%I
%S 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,
%T 1,0,21,70,315,924,1855,1854,1,0,28,112,630,2464,7420,14832,14833,1,0,
%U 36,168,1134,5544,22260,66744,133497,133496,1,0,45,240,1890,11088,55650,222480,667485,1334960,1334961
%N 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, n1, ..., 1, 0.
%C 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_, Mar 02 2007
%C 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.
%H Reinhard Zumkeller, <a href="/A098825/b098825.txt">Rows n = 0..125 of triangle, flattened</a>
%H P. Diaconis, <a href="https://www.jstor.org/stable/4355560">Group Representations in Probability and Statistics</a>, IMS, 1988; see p. 112.
%H Chris D. H. Evans, John Hughes and Julia Houston, <a href="https://doi.org/10.1348/000711002760554525">Significancetesting the validity of idiographic methods: A little derangement goes a long way</a>, British Journal of Mathematical and Statistical Psychology, 1 November 2002, Vol. 55, pp. 385390.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PartialDerangement.html">Partial Derangement</a>
%F T(0, 0) = 1; d(0) = T(0, 0); for k = n, n1, ..., 1, T(n, k) = c(n, k) d(nk) where c(n, k) = n! / [(k!) (nk)! ]; T(n, 0) = n!  [ T(n, n) + T(n, n1) + ... + T(n, 1) ]; d(n) = T(n, 0).
%F T(n,k) = A008290(n,nk).  _Vladeta Jovovic_, Sep 04 2006
%F Assuming a uniform distribution on S_n, the mean Hamming distance is n1 and the variance is 1 (Diaconis, p. 117).  _N. J. A. Sloane_, Mar 02 2007
%e Assume d(0), d(1), d(2) are given. Then
%e T(3, 3) = c(3, 3) d(0) = (1) (1) = 1
%e T(3, 2) = c(3, 2) d(1) = (3) (0) = 0
%e T(3, 1) = c(3, 1) d(2) = (3) (1) = 3
%e T(3, 0) = 3!  [ 1 + 0 + 3 ] = 6  4 = 2
%e d(3) = T(3, 0).
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, 0, 1;
%e 1, 0, 3, 2;
%e 1, 0, 6, 8, 9;
%e 1, 0, 10, 20, 45, 44;
%e 1, 0, 15, 40, 135, 264, 265;
%e 1, 0, 21, 70, 315, 924, 1855, 1854;
%e ...
%t 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}]] (* _Robert G. Wilson v_, Nov 04 2004 *)
%o (Haskell)
%o a098825 n k = a098825_tabl !! n !! k
%o a098825_row n = a098825_tabl !! n
%o a098825_tabl = map (zipWith (*) a000166_list) a007318_tabl
%o  _Reinhard Zumkeller_, Dec 16 2013
%Y Cf. A000166, A007318, A008290.
%Y T(2n,n) gives A281262.
%K nonn,tabl
%O 0,9
%A _Gerald P. Del Fiacco_, Nov 02 2004
%E More terms from _Robert G. Wilson v_, Nov 04 2004
