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

3,1

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).

Links

Table of n, a(n) for n=3..20.

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

Example

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.

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 *)

Crossrefs

Cf. A156992.

Sequence in context: A005548 A059410 A238103 * A361572 A099673 A105928

Adjacent sequences: A200975 A200976 A200977 * A200979 A200980 A200981

Keyword

nonn,easy

Author

Dennis P. Walsh, Nov 26 2011

Status

approved