Concrete mathematics : a foundation for computer science

5.1 BASIC IDENTITIES 157
I just hope I don’t of its variables. Everyone who’s manipulated binomial coefficients much has
fall into this trap
during the midterm.
fallen into this trap at least three times.
But the symmetry identity does have a big redeeming feature: It works
for all values of k, even when k < 0 or k > n. (Because both sides are zero in
such cases.) Otherwise 0 < k 6 n, and symmetry follows immediately from
(5.3):
n
n!
0k
= k!(n-k)!
Our next important
coefficients:
= (n-(n--l\! (n-k)! =
identity lets us move things in and out of binomial
(3 = I,(:::)) integer k # 0. (5.5)
The restriction on k prevents us from dividing by 0 here. We call (5.5)
an absorption identity, because we often use it to absorb a variable into a
binomial coefficient when that variable is a nuisance outside. The equation
follows from definition (5.1), because rk = r(r- 1 )E and k! = k(k- l)! when
k > 0; both sides are zero when k < 0.
If we multiply both sides of (5.5) by k, we get an absorption identity that
works even when k = 0:
k(l[) = r(;-i) , integer k.
This one also has a companion that keeps the lower index intact:
(r-k)(I) = r(‘i’), integer k.
We can derive (5.7) by sandwiching an application of (5.6) between two applications
of symmetry:
(r-k)(;) = (r-kl(rlk) (by symmetry)
= r(,.Ti! ,) (by (54)
(by symmetry)
But wait a minute. We’ve claimed that the identity holds for all real r,
yet the derivation we just gave holds only when r is a positive integer. (The
upper index r - 1 must be a nonnegative integer if we’re to use the symmetry
(5.6)
(5.7)