Inclusion Exclusion - Discrete Mathematics and its Applications - Lecture Slides, Slides of Discrete Mathematics

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Lecture 18

6.5 & 6.6 Inclusion-Exclusion

6.5 Inclusion-Exclusion

A A B

U

A  B | A  B | |= A | + | B | −| A B|

It’s simply a matter of not over-counting the blue area in the intersection.

General Case

(^1 21) ( ) ( ) 1

| ... | | | | | | | ... ( 1) | ... |

n n (^) i i (^) pairs ij i j (^) triples ijk i j k n i j k n

A A A A A A A A A A A A A

=

= − + −

• −

   

Proof: We show that each element is counted exactly once. Assume element ‘a’ is in r sets out of the n sets A1,...,An. -The first term counts ‘a’ r-times=C(r,1). -The second term counts ‘a’ -C(r,2) times (there are C(r,2) pairs in a set of r elements). -The k’th term counts ‘a’ -C(r,k) times (there are C(r,k) k-subsets in a set of r elements). -...

_- If k=r then there are precisely (-1)^(r+1) C(r,r) terms.

• For k>r ‘a’ is not in the intersection: it is counted 0 times. Total: C(r,1)-C(r,2)+...+(-1)^(r+1)C(r,r)_

Now use: (^) 0 to show that each element is counted exactly once 1 1

( 1) ( , ) 0 1 ( 1) ( , )

r (^) k k r k k

C r k C r k

=

=

− = ⇒ = −

∑ ∑ Docsity.com

Applications of Incl.-Excl.

We can use inclusion/exclusion to count the number of members of a set that do not have a bunch of properties: P1,P2,...,Pn.

Call N(Pi,Pj,Pk,...) the number of elements of a set that do have properties Pi, Pj, Pk,.... and N the total number of elements in the set.

By inclusion/exclusion we then have:

Theorem: Let Ai be the subset of elements of a set A that has property Pi. The number of elements in a set A that do not have properties P1,...Pn is given then by: 1 2 3 1 , 1 , , 1 1 2

n n n n i i j i j k n n i i ji j i j ki j k i j k i j k

A A A A A

N N P N P P N P P P N P P P

with N P P P A A A

= (^) < = (^) < <=

Examples

Compute the number of solutions to x1+x2+x3= where x1,x2,x3 non-negative integers and x1 <=3, x2<=4, x4<=6. P1: x1 > 3 P2: x2 > 4 P3: x3 > 6 The solution must have non of the properties P1,P2,P3. The solution of a problem x1+x2+x3=11 with constraints x1 > 3 is solved as follows:

x1 x2 x

7 more balls 4 balls in basket x already.

Total number of ways: C(7+3-1,7)=

Therefore: N-N(P1)-N(P2)-N(P3)+N(P1,P2)+N(P2,P3)+N(P1,P3)-N(P1,P2,P3) C(11+3-1,11) – C(7+3-1,7) – C(6+3-1,6) ... – 0. Docsity.com

Connection with De Morgan’s law

1 1

1 1 1 ,^1

1

| | | | | | | | ... ( 1) | ... |

n n i i^ i i n n n n (^) n i i^ i i^ i^ i^ i j i^ j^ n i j n i i

A A

A U A U A A A A A

A

= =

= = = <

=

= →

= − = − + + + −

=

∑ ∑ ^ ^ 

 

 

 (^) So we have 2 ways to solve the last example: x1+x2+x3 = 11 such that non of the following properties hold: P1: x1 > 3 P2: x2 > 4 P3: x3 > 6 or x1+x2+x3=11 such all of the following properties hold: Q1=NOT P1: 0<x1<= Q2=NOT P2: 0<x2<= Q3=NOT P3: 0<x3<=

Sometimes this is easier to compute.

Number of Onto-Functions

A B

f

x y

There is no element without incoming arrows

N-N(P1)-N(P2) ... + N(P1,P2) -...+(-1)^n N(P1,...,Pn).

N: number of function from A  B: n^m N(Pi): number of functions that do not have y1 in its range: (n-1)^m. There are n=C(n,1) such terms. N(Pi,Pj): (n-2)^m with C(n,2) terms.

Total: n^m – C(n,1)(n-1)^m + C(n,2)(n-2)^m – ...+(-1)^(n-1)C(n,n-1)1^m.

Derangements

A derangement: is a permutation that leaves non of the elements in its place.

Example: (3,2,1) is not a derangement, but (2,3,1) is one. Then: how many derangements are there of a set of n elements. This is the answer to the Hatcheck problem: If n people check in their hats and the employee randomly returns hats back to people: what is the probability no-one gets their own hat back:  P = number of derangements of n / total number of permutations = Dn / n!

Derangements

More generally we have:

N(Pi) = (n-1)! and there are C(1,n) such terms N(Pi,Pj) = (n-2)! and there are C(n,2) such terms.

Total: Dn = n! – C(n,1)(n-1)! + C(n,2)(n-2)! - ...+ (-1)^n C(n,n) (n-n)! = n! – [n!/1!(n-1)!] (n-1)! + n!/2!(n-2)!! ... = n!(1-1/1! + 1/2! -1/3!+...+(-1)^n 1/n!)

P(hatcheck) = 1-1/1!+1/2!-...(-1)^n1/n! = 0.368 if n is large.

so even if 1M people check in their hat there is a probability of 0.368 that no-one gets there hat returned correctly.

THE END...

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