Transcript Inclusion-Exclusion

Inclusion-Exclusion
Rosen 6.5 & 6.6
Longin Jan Latecki
basd on Slides by
Max Welling, University of California, Irvine
Vardges Melkonian, Ohio University, Athens
6.5 Inclusion-Exclusion
U
A
A
A B
B
| A B || A |  | B |  | A B |
It’s simply a matter of not over-counting the blue area in the intersection.
Example on
Inclusion/Exclusion Rule (2 sets)
• Question: How many integers from 1 through 100
are multiples of 3 or multiples of 7 ?
• Solution: Let A=the set of integers from 1 through 100
which are multiples of 3;
B = the set of integers from 1 through 100
which are multiples of 7.
Then we want to find n(A  B).
First note that A  B is the set of integers
from 1 through 100 which are multiples of 21 .
n(A  B) = n(A) + n(B) - n(A  B) (by incl./excl. rule)
= 33 + 14 – 4 = 43 (by counting the elements
of the three lists)
Now three Sets
area = 4-3=1
U
C
area = 2-1=1
B C
A C
A B C
area = 1
A
A B
B
Image a blue circle has area 4. The intersections between 2 circles have
area 2 and the intersection between three circles 1. What is the total area
covered?
| A B C || A |  | B |  | C |  | A B |  | B C |  | C
A=4+4+4 – 2 -2 -2 + 1 = 12 – 6 + 1 = 7.
A|  | A B C |
Example on
Inclusion/Exclusion Rule (3 sets)
• 3 headache drugs – A,B, and C – were tested on 40
subjects. The results of tests:
23 reported relief from drug A;
18 reported relief from drug B;
31 reported relief from drug C;
11 reported relief from both drugs A and B;
19 reported relief from both drugs A and C;
14 reported relief from both drugs B and C;
37 reported relief from at least one of the drugs.
Questions:
1) How many people got relief from none of the drugs?
2) How many people got relief from all 3 drugs?
3) How many people got relief from A only?
Example on
Inclusion/Exclusion Rule (3 sets)
S
C
A
B
We are given: n(A)=23, n(B)=18, n(C)=31,
n(A  B)=11, n(A  C)=19, n(B  C)=14 ,
n(S)=40, n(A  B  C)=37
Q1) How many people got relief from none of the drugs?
By difference rule,
n((A  B  C)c ) = n(S) – n(A  B  C) = 40 - 37 = 3
Example on
Inclusion/Exclusion Rule (3 sets)
Q2) How many people got relief from all 3 drugs?
By inclusion/exclusion rule:
n(A  B  C) = n(A  B  C)
- n(A) - n(B) - n(C)
+ n(A  B) + n(A  C) + n(B  C)
= 37 – 23 – 18 – 31 + 11 + 19 + 14 = 9
Q3) How many people got relief from A only?
n(A – (B  C))
(by inclusion/exclusion rule)
= n(A) – n(A  B) - n(A  C) + n(A  B  C)
= 23 – 11 – 19 + 9 = 2
The Principle of Inclusion-Exclusion
n
| A1
... An |  | Ai | 
A2
i 1

Aj | 
| Ai
pairs ( ij )
 (1)n 1 | Ai
Aj

| Ai
Aj
Ak | ...
triples ( ijk )
Ak
... An |
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)
r
(1) C (r , k )  0
Now use: 
k

k 0
r
1   (1) k 1 C ( r , k )
k 1
to show that each element is counted exactly once.
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:
| A |  | A1
A2
A3 ... An |
n
n
i 1
i , j 1
i j
N   N ( Pi )   N ( Pi , Pj ) 
with
N ( Pi , Pj , Pk ,...) | Ai
n

N ( Pi , Pj , Pk )  ...  (1) n N ( P1 , P2 ,..., Pn )
Aj
Ak
i , j , k 1
i j k
... |
A Picture
U
C
B C
A C
A B C
A
A B
B
|U |  | A B C |
Examples
Compute the number of solutions to x1+x2+x3=11
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:
7 more balls
4 balls in
basket x1
already.
x1
x2
x3
Total number of ways:
C(7+3-1,7)=36
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) – C(4+3-1,4) + … – 0.
Connection with De Morgan’s law
n
Ai 
i 1
Ai 
i 1
n
Ai | U | 
i 1

n
n
n
i 1
n
Ai | U |  | Ai |   | Ai
i 1
A j | ...  (1) n | A1 ...
An |
i, j
i j
n
Ai
i 1
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
Sometimes this is easier to compute.
P3: x3 > 6
or x1+x2+x3=11 such all of the following properties hold:
Q1=NOT P1: 0<x1<=3
Q2=NOT P2: 0<x2<=4
Q3=NOT P3: 0<x3<=6
Number of Onto-Functions
Onto or surjective functions:
A function f from A to B is onto if for every element b in B there is an
element a in A with f(a)=b.
If we have m elements in A and n in B, how many
onto functions are there?
x
y
f
We want all yi in the range of the function f.
Call Pi the property that yi is not in the
range of the the function f.
Then we are looking for the number of functions
that has none of the properties P1,...,Pn
A
B
There is no element without incoming arrows
Number of Onto-Functions
N(P1’P2’P3’) =
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.
x
y
f
Total:
n^m – C(n,1)(n-1)^m + C(n,2)(n-2)^m –
...+(-1)^(n-1)C(n,n-1)1^m.
m=6 and n=3
N(P1’P2’P3’) = 3^6 – C(3,1)2^6 + C(3,2)1^6 = 540
A
B
There is no element without incoming arrows