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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
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).
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
KEYWORD
nonn,easy
AUTHOR
Dennis P. Walsh, Nov 26 2011
STATUS
approved