Mathematics for Computer Science

“mcs” — 2017/3/3 — 11:21 — page 665 — #673
15.11. References 665
(e) Now use the previous parts to prove
X
\
jDj D
. 1/
ˇˇˇˇˇ
jI jC1 S iˇˇˇˇˇ
;¤I f1;:::;ng
i2I
(15.22)
(f) Finally, explain why (15.22) immediately implies the usual form of the Inclusion-
Exclusion Principle:
Homework Problems
jDj D
nX X
. 1/ iC1
iD1
I f1;:::;ng
jI jDi
\
S j : (15.23)
ˇ
ˇˇˇˇˇˇ
Problem 15.59.
A derangement is a permutation .x 1 ; x 2 ; : : : ; x n / of the set f1; 2; : : : ; ng such that
x i ¤ i for all i. For example, .2; 3; 4; 5; 1/ is a derangement, but .2; 1; 3; 5; 4/
is not because 3 appears in the third position. The objective of this problem is to
count derangements.
It turns out to be easier to start by counting the permutations that are not derangements.
Let S i be the set of all permutations .x 1 ; x 2 ; : : : ; x n / that are not
derangements because x i D i. So the set of non-derangements is
n[
S i :
(a) What is jS i j?
(b) What is ˇˇSi \ S j
ˇˇ where i ¤ j ?
(c) What is ˇˇSi1 \ S i2 \ \ S ikˇˇ where i1 ; i 2 ; : : : ; i k are all distinct?
iD1
(d) Use the inclusion-exclusion formula to express the number of non-derangements
in terms of sizes of possible intersections of the sets S 1 ; : : : ; S n .
j 2I
(e) How many terms in the expression in part (d) have the form
ˇ
ˇSi1 \ S i2 \ \ S ikˇˇ‹
(f) Combine your answers to the preceding parts to prove the number of nonderangements
is: 1 1
nŠ
1Š 2Š C 1 ˙ 1
:
3Š nŠ