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%I #24 Sep 26 2017 05:08:10
%S 1,0,0,1,1,0,4,0,1,10,24,27,16,12,0,1,297,672,736,480,246,64,24,0,1,
%T 13756,30480,32365,21760,10300,3568,970,160,40,0,1,925705,2016480,
%U 2116836,1418720,677655,243360,67920,14688,2655,320,60,0,1
%N Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers): Triangle T(n,k) = number of ways to get k matches for a deck with n cards, 2 of each kind.
%C This is a triangle of card matching numbers. A deck has n kinds of cards, 2 of each kind. The deck is shuffled and dealt in to n hands with 2 cards each. A match occurs for every card in the j-th hand of kind j. Triangle T(n,k) is the number of ways of achieving exactly k matches (k=0..2n). The probability of exactly k matches is T(n,k)/((2n)!/2^n).
%C Rows are of length 1,3,5,7,... = A005408(n). [Edited by _M. F. Hasler_, Sep 21 2015]
%C Analogous to A008290. - _Zerinvary Lajos_, Jun 10 2005
%D F. N. David and D. E. Barton, Combinatorial Chance, Hafner, NY, 1962, Ch. 7 and Ch. 12.
%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 174-178.
%D R. P. Stanley, Enumerative Combinatorics Volume I, Cambridge University Press, 1997, p. 71.
%H F. F. Knudsen and I. Skau, <a href="http://www.jstor.org/stable/2691467">On the Asymptotic Solution of a Card-Matching Problem</a>, Mathematics Magazine 69 (1996), 190-197.
%H Barbara H. Margolius, <a href="http://academic.csuohio.edu/bmargolius/homepage/dinner/dinner/cardentry.htm">Dinner-Diner Matching Probabilities</a>
%H B. H. Margolius, <a href="http://www.jstor.org/stable/3219303">The Dinner-Diner Matching Problem</a>, Mathematics Magazine, 76 (2003), 107-118.
%H S. G. Penrice, <a href="http://www.jstor.org/stable/2324927">Derangements, permanents and Christmas presents</a>, The American Mathematical Monthly 98(1991), 617-620.
%H <a href="/index/Ca#cardmatch">Index entries for sequences related to card matching</a>
%F G.f.: sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k) where n is the number of kinds of cards, k is the number of cards of each kind (here k is 2) and R(x, n, k) is the rook polynomial given by R(x, n, k)=(k!^2*sum(x^j/((k-j)!^2*j!))^n (see Stanley or Riordan). coeff(R(x, n, k), x, j) indicates the j-th coefficient on x of the rook polynomial.
%e There are 4 ways of matching exactly 2 cards when there are 2 different kinds of cards, 2 of each in each of the two decks so T(2,2)=4.
%e Triangle begins:
%e 1
%e "0", 0, 1
%e 1, '0', "4", 0, 1
%e 10, 24, 27, '16', "12", 0, 1
%e 297, 672, 736, 480, 246, '64', "24", 0, 1
%e 13756, 30480, 32365, 21760, 10300, 3568, 970, '160', "40", 0, 1
%e 925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, '320', "60", 0, 1
%e Diagonal " ": T(n,2n-2) = 0, 4, 12, 24, 40, 60, 84, 112, 144, ... equals A046092
%e Diagonal ' ': T(n,2n-3) = 0, 16, 64, 160, 320, 560, 896, 1344, ... equals A102860
%p p := (x,k)->k!^2*sum(x^j/((k-j)!^2*j!),j=0..k); R := (x,n,k)->p(x,k)^n; f := (t,n,k)->sum(coeff(R(x,n,k),x,j)*(t-1)^j*(n*k-j)!,j=0..n*k);
%p for n from 0 to 7 do seq(coeff(f(t,n,2),t,m)/2^n,m=0..2*n); od;
%t p[x_, k_] := k!^2*Sum[ x^j/((k-j)!^2*j!), {j, 0, k}];
%t R[x_, n_, k_] := p[x, k]^n;
%t f[t_, n_, k_] := Sum[ Coefficient[ R[x, n, k], x, j]*(t-1)^j*(n*k-j)!, {j, 0, n*k}];
%t Table[ Coefficient[ f[t, n, 2]/2^n, t, m], {n, 0, 6}, {m, 0, 2*n}] // Flatten
%t (* _Jean-François Alcover_, Sep 17 2012, translated from Maple *)
%Y Cf. A059056-A059071, A008290.
%K nonn,tabf,nice
%O 0,7
%A Barbara Haas Margolius (margolius(AT)math.csuohio.edu)
%E Additional comments from _Zerinvary Lajos_, Jun 18 2007
%E Edited by _M. F. Hasler_, Sep 21 2015
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