cp's OEIS Frontend

A200979 Number of ways to arrange n books on 4 consecutive bookshelves, leaving no shelf empty.

%I A200979 #13 Jan 03 2021 16:55:20
%S A200979 24,480,7200,100800,1411200,20321280,304819200,4790016000,79035264000,
%T A200979 1369944576000,24932991283200,475993469952000,9519869399040000,
%U A200979 199184959733760000,4353614119895040000,99262401933606912000,2357482045923164160000
%N A200979 Number of ways to arrange n books on 4 consecutive bookshelves, leaving no shelf empty.
%C A200979 To derive a(n) = n!*binomial(n-1,3), we note that there are n! ways to arrange the books in a row.
%C A200979 Then there are binomial(n-1,3) ways to place the arranged books on 4 consecutive shelves since there are binomial(n-1,3) compositions of n with 4 summands. Hence a(n) = n!*binomial(n-1,4).
%C A200979 We note that a(n) is the fourth column of triangle A156992, the number of ways to arrange n books on k consecutice bookshelves, leaving no shelves empty.
%F A200979 a(n) = n!*binomial(n-1,3) = n!*(n-1)*(n-2)*(n-3)/6 for n >= 4.
%F A200979 E.g.f.: x^4/(1-x)^4.
%F A200979 a(n) = A156992(n,4).
%e A200979 a(5) = 480 since there are 480 ways to arrange books b1, b2, b3, b4, and b5 on shelves s1, s2, s3, and s4. In this case, one shelf will hold two books, and the other three shelves will hold one each. There are 4 ways to choose the shelf for two books; there are 20 ways to choose the two books and place them in order on the two-book shelf; there are 6 ways to arrange the other three books on the other three shelves. Hence a(5) = 4*20*6 = 480.
%p A200979 seq(n!*binomial(n-1,3)n=4..20);
%K A200979 nonn
%O A200979 4,1
%A A200979 _Dennis P. Walsh_, Nov 26 2011