login
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
OFFSET
4,1
COMMENTS
To derive a(n) = n!*binomial(n-1,3), we note that there are n! ways to arrange the books in a row.
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).
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.
FORMULA
a(n) = n!*binomial(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).
EXAMPLE
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.
MAPLE
seq(n!*binomial(n-1, 3)n=4..20);
CROSSREFS
Sequence in context: A263476 A263470 A276816 * A361576 A052715 A052725
KEYWORD
nonn
AUTHOR
Dennis P. Walsh, Nov 26 2011
STATUS
approved