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A181386 Tetrahedron of terms C(r,n,m) representing the number of ways of choosing m disjoint subsets of r members from an original set of n members. 1

%I #36 Feb 20 2016 21:43:17

%S 1,1,1,1,1,2,1,1,1,1,3,3,1,1,1,1,1,1,4,6,4,1,1,3,1,1,1,1,5,10,10,5,1,

%T 1,6,3,1,1,1,1,1,1,6,15,20,15,6,1,1,10,15,1,4,1,1,1,1,1,7,21,35,35,21,

%U 7,1,1,15,45,15,1,10,1,1,1,1,1,1,1,8,28,56,70,56,28,8,1,1,21,105,105,1,20

%N Tetrahedron of terms C(r,n,m) representing the number of ways of choosing m disjoint subsets of r members from an original set of n members.

%C The start index for r is 1 but the start index for m and n is 0. For each value of r, the triangle T_r(n,m) has row n containing 1 + floor(n/r) terms.

%C From _Frank M Jackson_, Nov 20 2010: (Start)

%C C(r,mr,m) = C(r,mr-1,m-1).

%C C(1,m,m) = A000012, C(2,2m,m) = A001147,

%C C(3,3m,m), ..., C(10,10m,m) = A025035, ..., A025042.

%C C(2,26,10) = 150738274937250 and represents the number of possible plugboard settings for a WWII German Enigma Enciphering Machine.

%C C(r,2r,2) = A001700, C(r,3r,3) = A060542, C(r,4r,4) = A082368.

%C C(r,n,m) = C(r,mr-1,m-1)*binomial(n,rm),

%C and applied recursively gives the identity

%C C(r,n,m) = Binomial(n,r*m) * Product_{p=1..m} Binomial(r*(m-p+1)-1,r-1).

%C (End)

%C C(2,26,10) = A266365(10), where 26 is the size of the alphabet. - _Jonathan Sondow_, Dec 29 2015

%H T. Copeland, <a href="http://tcjpn.wordpress.com/2012/11/29/infinigens-the-pascal-pyramid-and-the-witt-and-virasoro-algebras/">Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras</a>

%H Tony Sale, <a href="http://www.codesandciphers.org.uk/enigma/steckercount.htm">Possible Plugboard Settings for a WWII German Enigma Enciphering Machine</a>

%F C(r,n,m) = n!/((n-r*m)!*m!*(r!)^m).

%e r=1, C(1,n,m) is

%e 1

%e 1, 1

%e 1, 2, 1

%e 1, 3, 3, 1

%e 1, 4, 6, 4, 1

%e 1, 5, 10, 10, 5, 1

%e r=2, C(2,n,m) is

%e 1

%e 1

%e 1, 1

%e 1, 3

%e 1, 6, 3

%e 1, 10, 15

%e r=3, C(3,n,m) is

%e 1

%e 1

%e 1

%e 1, 1

%e 1, 4

%e 1, 10

%t Flatten[Table[{n!/((n-r*m)!*m!*r!^m)}, {r, 1, 50}, {n, 0, 50}, {m, 0, Floor[n/r]}]]

%Y C(1,n,m) = T_1(n,m) = A007318, C(2,n,m) = T_2(n,m) = A100861, and C(2,26,m) = A266365.

%K nonn,tabf

%O 1,6

%A _Frank M Jackson_, Oct 16 2010

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