the principle of inclusion and exclusionlyuu/dm/2014/20140501.pdfthe principle of inclusion and...

The Principle of Inclusion and Exclusion

c⃝2014 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 385

If I am I because you are you,

and if you are you because I am I,

then I am not I,

and you are not you.

— Hassidic rabbi


The Example on P. 26 Revisited

How many ways are there to arrange TALLAHASSEE with

no adjacent As?

• Rearrange the characters as AAAEEHLLSST.

• AAAEEHLLSST has 11 characters, among which there

are 3 As.

• There are11!

3! 2! 1! 2! 2! 1!= 831, 600

ways to arrange the 11 characters by Eq. (2) on p. 16.

• But some of them are invalid.

c⃝2014 Prof. Yuh-Dauh Lyuu, National Taiwan University 87

The Example on P. 26 Revisited (continued)

• First, treat AA as a single, new character.

• The 10-character string (AA)AEEHLLSST can be

arranged in

10!

1! 1! 2! 1! 2! 2! 1!= 453, 600

ways.

• But some arrangements of (AA)AEEHLLSST may

contain 3 consecutive As.


The Example on P. 26 Revisited (concluded)

• To count them, as before, treat AAA as a single, new

character.

• The 9-character string (AAA)EEHLLSST can be

arranged in9!

1! 2! 1! 2! 2! 1!= 45, 360

ways.

• The desired number is hence

831, 600− 453, 600 + 45, 360 = 423, 360,

matching the earlier result.


The Setup

• Let S be a set with |S| = N .

• Let c1, c2, . . . , ct be conditions on the elements of S.

• N(a b c · · · ) denotes the number of elements of S that

satisfy

a ∧ b ∧ c ∧ · · · .

• N(c1 c2 · · · ct) denotes the number of elements of S that

satisfy none of the conditions ci.


The Principle of Inclusion and Exclusion (1854, 1883)a

N(c1 c2 · · · ct) = N −∑

1≤i≤t

N(ci)

+∑

1≤i<j≤t

N(cicj)

−∑

1≤i<j<k≤t

N(cicjck) + · · ·

+(−1)tN(c1c2 · · · ct)

=t∑

k=0

(−1)k∑

1≤i1<i2<···<ik≤t

N(ci1ci2 · · · cik).

aDa Silva and James Joseph Sylvester (1814–1897).


The Proof

• If x ∈ S satisfies none of the conditions ci, then x should

contribute one to N(c1 c2 · · · ct).– Indeed, it is counted once, in N .

• If x ∈ S satisfies 1 ≤ r ≤ t of the conditions ci, then x

should contribute zero to N(c1 c2 · · · ct).– It is counted once in N , r times in

∑1≤i≤t N(ci),

(r2

)times in

∑1≤i<j≤t N(cicj), . . ., and

(rr

)times in

N(ci1ci2 · · · cir ).– By the binomial theorem (p. 53), the total is

1− r +

(r

2

)−(r

3

)+ · · ·+ (−1)r

(r

r

)= (1− 1)r = 0,

as desired.


The Proof (concluded)

• We have exhausted all cases for x.a

aContributed by Mr. Cheng-Chang Liu (B01902009) on April 18, 2013.


Simplification of the Notation

• Define

S0 = N,

S1 = N(c1) +N(c2) + · · ·+N(ct),

S2 = N(c1c2) +N(c1c3) + · · ·+N(ct−1ct),

...

Sk =∑

1≤i1<i2<···<ik≤t

N(ci1ci2 · · · cik).

– Observe that Sk has(tk

)terms.

• The principle of inclusion and exclusion becomes

N(c1 c2 · · · ct) = S0 − S1 + S2 − · · ·+ (−1)tSt. (43)


A Corollary

• By DeMorgan’s law,

c1 ∨ c2 ∨ · · · ∨ ct = ¬(¬c1 ∧ ¬c2 ∧ · · · ∧ ¬ct).

• Hence the number of elements in S that satisfy at least

one of the conditions ci equals

N(c1 ∨ c2 ∨ · · · ∨ ct) = N −N(c1 c2 · · · ct).


A Corollary (concluded)

• By the inclusion-exclusion principle (43) on p. 394,

N(c1 ∨ c2 ∨ · · · ∨ ct) =∑

1≤i≤t

N(ci)

−∑

1≤i<j≤t

N(cicj)

+∑

1≤i<j<k≤t

N(cicjck)

− · · ·+ (−1)t−1N(c1c2 · · · ct)

= S1 − S2 + · · ·+ (−1)t−1St.


Number of Onto Functions Revisited

• Let A = {a1, a2, . . . , am} and B = {b1, b2, . . . , bn}.

• Let ci be the condition that bi is not in the range of

f : A → B.

• N(c1 c2 · · · cn) is the number of onto f : A → B.

• N(ci1ci2 · · · cik) is the number of

f : A → B − { bi1 , bi2 , . . . , bik }.

• Clearly

N(ci1ci2 · · · cik) = (n− k)m

for distinct i1, i2, . . . , ik.


Number of Onto Functions Revisited (concluded)

• By the principle of inclusion and exclusion,

N(c1 c2 · · · cn) =

n∑k=0

(−1)k∑

1≤i1<i2<···<ik≤n

N(ci1ci2 · · · cik)

=n∑

k=0

(−1)k(n

k

)(n− k)m

= n!S(m,n), (44)

confirming again Eq. (31) on p. 270.


A Useful Combinatorial Identity

Lemma 62 For m ≤ r ≤ n,(n−m

r −m

)=

m∑k=0

(−1)k(m

k

)(n− k

r

).

• In how many ways can we pick r numbers out of

{1, 2, . . . , n} and 1, 2, . . . ,m must be picked?

• The number is clearly (n−m

r −m

).


The Proof (continued)

• Alternatively, let ci denote the condition that i is not

picked, 1 ≤ i ≤ m.

• Clearly, N(c1 c2 · · · cm) is our goal.

• Now,

N(ci) =

(n− 1

r

),

N(cicj) =

(n− 2

r

), i ̸= j,

...



• By the inclusion-exclusion principle (43) on p. 394, the

desired number equals

N(c1 c2 · · · cm)

=n∑

k=0

(−1)k∑

1≤i1<i2<···<ik≤m

N(ci1ci2 · · · cik)

=

m∑k=0

(−1)k(m

k

)(n− k

r

).


Euler’s Phi Function

• Let ϕ(n) denote the number of positive integers

m ∈ {1, 2, . . . , n} such that gcd(m,n) = 1, where n ≥ 2.

– ϕ(p) = p− 1 for prime p.

– ϕ(1) = 1 by convention.

• It is a computationally hard problem without the

knowledge of n’s factorization.

– Related to the security of some cryptographical

systems such as RSA.


Euler’s Phi Function: The Formula

Theorem 63 Let n = pe11 pe22 · · · pett be the prime

factorization of n. Then ϕ(n) = n∏t

i=1(1−1pi).

• Let ci be the condition that a number from {1, 2, . . . , n}is divisible by pi.

• The desired number is

ϕ(n) = N(c1 c2 · · · ct).

• For distinct i1, i2, . . . , ik,

N(ci1ci2 · · · cik) =n

pi1pi2 · · · pik.



• By the principle of inclusion and exclusion,

ϕ(n)

= N(c1 c2 · · · ct)

=

t∑k=0

(−1)k∑

1≤i1<···<ik≤t

n

pi1pi2 · · · pik

= n

t∏i=1

(1− 1

pi

). (45)


Euler’s Phi Function: An Alternative Proof

• Let n = pe11 pe22 · · · pett be the prime factorization of n.

• Start with {1, 2, . . . , n}.

• The prime p1 will kill off every p1th number from the set.

• The prime p2 will kill off every p2th number from the

remaining set, etc.

• In the end, we end up with a set of size

n

(1− 1

p1

)(1− 1

p2

)· · ·

(1− 1

pt

)= ϕ(n).


An Example

• Suppose n = pe11 pe22 pe33 .

• Then

ϕ(n)

= n

(1− 1

p1

)(1− 1

p2

)(1− 1

p3

)= n

[1−

(1

p1+

1

p2+

1

p3

)+

(1

p1p2+

1

p1p3+

1

p2p3

)− 1

p1p2p3

].

• This may help convince you that Eq. (45) on 404 is

correct.


Application: ϕ(2n)

ϕ(2n)

= 2n∏

p | 2n,p prime

(1− 1

p

)

= 2n(1− 1

2

)= 2n−1.

Indeed, the only numbers in {1, 2, . . . , 2n} relatively prime

with 2 are the 2n/2 = 2n−1 odd numbers.


�� Q

��

��

��

��

��I+Q/

HXOHUSKL�QE �


Euler’s Phi Function Is Multiplicative

• Let n = m1m2, where gcd(m1,m2) = 1.

• Let m1 = pe11 pe22 · · · pess be the prime factorization of m1.

• Let m2 = pes+1

s+1 pes+2

s+2 · · · pett be the prime factorization of

m2.

• From the formula on p. 403,

ϕ(m1m2) = ϕ(n)

= n

t∏i=1

(1− 1

pi)

= m1

s∏i=1

(1− 1

pi) m2

t∏i=s+1

(1− 1

pi)

= ϕ(m1)ϕ(m2).


A Loose Lower Bound for the Phi Functiona

Theorem 64 (Hardy and Wright (1979))

ϕ(n) > n/(6 ln lnn) for n > 3.

aGodfrey Harold Hardy (1877–1947) and Edward Maitland Wright

(1906–2005).


Godfrey Harold Hardy (1877–1947)


Permutations without Fixed Points

• Write a permutation f on {1, 2, . . . , n} as 1 2 · · · n

f(1) f(2) · · · f(n)

• There are n! permutations.

• Permutation f has a fixed point at i if f(i) = i.

– i is invariant under f .


Number of Permutations without Fixed Points

What is the number of permutations without fixed points?

• Let SX be the number of permutations that fix all i ∈ X.

• By the principle of inclusion and exclusion, the desired

number isa ∑X⊆{1,2,...,n}

(−1)|X|SX .

• SX = (n− |X|)! as those numbers not in X form a

permutation.

aLet ci denote the condition that i is a fixed

point. Then the desired number is N(c1 c2 · · · cn) =∑nk=0(−1)k

∑1≤i1<i2<···<ik≤n N(ci1ci2 · · · cik ) =∑n

k=0(−1)k∑

{ i1,i2,...,ik } S{ i1,i2,...,ik } =∑

X⊆{1,2,...,n}(−1)|X|SX .



• The desired number is∑X⊆{1,2,...,n}

(−1)|X|(n− |X|)! =

n∑k=0

(−1)k(n

k

)(n− k)!

= n!

n∑k=0

(−1)k

k!(46)

≈ n!

e,

where e = 2.71828 . . . .

• A constant fraction of permutations have no fixed points!

• Or, if one picks a random permutation, with roughly

40% chance, that permutation will have no fixed points!


Derangements (Also p. 412)

• A derangement is a permutation of 1, 2, . . . , n in which

1 is not in the first place, 2 is not in the second place,

etc.a

• How many derangements of 1, 2, . . . , n are there?

• Let ci denote the condition that i is in the ith place.

• The desired number is N(c1 c2 · · · cn), which equals

dn ≡n∑

i=0

(−1)i(n

i

)(n− i)! = n!

n∑i=0

(−1)i1

i!≈ n!

e(47)

by the principle of inclusion and exclusion (43) on p. 394.

aJust a permutation without no fixed points!


A Combinatorial Identity for dn

• Let dk denote the number of derangements of 1, 2, . . . , k.

• By convention, d0 = 1.

• Any permutation of 1, 2, . . . , n can have n− k fixed

points for some k, with the rest being deranged.

• There are(

nn−k

)=

(nk

)choices for the fixed points.

• Hence

n! =n∑

k=0

(n

k

)dk. (48)

• Alternatively,

1 =

n∑k=0

dkk! (n− k)!

.



• With the help of Eq. (47) on p. 415, the following

complex identity is obtained:

n!

=n∑

k=0

(n

k

)k!

k∑i=0

(−1)i1

i!

= n!

n∑k=0

1

(n− k)!

k∑i=0

(−1)i1

i!.


An Example

One can numerically verify identity (48) on p. 416 with the

following data:

d0 = 1, d1 = 0,

d2 = 1, d3 = 2,

d4 = 9, d5 = 44,

d6 = 265,

d7 = 1854,

d8 = 14833,

d9 = 133496,

d10 = 1334961.


A Variation on Derangement

• How many permutations of 1, 2, . . . , n are there such

that i is not in the (i− 1)st place for 2 ≤ i ≤ n?

– For example, 12345 (but not 23451).

• Let ci denote the condition that i is in the (i−1)st place.

• Now N(ci) = (n− 1)!, N(cicj) = (n− 2)! with i ̸= j, etc.

• The desired number N(c2 c3 · · · cn) equals

n!−(n− 1

1

)(n− 1)! +

(n− 1

2

)(n− 2)!− · · ·

by the principle of inclusion and exclusion.


A Variation on Derangement (continued)

n−1∑i=0

(−1)i(n− 1

i

)(n− i)!

=

n−1∑i=0

(−1)i(n− 1)! (n− i)

i!

= n! +

n−1∑i=1

(−1)i(n− 1)! (n− i)

i!

= n! +

n−1∑i=1

(−1)in!

i!−

n−1∑i=1

(−1)i(n− 1)!

(i− 1)!

=

n−1∑i=0

(−1)in!

i!−

n−1∑i=1

(−1)i(n− 1)!

(i− 1)!


A Variation on Derangement (concluded)

=

n−1∑i=0

(−1)in!

i!+

n−2∑i=0

(−1)i(n− 1)!

i!

=

n−1∑i=0

(−1)in!

i!+

n−2∑i=0

(−1)i(n− 1)!

i!

+

[(−1)n

n!

n!+ (−1)n−1 (n− 1)!

(n− 1)!

]=

n∑i=0

(−1)in!

i!+

n−1∑i=0

(−1)i(n− 1)!

i!

= dn + dn−1 (49)

from Eq. (47) on p. 415.


A Simpler Proof

• Again, how many permutations of 1, 2, . . . , n are there

such that i is not in the (i− 1)st place for 2 ≤ i ≤ n?

• Consider a permutation of 1, 2, . . . , n, where

1. i is not in the (i− 1)st place for 2 ≤ i ≤ n.

2. 1 is not in the nth place.

• There are dn of such permutations as they are but

derangements with the location restrictions shifted.

• The 2nd condition that 1 is not in the nth place is extra.

• So we need to add to dn the number of permutations

that satisfy condition 1 but not condition 2.


A Simpler Proof (concluded)

• So consider permutations of 1, 2, . . . , n such that

1. i is not in the (i− 1)st place for 2 ≤ i ≤ n.

2. 1 is in the nth place.

• Remove 1 and rename i as i− 1 for 2 ≤ i ≤ n.

• The results are permutations of 1, 2, . . . , n− 1 such that

i is not in the ith place for 1 ≤ i ≤ n− 1.

• They are simply derangements of 1, 2, . . . , n− 1.

• Their count is dn−1, as desired.


Another Variation on Derangement

• Let A ⊆ { 1, 2, . . . , n }.

• How many permutations of 1, 2, . . . , n induce a

derangement of A?

– The original derangement has A = { 1, 2, . . . , n }.

• Assume A = { 1, 2, . . . ,m }, where m ≤ n, without loss

of generality.

• Let ci denote the condition that i is in the ith place.

• The desired number is N(c1 c2 · · · cm), which equals

m∑i=0

(−1)i(m

i

)(n− i)!

by the principle of inclusion and exclusion.


Integer Solutions of a Linear Equation with Upper Bounds

Theorem 65 The number of integer solutions of

x1 + x2 + · · ·+ xn = r, where 0 ≤ x1, x2, . . . , xn < b, is

⌊r/b⌋∑m=0

(−1)m(n

m

)(n+ r −mb− 1

r −mb

). (50)

• The number of integer solutions of x1+x2+ · · ·+xn = r,

where 0 ≤ x1, x2, . . . , xn, is N ≡(n+r−1

r

)(recall p. 82).

• Now impose upper bounds

0 ≤ x1, x2, . . . , xn < b.



• Let ci denote the condition that b ≤ xi.

• N(ci): The number of solutions satisfying ci equals(n+ r − b− 1

r − b

)as we are solving x1 + x2 + · · ·+ xn = r − b for

nonnegative integer solutions.

• N(cicj): The number of solutions satisfying ci ∧ cj with

i ̸= j equals (n+ r − 2b− 1

r − 2b

)as we are solving x1 + x2 + · · ·+ xn = r − 2b.

c⃝2014 Prof. Yuh-Dauh Lyuu, National Taiwan University 6


• And so on.

• Of course, there cannot be more than ⌊r/b⌋ conditions

ci.

• Our goal is

N(c1 c2 · · · c⌊r/b⌋).

• Recall the inclusion-exclusion principle (43) on p. 394:a

N(c1 c2 · · · c⌊r/b⌋) =⌊r/b⌋∑m=0

(−1)mSm.

aThe definition of Sm appeared on p. 394.



• The number of N(ci) terms in S1 is(n1

).

• The number of N(cicj) terms in S2 is(n2

).

• And so on.

• Finally,

N(c1 c2 · · · c⌊r/b⌋) =

⌊r/b⌋∑m=0

(−1)mSm

=

⌊r/b⌋∑m=0

(−1)m(n

m

)(n+ r −mb− 1

r −mb

).


An Example

• What is the number of positive integers x, where

x ≤ 999, whose sum of the digits equals 20?

– E.g., 389 and 776.

• Use xi to denote x’s ith digit.

• Now the problem is equivalent to x1 + x2 + x3 = 20,

where 0 ≤ xi < 10.

• Equation (50) on p. 425 says the answer is(30

)(3+20−1

20

)−

(31

)(3+20−10−1

20−10

)+(32

)(3+20−20−1

20−20

)= 36.


Generalized Principle of Inclusion and Exclusion Em

• Let Em denote the number of elements in S that satisfy

exactly m of the t conditions.

– The principle of inclusion and exclusion corresponds

to E0.


Generalized Principle of Inclusion and Exclusion Em

(concluded)

• Thena

Em = Sm −(m+ 1

1

)Sm+1 +

(m+ 2

2

)Sm+2

− · · ·+ (−1)t−m

(t

t−m

)St

=t∑

k=m

(−1)k−m

(k

k −m

)Sk

=t∑

k=m

(−1)k−m

(k

m

)Sk. (51)

aThe definition of Sk appeared on p. 394.


The Proof

• If x ∈ S satisfies fewer than m conditions, then x should

contribute zero to Em.

– Indeed, it contributes zero to

Sm, Sm+1, . . . , St.

• If x ∈ S satisfies exactly m conditions, then x should

contribute one to Em.

– Indeed, it contributes one to Sm and zero to

Sm+1, Sm+2, . . . , St.



• If x ∈ S satisfies m < r ≤ t of the conditions ci, then x

should contribute zero to Em.

• Indeed, it is counted(rm

)times in Sm,

(r

m+1

)times in

Sm+1, . . .,(rr

)times in Sr, and zero times for all terms

beyond Sr.

• The total count is

r∑k=m

(−1)k−m

(k

m

)(r

k

).



By Newton’s identity (p. 29),

r∑k=m

(−1)k−m

(k

m

)(r

k

)=

r∑k=m

(−1)k−m

(r

m

)(r −m

k −m

)

=r−m∑k=0

(−1)k(r

m

)(r −m

k

)

=

(r

m

) r−m∑k=0

(−1)k(r −m

k

)=

(r

m

)(1− 1)r−m = 0.


Permutations with m Fixed Points

• Recall from p. 412 that a bijective function f on

{1, 2, . . . , n} has a fixed point at i if f(i) = i.

• What is the number of permutations with m fixed

points?

• Let ci denote the condition that i is a fixed point.

• Then

Sk =

(n

k

)(n− k)! =

n!

k!. (52)



• From Eq. (51) on p. 431,

Em =n∑

k=m

(−1)k−m

(k

k −m

)Sk

=n∑

k=m

(−1)k−m

(k

k −m

)n!

k!

=n!

m!

n∑k=m

(−1)k−m 1

(k −m)!.

• For example, En−2 = n(n− 1)/2 (permutations with

n− 2 fixed points).


Generalized Principle of Inclusion and Exclusion Lm

• Let Lm denote the number of elements in S that satisfy

at least m of the t conditions.

• Then

Lm = Sm −(

m

m− 1

)Sm+1 +

(m+ 1

m− 1

)Sm+2

− · · ·+ (−1)t−m

(t− 1

m− 1

)St

=t∑

k=m

(−1)k−m

(k − 1

m− 1

)Sk. (53)


The Proof

• By definition,

Lm − Lm+1 = Em

for m < t by definition.

• Now we prove the identity by induction on m.

• First note that Et = Lt = St.

• Inductively, assume that

Lm+1 =

t∑k=m+1

(−1)k−(m+1)

(k − 1

m

)Sk.

• Also Em =∑t

k=m(−1)k−m(km

)Sk from (51) on p. 431.



• Finally, Lm equals

Lm+1 + Em

=

t∑k=m+1

(−1)k−(m+1)

(k − 1

m

)Sk +

t∑k=m

(−1)k−m

(k

m

)Sk

= Sm +

t∑k=m+1

(−1)k−m

[−

(k − 1

m

)+

(k

m

)]Sk

= Sm +

t∑k=m+1

(−1)k−m

(k − 1

m− 1

)Sk by Lemma 2 on p. 28

=t∑

k=m

(−1)k−m

(k − 1

m− 1

)Sk.


Permutations with Fixed Points

• Consider permutations on {1, 2, . . . , n}.

• Let ci stand for the condition that i is a fixed point.

• From Eq. (53) on p. 437, the number of permutations

with at least one fixed point is

L1 =t∑

k=1

(−1)k−1Sk.



• Recall that Eq. (52) on p. 435 says

Sk =n!

k!.

• Hencea

L1 =

n∑k=1

(−1)k−1Sk

= n!n∑

k=1

(−1)k−1 1

k!(54)

≈ n!

(1− 1

e

).

aAn alternative proof is via Eq. (46) on p. 414.


Checking for Consistency

• The sum of the number of permutations without fixed

points (E0) and those with fixed points (L1) should be

n!.

• Indeed, from Eq. (46) on p. 414 for E0 and Eq. (54) on

p. 441 for L1,

n!n∑

k=0

(−1)k

k!+ n!

n∑k=1

(−1)k−1

k!= n!.

• Note that E0 is dn, the number of derangements.


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