%I A098825
%S A098825 1,1,0,1,0,1,1,0,3,2,1,0,6,8,9,1,0,10,20,45,44,1,0,15,40,135,264,265,1,
%T A098825 0,21,70,315,924,1855,1854,1,0,28,112,630,2464,7420,14832,14833,1,0,36,
%U A098825 168,1134,5544,22260,66744,133497,133496,1,0,45,240,1890,11088,55650
%N A098825 Triangle read by rows: T(n,k) = number of partial derangements, that 
               is, the number of permutations of n distinct, ordered items in which 
               exactly k of the items are in their natural ordered positions, for 
               n >= 0, k = n, n-1, ..., 1, 0.
%C A098825 In other words, T(n,k) is the number of permutations of n letters that 
               are at Hammimg distance k from the identity permutation (Cf. Diaconis, 
               p. 112). - N. J. A. Sloane (njas(AT)research.att.com), Mar 02 2007
%C A098825 The sequence d(n) = 1, 0, 1, 2, 9, 44, 265, 1854, 14833, ... (A000166) 
               is the number of derangements, that is, the number of permutations 
               of n distinct, ordered items in which none of the items is in its 
               natural ordered position.
%D A098825 P. Diaconis, Group Representations in Probability and Statistices, IMS, 
               1988; see p. 112.
%D A098825 Chris D. H. Evans, John Hughes and Julia Houston. Significance-testing 
               the validity of idiographic methods: A little derangement goes a 
               long way, British Journal of Mathematical and Statistical Psychology, 
               1 November 2002, Vol. 55, pp. 385-390.
%H A098825 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               PartialDerangement.html">Partial Derangement</a>
%F A098825 T(0, 0) = 1; d(0) = T(0, 0); for k = n, n-1, ..., 1, T(n, k) = c(n, k) 
               d(n-k) where c(n, k) = n! / [(k!) (n-k)! ]; T(n, 0) = n! - [ T(n, 
               n) + T(n, n-1) + ... + T(n, 1) ]; d(n) = T(n, 0).
%F A098825 T(n,k) = A008290(n,n-k). - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 
               04 2006
%F A098825 Assuming a uniform distribution on S_n, the mean Hamming distance is 
               n-1 and the variance is 1 (Diaconis, p. 117). - N. J. A. Sloane (njas(AT)research.att.com), 
               Mar 02 2007
%e A098825 Assume d(0), d(1), d(2) are given. Then
%e A098825 T(3, 3) = c(3, 3) d(0) = (1) (1) = 1
%e A098825 T(3, 2) = c(3, 2) d(1) = (3) (0) = 0
%e A098825 T(3, 1) = c(3, 1) d(2) = (3) (1) = 3
%e A098825 T(3, 0) = 3! - [ 1 + 0 + 3 ] = 6 - 4 = 2
%e A098825 d(3) = T(3, 0)
%t A098825 t[0, 0] = 1; t[n_, k_] := Binomial[n, k]*k!*Sum[(-1)^j/j!, {j, 0, k}]; 
               Flatten[ Table[ t[n, k], {n, 0, 10}, {k, 0, n}]] (from Robert G. 
               Wilson v Nov 04 2004)
%o A098825 This program generates the number of partial derangements T(n, k) for 
               n = 0, 1, ..., 8 and k = n, n-1, ..., 0. Let T(0, 0) = 1 Let d(0) 
               = T(0, 0) For n = 1 to 8 Let PartialSum = 0 For k = n to 1 Step -1 
               Let c(n, k) = Factorial(n) / (Factorial(k) * Factorial(n-k)) Let 
               T(n, k) = c(n, k) * d(n-k) Let PartialSum = PartialSum + T(n, k) 
               Next k Let T(n, 0) = Factorial(n) - PartialSum Let d(n) = T(n, 0) 
               Next n
%Y A098825 Cf. A000166.
%Y A098825 Sequence in context: A128317 A084269 A051427 this_sequence A111460 A035327 
               A004444
%Y A098825 Adjacent sequences: A098822 A098823 A098824 this_sequence A098826 A098827 
               A098828
%K A098825 nonn,tabl
%O A098825 0,9
%A A098825 Gerald P. Del Fiacco (gerrydelfiacco(AT)hotmail.com), Nov 02 2004
%E A098825 More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 04 2004