Concrete mathematics : a foundation for computer science

194 BINOMIAL COEFFICIENTS
We can determine h(n, k) by noticing that it is the number of ways to
choose k lucky hat owners, namely (L), times the number of ways to arrange
the remaining n-k hats so that none of them goes to the right owner, namely
h(n - k, 0). A permutation is called a derangement if it moves every item,
and the number of derangements of n objects is sometimes denoted by the
symbol ‘ni’, read “n subfactorial!’ Therefore h(n - k, 0) = (n - k)i, and we
have the general formula
h(n,k) =
(Subfactorial notation isn’t standard, and it’s not clearly a great idea; but
let’s try it awhile to see if we grow to like it. We can always resort to ‘D,’ or
something, if ‘ni’ doesn’t work out.)
Our problem would be solved if we had a closed form for ni, so let’s see
what we can find. There’s an easy way to get a recurrence, because the sum
of h(n, k) for all k is the total number of permutations of n hats:
n! = xh(n,k) = t ($(n-k)i
k k
integer n 3 0.
(We’ve changed k to n - k and (,“,) to (L) in the last step.) With this
implicit recurrence we can compute all the h(n, k)‘s we like:
h(n, 0) h(n, 1) h(n,2) h(n,3) h(n,4) h(n,5) h(n, 6)
0 1
1 0 1
2 3 0 1
9 8 6 0 1
20 10 0 1
24645 2l 135 40 15 0 1
For example, here’s how the row for n = 4 can be computed: The two rightmost
entries are obvious-there’s just one way for all hats to land correctly,
and there’s no way for just three fans to get their own. (Whose hat would the
fourth fan get?) When k = 2 and k = 1, we can use our equation for h(n, k),
giving h(4,2) = ($h(2,0) = 6.1 = 6, and h(4,l) = (;)h(3,0) = 4.2 = 8. We
can’t use this equation for h(4,O); rather, we can, but it gives us h(4,O) =
(;)h(4,0), w h’ rc h is . true
but useless. Taking another tack, we can use the
The art of math-
ematics, as of life,
relation h(4,O) + 8 + 6 + 0 + 1 = 4! to deduce that h(4,O) = 9; this is the value is knowing which
of 4i. Similarly ni depends on the values of ki for k < n. truths are useless.