cou~rrl~IG

Its
Principles
and
Techniques (7)
by
Koh Khee Meng

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 17. The Principle of Inclusion and Exclusion
In Section 1 of[1], we introduced the Addition Principle (AP) which was expressed in
terms of sets as follows:

If Aand Bare disjoint finite sets, then lA u BI= !AI+ IB I ( 17.1)

In the above statement, A and B are assumed to be disjoint, written An B = <!>.

A and B have no elements in common. Can we express lA uBI in terms of
i.e.
!AI and IBI regardless of whether A and Bare disjoint? In counting the elements in
A u B, we may first count those in A and then those in B. In doing so, any element in
ABn (if there is) is counted exactly twice. Thus to get the exact count of u lA Bl,
the
number lA n Bl should be deducted. It follows that

(17.2)

This result can also be seen intutively with the help of the Venn diagram ofFigure 17.1.

Figure 17.1

Note that (17 .1) is a special case of ( 17 .2) as ( 17.1) follows from ( 17 .2) if we assume that
AnB = 4>.

Identity (17.2) is a simplest form of a principle called the Principle of Inclusion and
Exclusion (PIE), which is a very useful and powerful tool in enumeration. First of all, let us
show two applications of(17.2).

Example 17.1. Find the number ofintegers from the set {1,2, .. . ,1000} which are divisible by 3
or 5.

The integers which we are looking for are 3, 5, 6, 9, 10, 12, 15, 18, 20, ... , 999, 1000.
How many are there?

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 Let us try to present the solution more formally, and so we let

s = {1,2, ... ,1000},

A={xESix isdivisible by3},

and B = {x E SIx is divisible by 5}.

It is now clear that our task is to evaluate lA u Bl as Au B is the set of numbers in S which
are divisible by 3 or 5.

Before applying (17.2) to evaluate lA u Bl, we first introduce a useful notation. For a
real number r, let Lr J denote the greatest integer that is smaller than or equal to r. Thus

L3.15 J= 3, l
2
3
J= 6, L7 J=7. and so on. How many integers in { 1,2 ... , 10} are there which are

divisible by 3? There are three (namely, 3, 6, 9) and note that 'three' can be expressed as l 1~ J.
The number of integers in { 1,2, ... , 10} which are divisible by 5 is 'two' (namely 5 and 10), and

note that 'two' can be expressed as ll~ J

Indeed, in general, for any two natural numbers n, k

with k::;; n, the number of integers in the set {1,2, ... ,n} which are divisible by k can be

expressed as l%J
We can now return to our original problem of evaluating lA uBI. To apply (17.2), we
need to find IAI, IBI and lA 11 B I Using the notation introduced above, we see that
10 0
JAJ = l
3

J = 33 3 and JBJ = llO~O
J = 200. It remains to find lA n Bl. What does A n B

represent? Well, A n B is the set of integers in S which are divisible by both 3 and 5. How to
evaluate JA 11 Bj? It seems that this problem is as hard as that of evaluating lA u Bl. Luckily,
this is not so as there is a result in Arithmetic that can help us. Let a, b be any two positive
integers. It is known that an integer is divisible by both a and b when and only when it is
divisible by the LCM (least common multiple) of a and b. It thus follows that A 11 B is the set
of numbers in Swhich are divisible by the LCM of3 and 5. As the LCM of3 and 5 is 15, we
conclude that

A n B = {x E S I x is divisible by 15 }.

Thus JAnBJ=l 1 ~~ 0 J= 66.

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 Finally, by (17.2), we have

=333+200-66=467. 0

Example 17.2. Find the number ofpositive divisors of at least one ofthe numbers 5400 and
18000.

In Section 3 of[l], we discussed the problem of finding the number of positive divisors
of a natural number and stated the following relevant result: Let n ~ 2 be a natural number. If
n= p; p;1
... p;, where pi>p 2 , ... ,pk are distinct primes, then the number ofpositive
divisors ofn is (m 1 + l)(m 2 + 1) (mk + 1). We shall see that this result will play an important
role in solving our problem.

Let A ={x E NIx is a divisor of 5400} and B ={x EN I x is a divisor of 18000} .

Clearly, our task is to evaluate lA uBI. To apply (17.2), we need to count
IAI, IBI and IAnBI.

Observe that

and

Thus, applying the result stated above, we have

IAI=(3 + 1)(3 + 1)(2 + 1)=48

and IBI=(4 + 1)(2 + 1)(3 + 1) =60.

What does A n B represent? By definition, A n B is the set of common positive

divisors of5400 and 18000, and so it is the set ofpositive divisors ofthe LCM of5400 and
18000. Since LCM {5400,18000}=LCM {2 3 3 3 52 , 2 4 .3 2 53 }=2 3 32 5 2 , it follows that
IAnBI=(3 + 1)(2 + 1)(2+1) = 36.

Hence, by n7.2), we have

lA u Bl =I AI+ IBI-IA n Bl
=48+60-36
=72. 0

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Problem 17.1. Find the number of integers from the set {300, 301, .. .,1000} which are
multiples of 6 or 9.

Problem 17.2. How many positive integers n are there such that n is a divisor of at least one
ofthe numbers 10 40 , 20 30 ? (Putnam, 1983).

18. An Extension
Formula (17.2) provides an expression for lA u Bl. We shall now apply it to derive an
expression for lA u B u q where A, B and C are any three finite sets.
Observe that

IAuBucj = IAu(BuC)I
= IAI+IBuC 1-IAn(BuC)I (by (17.2)}
= IAI+IBuCI-I(AnB)u(AnC)I
= IAI+IBI+IC l-IB nCI-(IA nBI+IA ncj-I(A nB)n(A nC)I) (by (17.2))
=IAI+IBI+IC 1-(IAnBI+IA nCI+IBnC I)+ lA nB nC 1-
That is,
I
1.---IA_u_B_u_C_I=-IA-1-+1-BI_+_IC-1---,-~A-n_B_I_+I_A_n_C_I+-IB_n_C~I)+-IA_n_B_n_C_I--(1-8.---.,1)

We shall now show an application of(18 .1).

Example 18.1. Figure 18.1 shows a 4 by 8 rectangular grid with two specified corners p and q
and three specified segments uv, wx and yz.

y z

w X

u v

Figure 18.1

Find in the grid

(i) the number of shortest p - q routes;

(ii) the number of shortest p - q routes which pass through wx;
(iii) the number of shortest p - q routes which pass through at least one of
the segments uv, wx and yz;

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 (iv) the number of shortest p - q routes which do not pass through any of
the segments uv, wx and yz.

(i) The problem of counting the number of shortest p - q routes in a rectangular grid was
discussed in Example 6.1 [3]. Employing the idea developed there, it can be shown that
the number of shortest p- q routes in the grid ofFigure 18.1 is given by

(4:8) =c:J.
(ii) As shown in Figure 18.2, a shortest p- q route passing through wx consists of a shortest
p - w route (in a 2 by 3 grid), the segment wx and a shortest x - q route (in a 2 by 4 grid).
Thus the number of shortest p - q routes passing through wx is given by

w X

Figure 18.2

(iii) The counting is more complicated in this case. We introduce three subsets of the set of
shortest p- q routes required.

Let A (resp., B and C) be the set of shortest p- q routes which pass through uv ( resp., wx
and yz). We note that the answer we are looking for is not IAI +IBI
+ lcJ as the sets A, Care B,
not pairwise disjoint. The desired answer should be lA u B u Cl, and this gives us a chance to
apply (18 .1). To apply (18 .1), we need to evaluate each term on the right-hand side of(18.1).

First, applying the idea shown in the solution of part (i), we have

IBI -- (25) (62) (as shown in part (ii)),

and ICI = (3; 4) c: 3) 4G}=

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 Next, let us compute !An B!, !An q and !B n q. Observe that An B is 'the set of
sh?rtest p- q routes passing through both uv and wx. Any such shortest p- q route consists
of a shortest p - u route, the route uvwx and a shortest x - q route. Thus

!AnBI =G} 1 (~) =3(~} Likewise, we obtain IBnCj =G)(:)= 4G} And each

route in An C consists of a shortest p - u route, the segment uv, a shortest v- y route, the

segment yz and a shortest z - q route, which gives !A n q= G) G) (:) = 36 .

Finally, we evaluate !A n B n q. Each route in An B n C is a p - q route consisting

a shortest p - u route, the route uvwxyz and a shortest z - q route. Thus

IAnBnCj =G)(:)= 12.

We are now in a position to evaluate !A u B u q. By ( 18.1),

= 349 .

(iv) Before solving part (iv), let us introduce a set notation. SupposeD is a subset of a finite
setS. We denote by S \D the set of elements inS but not in D ; i.e.,

S\D = {xESix~D}.

It follows immediately that

IS \DI = ISI - IDI .. (18.2)

Let's return to part (iv). LetS be the set of shortest p- q routes in the grid of Figure
18.1. Then the problem in (iv) is to evaluate IS\ (Au B u C)l , which is equal to

lSI-lA u B u q by (18.2). By (i), we have lSI = c:) and by (iii), !Au B u C! = 349 .

Thus the desired answer for (iv) is c:J- 349 = 495- 349 = 146. 0

Problem 18.1. A group of students took examinations in Chemistry, Mathematics and Physics.
Among them, 12 passed Chemistry, 15 Mathematics, and 10 Physics; 8 passed both Chemistry
and Mathematics, 5 both Chemistry and Physics, and 6 both Mathematics and Phystcs. There
were 20 students who passed at least one of the three subjects. Find the number of students
who passed all three subjects.

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 Problem 18.2. Find the number ofintegers from the set {1,2, .. .,1000} which are
(i) divisible by at least one of 2, 3 and 5;
(ii) divisible by none of 2, 3 and 5.

Problem 18.3. The following figure shows a 5 by 9 rectangular grid with two specified corners
p and q and three specified segments ab, cd and ej Find in the grid
(i) the number of shortest p - q routes;
(ii) the number of shortest p - q routes that pass through at least one of the
segments ab, cd and ej,
(iii) the number of shortest p- q routes that do not pass through any ofthe
segments ab, cd and ef

e f

a b c

19. An Equivalent Form

In the solution of Example 18.1 (iv), we evaluated IS\ (Au B u C)l using ( 18.1) and
(18 .2). In this section, we shall derive an explicit expression for IS\ (Au B u C)l and show an
application of this formula.

In what follows, let S be a finite set which is 'very large' in the sense that all the sets that
we shall consider in a problem are subsets of S. In mathematics, we call such a set S a
universal set. For instance, in Example 17.1, the universal set is { 1,2, .. .,1000}; in Example
17.2, the universal set is the set of natural numbers; and in Example 18.1, the universal set is the
set of shortest p - q routes in the grid of Figure 18.1.

Let A ~ S . We write A for S \A, and call A the complement of A . In the study of
sets, there are two very important laws relating the set operations 'union', 'intersection' and
'complementation' . They are called De Morgan's laws and are stated below.

For A, B ~ S, AuB =An B

(19.1)
AnB=AuB .

Let A, B, C be any three subsets of S. We shall see that the set S \(Au B u C) that we
considered in Example 18.1 (iv) can be expressed as An B n C. Indeed,

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 S\(AuBuC)= AuBuC
= (AuB)uC
=AuBnC (by (19.1))
=A nB nC . (by (19.1))

It follows that jAnBnCJ = JS\(AuBuC)J = JSJ-JAuBuCJ Thus, by(18 .1), we

obtain

We have just seen how (19.2) was derived from (18 .1). It is not difficult to see also that (18.1)
can be derived from (19.2). We say that these two identities are equivalent.

Now, let X= { 1,2, ... ,m} and Y = { 1,2, ...,n}, where m, n E r.J . Recall (see Problem
11.5 [4]) that a mapping f: X~ Y is a rule that assigns to each element x of X a unique
element f (x) in Y. In this case, we say that x is mapped to f (x) under f, and call f (x) the
image of x under f . A mapping f : X ~ Y is l-1 (or injective) if f (i) :;:. f (j) in Y
whenever i :t:. j in X. A mapping f : X~ Y is onto (or surjective) if for each bEY, there is
a EX such that f(a) =b. Clearly, ifthere is a I- I mapping f : X~ Y, then m $ n; and if
there is an onto mapping f: X~ Y, then m ~ n. Four mappings are shown in Figure 19.1.
Observe that / 1 is neither 1 - 1 nor onto; f 2 is 1 - I but not onto; / 3 is onto but not I - l;
and / 4 is both 1 - 1 and onto.

Figure 19.1

The problems of counting the number of mappings and the number of I - 1 mappings
were proposed in Problem 11 .5 [4]. Let us reconsider these problems here. Let X= {I ,2,3}
and Y = { I,2,3,4,5} . How many mappings are there from X to Y? There are three elements in
X, and each ofthem can be mapped to one of the five elements in Y. Thus the number of
mappings from X to Y is given 5 5 5 = 53 . How many I - I mappings are there from X to
Y? The element' 1' in X can be mapped to one of the five elements in Y (5 choices). The
element '2' in X can be mapped to one of the remaining four elements in Y (4 choices; excluding
the image of '1 '). Finally, the element '3' in X can be mapped to one of the remaining 3

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 elements in Y (3 choices; excluding the images of '1' and '2'). Thus the number of 1 - 1
mappings from X to Y is given by 5 4 3. Indeed, in general, we have:

Suppose X= {1,2, ...,m} andY= {1,2, ... ,n} .

Then (i) the number of mappings from

X toY is given by nm; (19.3)

and (ii) the number of 1 - 1 mappings from X to Y

is given by n(n-1)(n -m + 1) where m ::;n. (19.4)

What can be said about the number of onto mappings from X to Y? It is interesting to
note that this problem is not as straight forward as those of counting the numbers of mappings
and 1- 1 mappings. In the following example, we shall see how identity (19.2) is used to tackle
this more difficult problem.

Example 19.1. Let X= {1,2,3,4,5} andY= {1,2,3}. Find the number of onto mappings from
XtoY.

LetS be the set of mappings from X to Y. We shall now introduce three subsets A, B, C
of S as follows. Let A be the set of mappings from X to Y \ {1}, B the set of mappings from X to
Y\{2}, and Cthe set ofmappings from X to Y\{3} . What do the sets A, B and C
represent? Well, A is the set of mappings from X to Y which contain '1' in Y as an image,
B is the set of mappings from X to Y which contain '2' in Y as an image, and C is the set of
mappings from X to Y which contain '3' in Y as an image. It follows that A n B n C is the set
of mappings from X to Y which contain ' 1', '2' and '3' in Y as images; that is, Ar1 Br1 C is the
set of onto mappings from X to Y. Thus our task here is to evaluate lA n Bn Cl. We can
therefore apply (19.2)!

Since Sis the set of mappings from { 1,2,3,4,5} to { 1,2,3 }, by (19.3), lSI = 35 . Since A
is the set of mappings from { 1,2,3,4,5} to {2,3}, by (19.3) again, IAI = 2 5 . Likewise,
IBI =ICI = 2 5 . As A nB is the set of mappings from {1,2,3,4,5} to {3 }, by (19.3) again,
lA r1 Bi = 15 = 1. Similarly, lA n q = IBn q = 1. Finally, observe that An B r1 C is the set
ofmappingsfromXto Y\{1,2,3} (=<j>). Thus AnBnC = <j>.and so IAnBncj = 0.

Now, by (19.1), we have

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We have seen in the above how Addition Principle (17.1) can be generalized to (17.2), and in
turn (17 .2) can be extended to (18.1); and moreover, we have derived an equivalent form ( 19 .2)
of(18.1). In the next issue ofthe Medley we shall introduce a more general form of(PIE)
which deals with any n subsets, where n ~ 2, and we shall see how it can be applied to solve
some interesting problems.

Problem 19.1. Seven distinct objects are to be put into three distinct boxes. Find the number
of ways this can be done if
(i) there is no restriction;
(ii) no box is empty.

Problem 19.2. LetS be the set of3-digit numbers abc such that a, b, c E {1,2, .. .,9} and a, b,
care pairwise distinct. (Thus, 489E S, 571ES, but 313 ~ S and 507 ~ S .) Find the number
of members abc inS such that a~ 3, b ~ 5 and c ~ 7.

Problem 19.3. Find the number of integer solutions to the equation x + y + z = 12 where
O~x~4, O~y~5 and O~z~6. (SeeSection8of[3].)

Problem 19.4. A 5-digit ternary number is a number x 1x 2 x 3 x 4 x 5 where X; = 0, 1 or 2 for

each i = 1, 2, ... , 5. Thus, 00000, 01001, 21022, 11002, etc, are 5-digit ternary numbers. Find
the number of5-digit ternary numbers in which each of the '0', '1' and '2' appears at least once.

Problem 19.5. Two scouts x1 , x 2 from School X, three scouts y 1 , y 2 , y 3 from School Y
and four scouts z 1 , z 2 , z3 , z 4 from School Z get together in a meeting. In how many ways
can they be arranged in a row if not all scouts from the same school are allowed to form a single
block in the row? (For instance, x 1 z 3 z 2 y, y 3 x 2 y 2 z 1 z4 is allowed, but
z1 z4 y 1 y 2 X2 x 1 y 3 z 3 z 2 and z 4 z 3 x1 y 3 y 1 y 2 x 2 z 1 z 2 are not allowed.)

Answers
Problem 17.1. 156 Problem 17.2. 2301

Problem 181. 2 Problem 18.2. (i) 734 (ii) 266

Problem 18.3. (i) c:) = 2002

(ii) (~)(:)+G)(~)+ c:). 3_ (;)(~)_G). 2 . 3

_(;)G). 3 + (;) . 2. 3 = 1089

(iii) 913

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 Problem 19.1. (i) 37 (ii) 3 7 - 3 . 2 7 + 3 =1806

Problem 19.2. 9. 8. 7-3.8. 7 + 3. 7-1 = 356

Problem 19.3.
c;J- (~)- GJ- GJ + GJ + GJ = 10

Problem 19.4.

Problem 19.5. 9! -(2!8!+3!7!+4!6!) + (2!3!6!+2!4!5!+3!4!4!)- 2!3!4!3!

References
[1] K. M. Koh and B. P. Tan, Counting-Its Principles and Techniques (1), Mathematical
Medley Vol. 22 March (1995) 8- 13 .

[2] K. M. Koh and B . P. Tan, Counting-Its Principles and Techniques (2), Mathematical
Medley Vol. 22 September (1995) 47-51.

[3] K. M. Koh and B. P. Tan, Counting-Its Principles and Techniques (3), Mathematical
Medley Vol. 23 March (1996) 9- 14.

[4] K. M . Koh and B . P. Tan, Counting-Its Principles and Techniques (4), Mathematical
Medley Vol. 23 September ( 1996) 44 - 50.

[5] K. M . Koh and B . P. Tan, Counting-Its Principles and Techniques (5), Mathematical
Medley Vol. 24 March (1997) 8- 12.

[6] K. M. Koh and B. P . Tan, Counting-Its Principles and Techniques (6), Mathematical
Medley Vol. 24 September (1997) 46-48 .

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