Principle of Inclusion and Exclusion

Principle of

Inclusion and Exclusion

•••• An Illustrative Example

Determine the number of integers n, 1≤≤≤≤n≤≤≤≤100, which are not divisible by 2, 3, 5.

Let S={1, 2, …, 100} and N=|S|=100.

Define three conditions as follows: c₁ : divisible by 2;

c₂ : divisible by 3;

c₃ : divisible by 5.

The answer is denoted by N(c₁c₂c ), which is ₃ evaluated below.

N(c₁) = _└100/2_┘ = 50; N(c₂) = _└100/3_┘ = 33;

N(c₃) = _└100/5_┘ = 20.

N(c₁c₂) = _└100/6_┘ = 16; N(c₁c₃) = _└100/10_┘ = 10;

N(c₂c₃) = _└100/15_┘ = 6.

N(c₁c₂c₃) = _└100/30_┘ = 3.

N(c₁c₂c ) = N −₃ −−− (N(c₁)++++N(c₂)++++N(c₃)) ++++ (N(c₁c₂)++++N(c₁c₃)++++N(c₂c₃)) −−−− N(c₁c₂c₃)

= 100 −−−− (50++++33++++20) ++++ (16++++10++++6) −−−− 3

= 26.

•••• The Principle

S : a set; N = |S|

c₁, c₂, …, c_t : conditions

N(c_i) : the number of elements in S that satisfy c_i

N(c_ic_j) : the number of elements in S that satisfy c_i and c_j (and perhaps some others)

N(c_i) : the number of elements in S that do not satisfy

c_i (N(c_i) = N−−−−N(c_i))

N(c_i c_j) : the number of elements in S that do not satisfy either of c_i and c_j

Theorem. The number of elements in S that satisfy none of c₁, c₂, …, c_t is equal to

N(c₁c₂…c_t)

=

− − − −

∑ ∑

∑ ∑

≤≤

≤≤

≤≤≤

≤i t 1

N(c_i)

+

∑ ∑ ∑

∑

≤

≤

≤

≤

<

<<

<

≤

≤≤

≤i j t 1

N(c_ic_j)

− − − −

∑ ∑

∑ ∑

≤

≤≤

≤

<

<<

<

<

<

<

<

≤

≤

≤

≤i j k t 1

N(c_ic_jc_k)

+

•••• •••• ••••

+

(−−−−1)^tN(c₁c₂…c_t).

When t=2, c₁ →→→→ A and c₂ →→→→ B.

N(c₁ c ) = ₂ |S| −−−− |A∪∪∪B| ∪

= |S| −−−− (|A| + |B| −−−− |A∩∩∩B|) ∩

= |S| −−−− (|A| + |B|) + |A∩∩∩∩B|

When t=3, c₁ →→→→ A, c₂ →→→→ B, and c₃ →→→ C. →

N(c₁c₂c ) = ₃ |S| −−− |A∪− ∪∪∪B∪∪∪∪C|

= |S| −−− (|A| + |B| + |C| −− −− |A∩− ∩∩∩B| −−− |A∩− ∩∩∩C| −−−−

|B∩∩∩∩C| + |A∩∩∩B∩∩ ∩∩C|) ∩

= |S| −−− (|A| + |B| + |C|) +−

(|A∩∩∩∩B| + |A∩∩∩C| + |B∩∩ ∩∩∩C|) −−−−

|A∩∩∩∩B∩∩∩∩C|

This theorem can be proved by induction on t.

Here we prove it by a combinatorial method.

Observation:

|A∪∪∪∪B| = |A| + |B| −−−− |A∩∩∩B| ∩ a 1 = 1 + 0 −−−− 0 b 1 = 1 + 0 −−−− 0 c 1 = 1 + 1 −−−− 1 d 1 = 0 + 1 −−−− 0 e 0 = 0 + 0 −−−− 0 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −

4 = 3 + 2 −−−− 1

Proof : For each x ∈∈∈∈ S, we show that x contributes the same count to each side of the equation.

N(c₁ c₂…c_t)

=

− − − − ^{∑ ∑ ∑ ∑}

≤≤

≤≤

≤≤

≤≤i t 1

N(c_i)

+ ^{∑ ∑ ∑ ∑}

≤

≤≤

≤

<

<<

<

≤

≤≤

≤i j t 1

N(c_ic_j)

− − − −

∑

∑

∑

∑

≤

≤

≤

≤

<

<

<

<

<

<<

<

≤

≤

≤

≤i j k t 1

N(c_ic_jc_k)

+

+

(−−−−1)^tN(c₁c₂…c_t).

Case 1. x satisfies none of the conditions.

x is counted once in N(c₁c₂…c_t) and N, but not in the other terms.

Case 2. x satisfies r of the conditions.

(1) x contributes nothing to N(c₁c₂…c_t ) (2) x is counted once in N.

(3) x is counted (

1

r) times in

∑ ∑ ∑ ∑

≤

≤

≤

≤

≤

≤≤

≤i t 1

N(c_i).

(4) x is counted (

2

r) times in

∑ ∑ ∑ ∑

≤

≤

≤

≤

<

<

<

<

≤

≤≤

≤i j t 1

N(c_ic_j).

•

•

•

x is counted (

r

r) times in

∑ ∑ ∑ ∑

i1

c c …i2

ir

c ).

⇒

⇒⇒

⇒ left-hand side : 0.

right-hand side : 1 −−−− (

1 r) ++++ (

2

r) −−−− … ++++ (−−−−1)^r(

r r)

= (1++++(−−−−1))^r = 0.

Corollary. The number of elements in S that satisfy at least one of the conditions is N(c₁ or c₂ or … or c_t) = N −−−− N(c₁c₂…c_t).

Let S₀ = N.

S₁ =

∑ ∑ ∑ ∑

≤

≤

≤

≤

≤

≤≤

≤i t 1

N(c_i).

S₂ =

∑ ∑ ∑ ∑

≤

≤≤

≤

<

<<

<

≤

≤

≤

≤i j t 1

N(c_ic_j).

•

•

•

S_k =

∑ ∑ ∑ ∑

i1

c c …i2

ik

c ).

Theorem. The number of elements in S that satisfy exactly m of c₁, c₂, …, c_t is

E_m = S_m −−−− (

1 +1 + + +

m )S_m+1 + (

2 +2 + + +

m )S_m+2 −−−− … +

(−−−−1)^{t−−−m−} (

m t

t

−

−

−

− )S_t.

Proof. For each x∈∈∈∈S, we show that x contributes the same count to each side of the equation.

Case 1. x satisfies fewer than m conditions.

x is not counted on either side of the equation.

Case 2. x satisfies exactly m of the conditions.

x is counted once in E_m and once in S_m, but not in S_m+1, S_m+2, …, S_t.

Case 3. x satisfies r (m<<<<r≤≤≤≤t) of the conditions.

x is counted (

m

r ) times in S_m, (

+1 + + + m

r ) times in

S_m+1, …, (

r

r) times in S_r.

But, x contributes nothing to the other terms.

⇒

⇒⇒

⇒ left-hand side : 0.

right-hand side :

(

m

r ) −−− −

(

1 +1 + ++ m

)

+1 ++ + m

r ) +

(

2 +2 + ++ m

)

+2 + + + m

r ) −−−− … +

(−−−−1)^{r−−−−m}(

m r

r

−

−

−

− )(

r r)

=

. .

.

= 0.

Theorem. The number of elements in S that satisfy at least m of the t conditions is

L_m = S_m −−−−(

−1

−

−

− m

m )S_m+1 + (

1 1

−

−

−

− + + + + m

m )S_m+2 −−−− … +

(

)

1 1

−

−−

−

−

−−

− m

t )S_t.

An inductive proof of this theorem was outlined as problem 8 on page 401 of Grimaldi’s book.

For your reference, the following shows an alternative proof that is based on a combinatorial method.

For each x∈S, we show that x contributes the same count to each side of the equation.

Case 1. x satisfies fewer than m conditions.

x is not counted on either side of the equation.

Case 2. x satisfies r (m≤r≤t) of the t conditions.

x is counted once in L_m, ^r

m

  

  times in S_m,

1 r m+

 

 

  times in S_m+1, …,

r r

  

  times in S_r. But, x contributes nothing to any of the other terms in the equation.

Left-hand side: 1.

Right-hand side:

r m

  

  −

1 1

m r

m− m+

   

   

    + ¹

1 2

m r

m m

+

− +

   

   

    −…+(−1)^{r−m 1}

1

r r

m r

−

−

   

   

   

= ^{( 1)}

0 1

r m k

m k m

k r m

− +

∑ + −

≤ ≤ −

   

   

   (−1)^k

= ^{( 1)}

0

r m k

m k

k r m k

− +

∑ +

≤ ≤ −

   

   

   (−1)^k (assume m−1+k≥0)

= 0

r m k k r m

m

∑ k

≤ ≤ − +

  − 

   

    (refer to Concrete Mathematics, 2^nd edition, by Graham, Knuth, and Patashnik, pp. 164, Eq. (5.14))

= ^r

m k k

m

∑ k

+

  − 

   

    (

k

−m

 

 

  =0 as k<0, and

m k

r

+

 

 

  =0 as k>r−m)

= r m

r m

 − 

 

 −  (refer to Concrete Mathematics, 2^nd edition, by Graham, Knuth, and Patashnik, pp. 169, Eq. (5.23))

= 1.

Since m≥0 and k≥0, we have m=0 and k=0 if m−1+k<0.

When m=0 and k=0, the right-hand side is 0

 r

   =1.

Ex. Compute the number of integer solutions to

x₁ + x₂ + x₃ + x₄ = 18, where 0≤≤≤≤x_i ≤≤≤≤7 for 1≤≤≤≤i≤≤≤≤4.

Let S be the set of integer solutions to

x₁ + x₂ + x₃ + x₄ = 18, where x_i ≥≥≥≥0 for 1≤≤≤≤i≤≤≤≤4.

Also, let c_i denote the constraint of x_i ≥≥≥≥8.

Then, N(c₁c₂c₃c₄) is the answer.

N = H(4, 18) = C(4+18−−−−1, 18) = C(21, 18).

N(c_i) = H(4, 10) = C(13, 10).

N(c_ic_j) = H(4, 2) = C(5, 2).

N(c_ic_jc_k) = 0.

N(c₁c₂c₃c₄) = 0.

N(c₁c₂c₃c₄) = C(21, 18) −−−− C(4, 1)××××C(13, 10) + C(4, 2)××××C(5, 2) −−−− 0 + 0

= 246.

Ex. Compute the number of ways to permute a, b, c, …, y, z so that none of car, dog, pun, and byte occurs.

Let S be the set of all permutations of the 26 letters.

Let c₁, c₂, c₃ and c₄ denotethe conditions that the permutations contain car, dog, pun, and byte, respectively.

Then, N(c₁c₂c₃c₄) is the answer.

N = 26!.

N(c₁) = N(c₂) = N(c₃) = 24!; N(c₄) = 23!.

N(c₁c₂) = N(c₁c₃) = N(c₂c₃) = 22!; N(c_ic₄) = 21!.

N(c₁c₂c₃) = 20!; N(c_ic_jc₄) = 19!.

N(c₁c₂c₃c₄) = 17!.

N(c₁c₂c₃c₄) = 26! −−− (3− ××××24!+23!) + (3××××22!+ 3××××21!) −−− (20!− +3××××19!) + 17!.

Ex. Given a positive integer n≥≥≥≥2, the Euler’s phi function, denoted by φφφφ(n), is the number of integers m so that

1≤≤≤≤m≤≤≤≤n and gcd(m, n)=1 (m, n are relatively prime).

For example, φφφφ(2)=1, φφφφ(3)=2, φφφφ(4)=2, and φφφφ(5)=4.

Consider n = 1260 = 2²××××3²××××5××××7 (質因數分解質因數分解質因數分解質因數分解).

Let S = {1, 2, …, 1260}.

φ φφ

φ(1260) can be computed as follows.

Let c₁, c₂, c₃, c₄ denotethe conditions that the integers m are divisible by 2, 3, 5, 7, respectively.

Then, φφφφ(1260) = N(c₁c₂ c₃c₄).

N = 1260.

N(c₁) = 1260/2 = 630; N(c₂) = 1260/3 = 420;

N(c₃) = 1260/5 = 252; N(c₄) = 1260/7 = 180.

N(c₁c₂) = 1260/6 = 210; N(c₁c₃) = 1260/10 = 126;

N(c₁c₄) = 1260/14 = 90; N(c₂c₃) = 1260/15 = 84;

N(c₂c₄) = 1260/21 = 60; N(c₃c₄) = 1260/35 = 36.

N(c₁c₂c₃) = 1260/30 = 42; N(c₁c₂c₄) = 1260/42 = 30;

N(c₁c₃c₄) = 1260/70 = 18; N(c₂c₃c₄) = 1260/105 = 12.

N(c₁c₂c₃c₄) = 1260/210 = 6.

N(c₁c₂c₃c₄) = 1260 −−− (630− +420+252+180) + (210+126+90+84+60+36) −−− − (42+30+18+12) + 6

= 288.

In general, suppose n = p₁^e1 p₂^e2… p_t^et (質因數分解質因數分解質因數分解質因數分解).

Then, φφφφ^{(n) =} n××××(1−−−−

1

1

p )××××(1−−−−

2

1

p )×××× … ××××(1−−−− ¹ pt ).

For example, when n = 1260, φ

φφ

φ^{(1260) = 1260}××××(1−−−− ¹

2)××××(1−−−− ¹

3)××××(1−−−− ¹

5)××××(1−−−− ¹ 7)

= 288.

Another description of Euler’s formula:

Suppose that n is a positive integer and p₁, p₂, …, p_r are distinct prime numbers, where n>p_i for all 1≤≤≤≤i≤≤≤≤r.

Let F(n) = |{m| 1≤≤≤≤m≤≤≤≤n is an integer and gcd(m, p_i)=1 for all 1≤≤≤≤i≤≤≤≤r}|.

Then, F(n) is equal to f(n) = n××××(1−−−−

1

1

p )××××(1−−−−

2

1

p )×××× … ××××(1−−−− ¹ pr ), if f(n) is an integer.

Ex. Suppose that n=23, p₁ =2, and p₂ =3 (i.e., r=2).

There are 8 integers in {1, 2, …, 23} that are relatively prime to both 2 and 3. They are 1, 5, 7, 11, 13, 17, 19, and 23.

f(n) = n××××(1−−−− ¹

2)××××(1−−−− ¹ 3) =

3 n. f(23) is not an integer

We first compute f(21)=7 (or f(24)=8), and then obtain F(23)=F(21)+1=f(21)+1=8 (or F(23)= F(24)=f(24)=8), where 1 indicates the integer 23.

Ex. Six married couples are to be seated at a circular table. In how many ways can they be arranged so that no wife sits next to her husband?

Let c_i denotethe condition that couple i are neighboring, where 1≤≤≤≤i≤≤≤≤6.

The answer is N(c₁c₂…c₆).

N = ^12!

12 = 11!.

N(c_i) = ^11!

11 ××××2 = 2××××10!.

N(c_ic_j) = ^10!

10 ××××2² = 2²××××9!.

Similarly, N(c_ic_jc_k) = 2³××××8!, N(c_ic_jc_kc_l) = 2⁴××××7!, N(c_ic_jc_kc_lc_r) = 2⁵ ××××6!, N(c_ic_jc_kc_lc_rc_s) = 2⁶ ××××5!.

N(c₁c₂…c₆) = 11! −−− C(6, 1)− ××××2××××10! + C(6, 2)××××2²××××9!

−−

−− C(6, 3)××××2³××××8! + C(6, 4)××××2⁴××××7!

−−

−− C(6, 5)××××2⁵ ××××6! + C(6, 6)××××2⁶××××5!

= 12,771,840.

Ex. How many ways are there to connect five vertices a, b, c, d and e so that none of them is isolated?

(O) (O) (X) (X)

Let c₁, c₂, c₃, c₄ and c₅ denotethe conditions that vertices a, b, c, d and e are isolated, respectively.

The answer is N(c₁c₂…c₅).

N = 2¹⁰ (at most 10 edges among five vertices).

N(c_i) = 2⁶. N(c_ic_j) = 2³. N(c_ic_jc_k) = 2¹.

N(c_ic_jc_kc_l) = N(c_ic_jc_kc_lc_r) = 2⁰.

N(c₁c₂…c₅) = 2¹⁰ −−− C(5, 1)− ××××2⁶ + C(5, 2)××××2³

−−

−− C(5, 3)××××2¹ + C(5, 4)××××2⁰

−−

−− C(5, 5)××××2⁰

= 768.

Ex. For the example above, how many ways are there to connect five vertices a, b, c, d and e so that exactly two of them are isolated?

The answer is E₂ = S₂ −−− C(3, 1)− ××××S₃ + C(4, 2)××××S₄

−−

−− C(5, 3)××××S₅

= C(5, 2)××××2³ −−−− C(3, 1)××××C(5, 3)××××2¹ + C(4, 2)××××C(5, 4)××××2⁰

−−

−− C(5, 3)××××C(5, 5)××××2⁰

= 80 −−−− 60 + 30 −−−− 10

= 40.

•••• Derangements

An arrangement of 1, 2, …, n is called a derangement, if 1 is not at the first place (its natural position), 2 is not at the second place (its natural position), …, and n is not at the nth place (its natural position).

Consider n=10, and let c_i denote the condition that integer i is at the ith place, where 1≤≤≤≤i≤≤≤≤10.

The number of derangements of 1, 2, …, 10 is N(c₁c …₂ c ) =₁₀ 10! −−−− (

1

10)9! +++ (+

2

10)8! −−−− … ++++(

10

10)0!

= 1334960.

Ex. There are seven books undergoing a two-round review process of seven reviewers. Each book is reviewed by two distinct reviewers. In how many ways can these books be reviewed?

There are 7! ways for the first-round review.

There are d₇ ways for the second-round review, where d₇ is the number of derangements of 1, 2,

…, 7.

The answer is 7!××××d₇.

×

× ×

×

•••• Rook Polynomials

Problem : Given a chessboard C of arbitrary shape and size, determine the number r_k of ways of placing k nontaking rooks on C.

… …

… ××××

… … …

…

C :

r₁ = 6, r₂ = 8, r₃ = 2 r_k = 0 for k≥≥≥≥4

Let r₀ = 1.

rook polynomial R(C, x) =

∑ ∑ ∑ ∑

=

==

=0 i

r_ixⁱ

= 1+6x+8x²+2x³

C₂ : C₁ :

C :

R(C, x) = 1+11x+40x²+56x³+28x⁴+4x⁵ R(C₁, x) = 1+4x+2x²

R(C₂, x) = 1+7x+10x²+2x³

C₁ and C₂ are “disjoint” (no square in the same row or column)

⇒⇒

⇒⇒ R(C, x) = R(C₁, x)

⋅⋅⋅⋅

Count r₃ in C as follows.

Case 1. All three rooks are placed on C₂ : 2 ways Case 2. Two rooks are placed on C₂ and one rook is

placed on C₁ : 10××××4 = 40 ways

Case 3. One rook is placed on C₂ and two rooks are placed on C₁ : 7××××2 = 14 ways

r₃ = 2+40+14 = 56

Compute the coefficient r₃ of x³ in R(C₁, x)

⋅⋅⋅⋅

⋅⋅⋅⋅

= (1+4x+2x²)

⋅⋅⋅⋅

= . . . + 1

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

= . . . + (1

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

If C is a chessboard made up of pairwise disjoint subboards C₁, C₂, …, C_n, then

R(C, x) = R(C₁, x)

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

Let r_k(C) denote the number of ways of placing k nontaking rooks on the chessboard C.

C:

r_k(C) consists of the following two parts.

∗∗∗

∗

C_s :

C_e : (C_e is obtained from C by deleting the designated square.)

1. A rook on the square designated by “∗∗∗∗”.

(C_s is obtained from C by deleting the row and the column containing the designated square.)

r_{k−−−1−} (C_s) is included in r_k(C).

2. No rook on the designated square.

r_k(C_e) is included in r_k(C).

Therefore, r_k(C) = r_{k−−−1−} (C_s) + r_k(C_e), and R(C, x) = xR(C_s, x) + R(C_e, x).

= x

⋅⋅⋅⋅

∗

∗∗

∗

+

∗

∗∗

∗

= x²

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

+ ∗∗∗∗

= x²

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

= (x²+ x)

⋅⋅⋅⋅

⋅⋅⋅⋅

Ex.

= (x²+x)

⋅⋅⋅⋅

⋅⋅⋅⋅

= 1+6x+10x²+4x³.

∗∗∗

∗

C : Ex.

Determine the number of ways of placing four nontaking rooks on the unshaded area of C.

Let c_i be the condition that a rook is placed on the shaded area of row i.

N(c₁c₂ c₃ c₄) = S₀ –S₁ +S₂ –S₃ +S₄

= 5!–r₁

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

where R(the shaded area of C, x)

= (1+3x+x²)

⋅⋅⋅⋅

= 1+7x+16x²+13x³+3x⁴ = 1+r₁x+r₂x²+r₃x³+r₄x⁴.

∴

∴

∴

∴ N(c₁ c₂ c₃c₄) = 5!–7

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

⋅⋅⋅⋅

= 25.

Ex. Four people, denoted by R₁, R₂, R₃ and R₄, are assigned to five tables, denoted by T₁, T₂, T₃, T₄ and T₅, in a wedding reception. In how many ways can they be assigned to four distinct tables, subject to the following four restrictions:

(a) R1 is not assigned to T₁ or T₂; (b) R2 is not assigned to T₂;

(c) R3 is not assigned to T₃ or T₄; (d) R4 is not assigned to T₄ or T₅.

The answer is 25, as computed in the example above.

Ex. Two dice, denoted by R and G, are rolled six times.

Under the condition of (R, G)∉∉∉∉{(1, 2), (2, 1), (2, 5), (3, 4), (4, 1), (4, 5), (6, 6)}, what is the probability that all six values 1, 2, …, 6 occur for both R and G?

Consider the following left chessboard, where the row (column) labels represent the outcome on R (G).

The right chessboard is obtained by relabeling the rows and columns, where the seven shaded squares constitute four pairwise disjoint subboards.

Therefore, r(C, x) = (1+4x+2x²)××××(1+x)³

= 1+7x+17x²+19x³+10x⁴+2x⁵.

The probability is P = ^6! ₆ 29

×S

×

×

× , where there are 6! ways for R to occur 1, 2, …, 6 and S is the number of ways for G to occur 1, 2, …, 6 under the condition.

For example, (1, ?), (2, ?), (3, ?), (4, ?), (5, ?), (6, ?) and (4, ?), (5, ?), (6, ?), (1, ?), (2, ?), (3, ?) are two ways for R to occur 1, 2, 3, 4, 5, 6.

Then, each way for the former to occur 1, 2, 3, 4, 5, 6 for G, e.g., (1, 6), (2, 2), (3, 3), (4, 4), (5, 5), (6, 1),

uniquely corresponds to a way for the latter to occur 1, 2, 3, 4, 5, 6 for G, e.g., (4, 4), (5, 5), (6, 1), (1, 6), (2, 2), (3, 3), and vice versa.