Module #15 - Combinatorics 1 Chapter 6 The Inclusion-Exclusion Principle and Applications

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Module #15 - Combinatorics

Chapter 6

The Inclusion-Exclusion Principle and Applications

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Summary

• The inclusion-exclusion principle

• Combinations with repetition

• Derangements

• Permutations with forbidden positions

• Another forbidden position problem

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Section 6.1

The inclusion-exclusion principle

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Examples for review

Ex.1 Count the permutations of {1,2,…,n} in which 1 is not in the first position.

Direct: (n-1) x (n-1)!

Indirect: If 1 is in the first position, the permutation number is (n-1)!. Hence the answer is n! – (n-1)!

Ex.2 Count the number of integers between 1 and 600 which are not divisible by 6.

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Complement

The complement of a subset A of a set S is the set consisting of those objects in S which are not in A. Denoted by

Let P1, P2, …, Pm be m properties referring to the objects in S and let Ai ={x: x in S and x has property Pi}, (i=1,2,…,m).

Then, Ai ∩ Aj is the subset of objects which have both properties Pi and Pj (and possibly others). The subset of objects having none of the properties is

}.,:{ AxSxxASA

.21 mAAA

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The inclusion-exclusion principle (IEP)

The number of objects in S which have none of the properties P1, P2, …., Pm is given by

where the first sum is over all 1-combination {i} of {1,2,3…,m}, the second sum is over all 2-combinations {i, j} of {1,2,..,m}, the third sum is over all 3-combinations {i, j, k} of {1,2,…,m}, and so on.

mm

kjijiim

AAA

AAAAAASAAA

21

21

)1(

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Special Case

• Generally, the number of terms on the right side of the equality is

321

323121

321321

212121

)(

)(

.

AAA

AAAAAA

AAASAAA

AAAASAA

m2m

m

2

m

1

m

0

m

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A corollary of IEP

The number of objects of S which have at least one of the properties P1, P2, …, Pm is given by

where the summations are as specified in IEP.

,)1( 211

21

mm

kjijiim

AAA

AAAAAAAAA

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Examples

Find the number of integers between 1 and 1000, inclusive, which are divisible by none of 5, 6, and 8.

Let P1 be the property that an integer is divisible by 5, P2 the property that an integer is divisible by 6, and P3 the property that an integer is divisible by 8. Let S be the set consisting of the first 1000 positive integers. For i=1, 2, 3 let Ai be the set consisting of those integers in S with property Pi. We wish to find the number of integers in

………………………

321 AAA

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Exercise

• Count the number of integers between 1 and 1000, inclusive, which are not divisible by 4, 5 or 6.

• How many permutations of the letters M, A, T, H, I, S, F, U, N are there such that none of the words MATH, IS and FUN occur as consecutive letters?

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Special case for IEP

Assume that the size of the set Ai1 ∩ Ai2 ∩ … ∩ Aik that occurs in the IEP depends only on k and not on which k sets are used in the intersection. Then

where ak = |Ai1 ∩ Ai2 ∩ … ∩ Aik|. This is because the kth summation that occurs in the IEP contains C(m, k) summands each equal to ak.

mm

kk

m

aak

m

am

am

am

aAAA

)1)1(

321 321021

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Example

• How many integers between 0 and 99,999 (incusive) have among their digits each of 2,5 and 8?

• Let S be the integers between 0 and 99,999. P1

(resp., P2 and P3) be the property that an integer does not contain the digit 2 (resp., 5 and 8). Let A i be the set consisting of those integers in S with property Pi. We wish to count the number of integers in

321 AAA

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Example (cont’d)

• S (resp., Ai) is the 5-permutations of the multiset in which each digit 0, 1, 2, …, 9 (resp., 0, 1, 3, …, 9; 0, 1, 2, 3, 4, 6, 7, 8, 9; 0, 1, 2, …, 7, 9) has repetition number 5 or greater. Hence,

• a0 = 105, a1 = 95, a2=85, a3 = 75 and the answer is 105 – 3×9 ５＋３ ×85 – 75.

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Section 6.2

Combinations with repetition

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Review

The number of r-combinations of a set of n distinct elements is C(n, r) = n!/r!(n-r)!

The number of r-combinations of a multiset with k distinct objects each with an infinite repetition number equals

C(r+k-1, r).

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Review: r-combinations

• An r-combination of elements of a set S is simply a subset TS with r members, |T|=r.

• The number of r-combinations of a set with n=|S| elements is

• Note that C(n,r) = C(n, n−r)– Because choosing the r members of T is the same

thing as choosing the n−r non-members of T.

)!(!

!

!

)!/(!

),(

),(),(

rnr

n

r

rnn

rrP

rnP

r

nrnC

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Review: Combinations of Multisets

• If S is a multiset, then an r-combination of S is an unordered selection of r of the objects of S. Thus an r-combination is itself a multiset, a submultiset of S.

• Example. If S = {2{a}, 1{b}, 3{c}}, then the 3-combinations of S are {2{a}, 1{b}}, {2.a, 1.c}, {1.a, 1.b, 1.c}, {1.a, 2.c}, {1.b,2.c}, {3.c}.

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Review: r-combinations

Let S be a multiset with objects of k different types where each has an infinite repetition number. Then the number of r-combinations of S equals

.1

11

k

kr

r

kr

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About repetition number

• Suppose T is a multiset and an object x of T of the a certain type has repetition number which is greater than r. Then the number of r-combinations of T equals the number of r-combinations of the multiset obtained from T by replacing the repetition number of x by r.

• Furthermore, any repetition number which is greater than r can be replaced by r.

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Example

• The number of 8-combinastions of the multiset {3{a}, ∞{b}, 6{c}, 10{d}, ∞{e}} is the same as the number of 8-combinations of the multiset {3{a}, 8{b}, 6{c}, 10{d}, 8{e}}.

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• Example 1: Determine the number of 10-combinations of the multiset T = {3{a}, 4{b}, 5{c}}.

• Hint: Let T* = {∞{a}, ∞{b}, ∞{c}}, P1 (resp., P2, and

P3) be the property that a 10-combination of T* has more than 3 a’s (resp., 4 b’s and 5 c’s) and A1 (resp., A2 and A3) be the 1—combinations of T* which have property P1 (resp., P2 and P3). We wish to determine the size of the set 321 AAA

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Hint (cont’d)

• The set A1 consists of all 10-conbinations of T* in which a occurs at least 4 times. If we take any one of these 10-combinations in A1 and remove 4 a’s, we are left with a 6-combination of T*. Conversely, if we take a 6-combination of T* and add 4 a’s to it, we get a 10-combination of T* in which a occurs at least 4 times. Thus the number of 10-combinations in A1 equals the number of 6-combinations of T*.

• CONTINUE BY YOURSELF!

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• Example 2: What is the number of integer solutions of the equation x1 + x2 +x3 +x4 = 18 which satisfy 1≤x1≤5, -2≤x2≤4, 0≤x3≤5, 3≤x4≤9?

• Hint: The equation equals to y1 + y2 +y3 +y4 = 16 which satisfy 0≤y1≤4, 0≤y2≤6, 0≤y3≤5, 0≤y4≤6? The following solution similar to example 1.

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Second solution

• Let S be the set of all non-negative integral solutions of the above equation, P1 (P2, P3 and P4) be the property that y1 ≥ 5 (y2 ≥ 7, y3≥6 and y4≥7) and Ai be the subset of S consisting of he solutions satisfying property Pi(i = 1, 2, 3, 4). Let z1 = y1 – 5, z2 = y2 ,z3 = y3 and z4 = y4. Then the number of solutions in A1 is the same as the number of nonnegative integral solutions of z1 +z2+z3+z4=11. Hence, |A1| = C(11+4-1, 11).

• CONTINUE BY YOURSELF.

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Section 6.3

Derangements

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Derangements

• A derangement of {1, 2, …, n} is a permutation i1i2…in of {1, 2, …, n} such that i1 ≠1, i2≠2, …, in≠n (i.e., no integer is in its natural position).

• We denote by Dn the number of derangements of {1, 2, …, n}.

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Examples

• For n =1, there are no derangements.

• For n =2, the only derangement is 2 1

• For n =3, there are two derangements:

2 3 1 and 3 2 1.

• For n = 4, there are 9 derangements:

2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321.

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Formulas for Counting Dn

• For n ≥ 1

.2)1(

)2()1(

.3),)1((

.3),)(1(

.!

1)1(

!3

1

!2

1

!1

11(!

122

211

12

n

DD

nDnDnDD

nDDnDn

nD

n

n

nnnn

nnn

nn

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Examples

• At a party there are n men and n women. In how many ways can the n women choose male partners for the first dance? How many ways are there for the second dance if everyone has to change partners?

• Answer: for the first dance there are n! possibilities. For the second dance, the number of possibilities is Dn.

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Exercise

• Suppose the n men and n women at the party above check their hats before the dance. At the end of the party their hats are returned randomly. In how many ways can they be returned if each man gets a male hat and each women gets a female hat, but no one gets the hat he or she checked?

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Section 6.4

Permutations with forbidden positions

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Definition of P(X1, X2,…,Xn)

• Let X1, X2, …, Xn be (possibly empty) subsets of {1, 2, …, n}. We denote by P(X1, X2,…,Xn) the set of all permutations i1i2…in of {1, 2, …, n} such that i1 is not in X1, i2 is not in X2 … in is not in Xn.

• Let p(X1, X2,…,Xn) = | P(X1, X2,…,Xn) |

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Examples• Let n = 4 and let X1={1, 2}, X2 ={2, 3}, X3 ={3, 4}

and X4 ={1, 4}. Then P(X1, X2, X3, X4) consists of all permutations i1i2 i3i4 of {1,2 ,3 ,4} such that i1 ≠ 1,2; i2 ≠ 2,3; i3 ≠ 3,4; i4 ≠ 1,4. Hence, P(X1, X2, X3, X4) = {3412, 4123} and p(X1, X2,…,Xn) = 2.

• Let Xk = {k} (k =1, 2, …,n). Then the set P(X1, X2,…,Xn) equals the set of all permutations i1i2…in of {1, 2, …, n} for which ik ≠ k. We conclude that P(X1, X2,…,Xn) is the set of derangements of {1,2, …, n} and we have p(X1, X2,…,Xn) = Dn.

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Placement of non-attacking rooks in chessboard

• The permutation i1i2…in of {1, 2, …, n} corresponds to the placement of n rooks on the board in the square with coordinates (1, i1), (2, i2), …, (n, in).

• The permutations in P(X1, X2,…,Xn) correspond to placements of n non-attacking rooks on an n-by-n board in which there are certain squares in which it is forbidden to put a rook.

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An example

• Let n =5, X1 ={1, 4}, X2 ={3}, X3 = {4}, X4 ={1, 5}, X5 ={2, 5}. Then P(X1, X2, X3, X4, X5) are in one-to-one correspondence with the placement of 5 non-attacking rooks on the board with forbidden positions as shown.

x

x x

x

x

x x

54321

5

4

3

2

1

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Placement of rooks in chess board

• The number of ways to place n non-attacking, indistinguishable rooks on an n-by-n board with forbidden positions equals

n! – r1(n-1)! + r2(n-2)! - ... + (-1)k rk(n-k)!+…+(-1)rrn.

where rk is the number of ways to place k non-attacking rooks on the n-by-n board where each of the k rooks is in a forbidden position (k=1, 2, …, n).

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An example

• Determine the number of ways to place 5 non-attacking rooks on the following 5-by-5 board, with forbidden positions as shown.

x x

x x

x x

x

54321

5

4

3

2

1

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Solution

• We will decide the values of r1, r2, …, r5. r1 is the number of forbidden positions 7. Before evaluating r2, r3,…,r5, we note that the set of forbidden positions can be partitioned into two “independent” parts, one part F1 containing three positions and the other part F2 containing four.

x x

x x

x x

x

54321

5

4

3

2

1

“Independent” means squares in different parts do not belong to a common row or column.

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Evaluate r2

• We now evaluate r2, the number of ways to place 2 non-attacking rooks in forbidden positions. The rooks may be both in F1, both in F2 or one in F1 and one in F2. Hence, r2 = 1+2+3x4 =15.

• Continue to evaluate r3, r4, r5

by yourself.

x x

x x

x x

x

54321

5

4

3

2

1

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Section 6.5

Another forbidden position problem

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Problem description

• Suppose a class of 8 boys take a walk every day. They walk in a line of 8 so that every boy except the first is preceded by another. In order that a child not see the same person in front of him, on the second day they decide to switch positions so that no boy is preceded by the same boy who preceded him the first day. In how many ways can they switch positions?

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Illustration of the problem

• If we assign to the boys the numbers 1, 2, …, 8 with the last boy in the column of the first day receiving the number 1, …, and the first boy receiving the number 8 as in 1 2 3 4 5 6 7 8, then we are asked to determine the number of permutations of the set {1 2 3 4 5 6 7 8} in which the patterns 12, 23, 34 ,….,78 do not occur. E.g., 31542876 is an allowable premutation while 84312657 is not.

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Formulas for Counting Qn

• For n ≥1

Qn = n! –C(n-1, 1)(n-1)! + C(n-1, 2) (n-2)! – C(n-1, 3)(n-3)! + …+(-1)n-1(C(n-1, n-1)1!.

Q1=1, Q2=1, Q3=3 and Q4=11.

Qn = Dn + Dn-1, (n≥2).

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Assignments

• Find the number of integers between 1 and 10000 inclusive which hare neither prefect squares nor perfect cubes.

• What is the number of integer solutions of the equation x1 + x2 +x3 +x4 = 20 which satisfy 1≤x1≤6, 0≤x2≤7, 4≤x3≤8, 2≤x4≤6?

• Determine the number jof permutations of {1, 2, …, 8} in which no even integer is in its natural position.

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Assignments (cont’d)

• Determine a general formula for the number of permutations of the set {1, 2, …, n} in which exactly k integers are in their natural positions.

• Eight girls are seated around a carousel. In how many ways can they change seats so that each has a different girl in front of her?

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Assignments (cont’d)

• What is the number of ways to place six non-attacking rooks on the 6-by-6 boards with forbidden positions as shown?

x x

x x

x x

x x

x x

x x

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Module #15 - Combinatorics 1 Chapter 6 The Inclusion-Exclusion Principle and Applications