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Number of ways to arrange n books on 3 consecutive shelves leaving none of the shelves empty.
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%I #27 Jan 03 2021 15:55:58

%S 6,72,720,7200,75600,846720,10160640,130636800,1796256000,26345088000,

%T 410983372800,6799906713600,118998367488000,2196892938240000,

%U 42682491371520000,870722823979008000,18611700362551296000,416026243398205440000

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

%C 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.

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

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

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

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

%F a(n) = A001754(n)*3!. - _Geoffrey Critzer_, Sep 02 2013

%e 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.

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

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

%Y Cf. A156992.

%K nonn,easy

%O 3,1

%A _Dennis P. Walsh_, Nov 26 2011