A115791 Number of different ways to select n elements from five sets of n elements under the precondition of choosing at least one element from each set.

0, 0, 0, 0, 3125, 97200, 1932805, 31539200, 461828790, 6332578125, 83235183075, 1063505908080, 13327125965725, 164758298214965, 2017489363833125, 24538128923443200, 297028957324770140, 3583456866615114630

Offset

1,5

Comments

The number of different ways to select n elements from five sets of n elements under the precondition of choosing at least one element from each set.

Links

Table of n, a(n) for n=1..18.

Formula

a(n) = binomial(5*n,n)-5*binomial(4*n,n)+10*binomial(3*n,n)-10*binomial(2*n,n)+5; ; also: a(n)=sum{binomial(n,i)*binomial(n,j)*binomial(n,k)*binomial(n,l)*binomial(n,m)||i,j,k,l,m=1...(n-4),i+j+k+l+m=n}. General formula for N sets with m elements each: the number of different ways to select k elements from j different sets: G(N,m,j,k) = binomial(N,j)*sum(binomial(j,i)*binomial(i*m,k)*(-1)^i*(-1)^j|i=1...j); Recursion formula: G(N,m,j,k) = binomial(N,j)*binomial(j*m,k) - sum(binomial(N-i,j-i)*G(N,m,i,k)|i=1...j-1);

Example

a(6)=binomial(30,6)-5*binomial(24,6)+10*binomial(18,6)-10*binomial(12,6)+5=97200;

Mathematica

Table[Binomial[5n, n]-5Binomial[4n, n]+10Binomial[3n, n]-10Binomial[2n, n]+5, {n, 20}] (* Harvey P. Dale, Nov 06 2011 *)

Crossrefs

Cf. A115111, A115112, A115246.

Sequence in context: A084649 A169754 A016817 * A016853 A016973 A017045

Adjacent sequences: A115788 A115789 A115790 * A115792 A115793 A115794

Keyword

nonn

Author

Hieronymus Fischer, Jan 31 2006

Status

approved