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A200979 Number of ways to arrange n books on 4 consecutive bookshelves, leaving no shelf empty. 0
24, 480, 7200, 100800, 1411200, 20321280, 304819200, 4790016000, 79035264000, 1369944576000, 24932991283200, 475993469952000, 9519869399040000, 199184959733760000, 4353614119895040000, 99262401933606912000, 2357482045923164160000 (list; graph; refs; listen; history; text; internal format)



To derive a(n)=n!C(n-1,3), we note that there are n! ways to arrange the books in a row.

Then there are C(n-1,3) ways to place the arranged books on 4 consecutive shelves since there are C(n-1,3) compositions of n with 4 summands. Hence a(n)=n!C(n-1,4).

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.


Table of n, a(n) for n=4..20.


a(n) = n!C(n-1,3) = n!(n-1)(n-2)(n-3)/6 for n>=4.

E.g.f: x^4/(1-x)^4.

a(n) = A156992(n,4).


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.


seq(n!*binomial(n-1, 3)n=4..20);


Sequence in context: A263476 A263470 A276816 * A052715 A052725 A006548

Adjacent sequences:  A200976 A200977 A200978 * A200980 A200981 A200982




Dennis P. Walsh, Nov 26 2011



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Last modified March 26 01:08 EDT 2019. Contains 321479 sequences. (Running on oeis4.)