OFFSET
0,7
COMMENTS
Given n players there are a(n) different ways of arranging those players in a 3 against 3 contest.
Number of ways to select two disjoint subsets of size 3 from a set of n elements. - Joerg Arndt, Mar 29 2016
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
FORMULA
a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)/72.
a(n) = binomial(n,3) * binomial(n-3,3) / 2. - Joerg Arndt, Mar 29 2016
From Colin Barker, Mar 29 2016: (Start)
a(n) = 10*A000579(n).
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7) for n>6.
G.f.: 10*x^6 / (1-x)^7.
(End)
EXAMPLE
When there are 6 players, there are 10 different 3 against 3 matches that can be played: ABC v DEF, ABD v CEF, ABE v CDF, ABF v CDE, ACD v BEF, ACE v BDF, ACF v BDE, ADE v BCF, ADF v BCE, AEF v BCD.
MATHEMATICA
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 0, 0, 0, 0, 0, 10}, 40] (* Harvey P. Dale, Sep 17 2016 *)
PROG
(PARI) concat(vector(6), Vec(10*x^6/(1-x)^7 + O(x^50))) \\ Colin Barker, Mar 29 2016
(PARI) a(n)=binomial(n, 3)*binomial(n-3, 3)/2 \\ Charles R Greathouse IV, May 22 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Elliott Line, Mar 29 2016
STATUS
approved