A200978 Number of ways to arrange n books on 3 consecutive shelves leaving none of the shelves empty.

6, 72, 720, 7200, 75600, 846720, 10160640, 130636800, 1796256000, 26345088000, 410983372800, 6799906713600, 118998367488000, 2196892938240000, 42682491371520000, 870722823979008000, 18611700362551296000, 416026243398205440000

Offset: 3

Author

Dennis P. Walsh, Nov 26 2011

Keywords

• nonn
• easy

Comments

To derive a(n), we note that there are n! ways to arrange n books in a row and there are binomial(n-1,2) ways to place the n arranged books on 3 consecutive shelves (since binomial(n-1,2) is the number of compositions of n with 3 summands). Hence a(n) = n!*binomial(n-1,2) for n >= 3.

The number of ways to arrange n books on two nonempty bookshelves is given by A062119(n).

Examples

			a(4)=72 since there are 72 ways to arrange books b1, b2, b3, and b4 on 3 consecutive shelves s1, s2, and s3. Note that there are 24 arrangements with two books on shelf s_i (i=1,2,3) and one book on each of the other two shelves. (For instance, there are 12 ways to select and permute the two books for s1 and 2 ways to select the single books for s2 and s3.) Hence there are 3(24)=71 book arrangements.
		

Crossrefs

Cf. A156992.

Programs

Maple

seq(n!*C(n-1,2),n=3..20);

Mathematica

nn=20;Drop[Range[0,nn]!CoefficientList[Series[(x/(1-x))^3,{x,0,nn}],x],3] (* Geoffrey Critzer, Sep 02 2013 *)

Formula

a(n) = n!*binomial(n-1,2) = n!*(n-1)*(n-2)/2, n >= 3.

a(n) = A156992(n,3).

E.g.f.: x^3/(1-x)^3.

a(n) = A001754(n)*3!. - Geoffrey Critzer, Sep 02 2013