) n-0-l) choices choices n—2 choices ... 1 rth Counting 20 We wish to choose r elements from {1, 2 , . . ,n} to fill the r spaces, where the ordering of elements matters. There are n choices for the 1st space. After fixing one in the 1st space, there are n — 1 choices remaining for the 2nd space. After fixing one in the 2nd space, there are n — 2 choices left for the 3rd space, and so on. After fixing one in the (r — l)th space, there are n — (r — 1) choices left for the rth space.