By Albert Nijenhuis, Herbert S. Wilf

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K + 3J (k+l)(k n\. , j ; there are k + 1 such choices. As we have a total of n! permutations, Prob{y has k descendants} = (k + 2)(k + 3) We shall consider a similar construction for the case when k exceeds j —2 but is not large enough for j + k + I to exceed n (that is, x — 1 may not be feasible). , x + k + 1} including j to be in the subtree rooted at j . By the condition j + k + 1 < n, the integer x+k + l is always feasible. The integer JC — 1 is feasible only if x > 2. Each of the j — 1 choices x = 2, 3 , .

By the condition j + k + 1 < n, the integer x+k + l is always feasible. The integer JC — 1 is feasible only if x > 2. Each of the j — 1 choices x = 2, 3 , . . 4). ,&} to be considered in the construction. With x = 1, the number x — 1 is not Binary Trees 17 feasible (but x + k + 1 is). 4) gives n \ x 1 x 1 x k\ x(n-k-2)\ k + 2j = 1 n\ ( £ + ! ) ( £ + 2) permutations in this case. It follows that 20' - 1) Prob{y has k descendants) = — h (k+\)(k + 2) (k + l)(k + 2)(k + 3) The two other cases are when x — 1 is feasible but x + k + 1 may not be, and the case when both x — 1 and x + k + 1 may not be feasible.

The absolute rank of X; is denoted by Absrank(Xi). The range of Absrank(Xi) is { 1 , . . , n\. , X, only. That is, Seqrank(Xj) = j , if X,· is the 7'th smallest among X l , . . , X,·. So, the range of Seqrank(Xj) is (1, . . , i), for each 1 e | 1 , n\. 38 Sorting and Associated Concepts We assume throughout that the elements of our sample are distinct, as duplicate keys may only occur with probability 0 when the keys are taken from a continuous distribution. That is, the computation of any probability is not altered by ignoring the null set (set with probability zero) of sample space outcomes in which two or more keys are equal.