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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 (list; graph; refs; listen; history; text; internal format)
 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(n-1,2) ways to place the n arranged books on 3 consecutive shelves (since C(n-1,2) is the number of compositions of n with 3 summands). Hence a(n)=n!C(n-1,2) for n>=3. The number of ways to arrange n books on two nonempty bookshelves is given by A062119(n). LINKS FORMULA a(n) = n!C(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 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(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 * A099673 A105928 A186665 Adjacent sequences:  A200975 A200976 A200977 * A200979 A200980 A200981 KEYWORD nonn,easy AUTHOR Dennis P. Walsh, Nov 26 2011 STATUS approved

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Last modified March 20 15:43 EDT 2019. Contains 321345 sequences. (Running on oeis4.)