9. Aplicacions dels làsers
9.2. Aplicacions Mèdiques
. 1/j.0 j /0D 1 0 0
!
. 1/000D 1
with the convention 00D 1. This is correct since the number of partitions of the empty set into zero parts equals 1 (the empty partition). The other two cases are left for you to check in the following Question.
Question 99 Forn > 1, what value should S.n; 0/ take? Does the formula agree? For k > 1, what value should S.0; k/ take? Does the formula agree?
Another example and a warning
The problem of counting ciphers is more commonly known as the problem of derange-ments: How many permutations of Œn have no fixed points? (A fixed point of a function f is a value i for which f .i /D i.) It is also known as the hat-check problem: in how many ways can the hats of n people be re-distributed so that each person receives exactly one hat but no person receives their own hat? The cipher problem is equivalent to either of these problems with nD 26.
Consider modifying the hat-check problem by removing the requirement that each per-son receives exactly one hat. That is, allow any perper-son to receive any number of hats but still require that no person receives their own hat. This involves counting functions rather than permutations, and an application of inclusion-exclusion might use
U WD set of all possible functions Œn ! Œn
fi WD “the function fixes element i”, for all i 2 Œn.
As usual we want ND.;/. For any j -subset J of the properties it follows that N>.J / D nn j. This means that the answer is
ND.;/ D Xn jD0
n j
!
. 1/jnn j:
This is correct but if we apply the binomial theorem to this sum we get Xn
jD0
n j
!
. 1/jnn j D . 1 C n/nD .n 1/n:
There should be a simple explanation for this simple answer, and there is: for every i 2 Œn, there are n 1choices (anything except i ) for the value of f .i /.
This problem warns us to seek the simplest solutions first before trying more compli-cated methods!
The more general formula
The basic principle of inclusion-exclusion applies to counting objects that satisfy none of the properties. How might we count the objects that satisfy some of the properties?
Theorem 3.1.5 (general principle of inclusion-exclusion) Let U be a universe of objects, and letP D fp1; p2; : : : ; png be a set of properties that the objects may or may not have.
IfS is any subset of P , then the number of objects in U with the properties in S and no others is
ND.S /D X
JWSJ P
. 1/jJ j jSjN>.J /:
In this case, the sum is over all subsets of P that contain the elements of S . The proof of the general principle is in Exercise 16 and uses essentially the same technique as the proof of the basic principle.
Example: counting divisors again
How many integers in Œ100 are divisible by 2 but not by 3 or 5?
This question still has U D Œ100 and P D fd2; d3; d5g as defined earlier, but now we seek ND.d2/. So we apply the formula in the theorem with S D fd2g. The sum will then be over the four subsets
fd2g; fd2; d3g; fd2; d5g; fd2; d3; d5g:
The formula gives
ND.d2/D N>.d2/
N>.d2d3/C N>.d2d5/
C N>.d2d3d5/:
We have already computed these values. The answer is 50 .16C 10/ C 3 D 27.
Question 100 How many integers inŒ100 are divisible by 3 but not by 2 or 5? How many are divisible by 2 and 3 but not by 5?
Summary
Inclusion-exclusion is tailor-made for counting problems that fit the universe/properties framework. The properties generally describe “bad” traits, and the inclusion-exclusion for-mula counts those objects with none of the bad traits. In applying the forfor-mula, some prob-lems allow shortcuts because the value of N>.J /only depends on the size of J . This was the case in both the cipher and onto function examples. Other problems, such as the one involving counting integers not divisible by 2, 3, or 5, do not allow such shortcuts. In that example we had to compute each value of N>.J /separately.
Exercises
1. How many integers in Œ10000 are not divisible by 2, 3, or 5? How many are not divisible by 2, 3, 5, or 13?
2. How many integers in Œ100 are not divisible by 4, 6 or 7?
3. Use inclusion-exclusion to prove the formula
jA1[ A2[ A3j D jA1j C jA2j C jA3j
jA1A2j jA1A3j jA2A3j C jA1A2A3j:
(The notation A1A2means A1\ A2.)
3.1. Inclusion-exclusion 93 4. Find the number of 13-card hands drawn from a 52-card deck that...
(a) have at least one card in each suit.
(b) are void in exactly one suit. (“Void in spades” means there are no spades in the hand.)
5. After a day of skiing a family of six washes all of their ski-wear including their gloves.
The next day, each family member grabs two gloves from the pile.
(a) Assume that everyone grabs one left-hand and one right-hand glove. In how many ways can they do this so that no one has both of their own gloves?
(b) Answer part (a) assuming instead that each family member grabs any two gloves.
6. Answer the hat-check problem (i.e., the problem of derangements) for general n. This number is known as Dn.
7. Let Dnbe as defined in the previous exercise.
(a) Calculate lim
n !1
Dn
nŠ. Interpret your result.
(b) Prove that for any n, Dnequals the closest integer to nŠ=e.
8. In how many ways can you distribute 20 identical objects to 10 distinct recipients so that each recipient receives at most five objects? How many ways if each receives at least one but at most five objects?
9. Generalize the previous problem: In how many ways can you distribute k identical objects to n distinct recipients so that each recipient receives at most r objects?
10. How many functions Œ6 ! Œ7 have at most two arrows pointing to each element of the codomain?
11. When k < n, what is the value of the sum Xn jD0
n j
!
. 1/j.n j /k? Explain combi-natorially.
12. Derive an identity for nk
via inclusion-exclusion by counting the k-multisets of Œn in which each element of Œn appears at most once. Use pi D “element i appears more than once in the multiset” as the i -th property, for 1 6 i 6 n.
13. Suppose that in an inclusion-exclusion problem, there exists a function f such that N>.J /D f jJ j
for any subset J of P . Prove:
ND.;/ D Xn jD0
n j
!
. 1/jf .j /:
14. Give a combinatorial proof of the identityPn kD0
n j
. 1/j D 0 wherein the left side is computed using inclusion-exclusion.
15. Prove combinatorially, using inclusion-exclusion, the identity that results from letting xD 1 and y D 2 in the binomial theorem (Theorem 2.2.2, p. 63).
16. Prove Theorem 3.1.5 by adapting the technique we used to prove Theorem 3.1.2.
17. A taxi drives from the intersection labeled A to the intersection labeled B in the grid of streets shown below. The driver only drives north (up) or east (right).
A
B
congested intersection
=
Traffic reports indicate that there is heavy congestion at the intersections identified.
How many routes from A to B can the driver take that...
(a) avoid all congested intersections?
(b) pass through at most one congested intersection?
18. A 4 word search puzzle is a 4 array of capital letters. How many 4-by-4 word searches have the word MATH appearing at least once either horizontally, vertically, or diagonally? Here are examples of four different such puzzles:
F H M A M A T H M M M M M A T H
M A T H A T H M G A A P M A T H
G Z Z Q T S M E Z R T Y M A T H
F A Y U H E E N K L H H M A T P
Assume MATH appears left-to-right, top-to-bottom, or top-left-to-bottom-right only.
Travel Notes
Several 19th century mathematicians have been associated with discovering the inclusion-exclusion formula, including Daniel da Silva, Abraham de Moivre, and J. J. Sylvester, but it was da Silva who first published it in 1854.
The hardest part about inclusion-exclusion is the notation. The use of N>and ND is fairly common but not universal. The use of the subset sumP
JWJ P avoids undue use of
“ ” in something like ND.;/ D N>.;/ X
i
N>.pi/CX
i6Dj
N>.pipj/ X
i;j;k different
N>.pipjpk/ C C . 1/nN>.p1p2 pn/:
In Sections 8.5 and 8.6, we study a powerful generalization of inclusion-exclusion called the principle of M¨obius inversion. In the foundational paper concerning M¨obius inversion, Rota (1964) begins by declaring that “One of the most useful principles of enumeration in discrete probability and combinatorial theory is the celebrated principle of inclusion-exclusion. When skillfully applied, this principle has yielded the solution to many a combinatorial problem.”
3.2 Mathematical induction
The reader familiar with induction can either omit this section or skim the examples.
In this section we highlight how to use induction to complement our combinatorial proof techniques. Sometimes you might first discover the truth of an identity using induc-tion, and then later realize a combinatorial proof. Other times, induction ends up being the only thing that works.
3.2. Mathematical induction 95