

A200978


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


0



6, 72, 720, 7200, 75600, 846720, 10160640, 130636800, 1796256000, 26345088000, 410983372800, 6799906713600, 118998367488000, 2196892938240000, 42682491371520000, 870722823979008000, 18611700362551296000, 416026243398205440000
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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 C(n1,2) ways to place the n arranged books on 3 consecutive shelves (since C(n1,2) is the number of compositions of n with 3 summands). Hence a(n)=n!C(n1,2) 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!C(n1,2) = n!(n1)(n2)/2, n>=3
a(n) = A156992(n,3).
E.g.f: x^3/(1x)^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 si (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(n1, 2), n=3..20);


MATHEMATICA

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


CROSSREFS

Cf. A156992.
Sequence in context: A005548 A059410 A238103 * A099673 A105928 A186665
Adjacent sequences: A200975 A200976 A200977 * A200979 A200980 A200981


KEYWORD

nonn,easy


AUTHOR

Dennis P. Walsh, Nov 26 2011


STATUS

approved



