Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 666 — #674
666
Chapter 15
Cardinality Rules
Conclude that the number of derangements is
nŠ 1
1
1Š C 1 2Š
1
3Š C ˙ 1
nŠ
(g) As n goes to infinity, the number of derangements approaches a constant fraction
of all permutations. What is that constant? Hint:
e x D 1 C x C x2
2Š C x3
3Š C
:
Problem 15.60.
How many of the numbers 2; : : : ; n are prime? The Inclusion-Exclusion Principle
offers a useful way to calculate the answer when n is large. Actually, we will use
Inclusion-Exclusion to count the number of composite (nonprime) integers from 2
to n. Subtracting this from n 1 gives the number of primes.
Let C n be the set of composites from 2 to n, and let A m be the set of numbers in
the range m C 1; : : : ; n that are divisible by m. Notice that by definition, A m D ;
for m n. So
C n D
(a) Verify that if m j k, then A m A k .
n[
1
iD2
A i : (15.24)
(b) Explain why the right-hand side of (15.24) equals
[
A p : (15.25)
primes p p n
(c) Explain why jA m j D bn=mc 1 for m 2.
(d) Consider any two relatively prime numbers p; q n. What is the one number
in .A p \ A q / A pq ?
(e) Let P be a finite set of at least two primes. Give a simple formula for
\
A :
ˇ pˇˇˇˇˇˇ
p2P