A271040 Number of different 3 against 3 matches given n players.

0, 0, 0, 0, 0, 0, 10, 70, 280, 840, 2100, 4620, 9240, 17160, 30030, 50050, 80080, 123760, 185640, 271320, 387600, 542640, 746130, 1009470, 1345960, 1771000, 2302300, 2960100, 3767400, 4750200, 5937750, 7362810, 9061920, 11075680, 13449040, 16231600, 19477920

Offset: 0

Author

Elliott Line, Mar 29 2016

Keywords

• nonn
• easy

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

Examples

			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.
		

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).

Crossrefs

Cf. A050534, the analogous situation for 2 against 2 matches.

Programs

Mathematica

LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,0,0,0,10},40] (* Harvey P. Dale, Sep 17 2016 *)

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

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)