- For two seperate tasks,
- if Task A can be done in n ways, and
- Task B can be done in m ways.
- Then A and B can be done n + m ways
- In a list, AB If there are
- n ways of choosing A, and
- m ways of choosing B, then
- There are n x m ways of choosing AB.
How many ways can we pick a license plate with
a. 3 Letters, then 4 numbers
* 26 x 26 x 26 x 10 x 10 x 10 x 10 = 26^3 x 10^4
b. 2 Even numbers, 3 vowels, 1number, then 2 letters?
* 2 Even Numbers = 5 x 5
* 3 Vowels = 5 x 5 x 5
* 1 Number = 10
* 2 Letters = 26 x 26
* Answer = 5^5 x 10 x 26^2
n! = (n)(n-1)(n-2) ... (3)(2)(1)
- 4! = 4 x 3 x 2 x 1 = 24
- 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
- 0! = 1
The number of permutations of length k from a list of n elements is:
n!/(n-k)! - With no repetition
On the calculator p(n,k)
-
How many ways can we list 4 number out of a list consisting of 7 numbers w/o repetition?
n = 7 k = 4 P(7,4) = 7!/(7-4)! = 7!/3! = 840 -
How many permutation can we make with the word BASE?
n = 4 k = 4 P(4,4) = 4!/(4-4)! = 4!/0! = 4!/1 -
How many permutations can we make with the word BALL?
B A L L 4!/2! = 4x3 = 12 -
How many arrangements can we make with the word DATABASES?
D A T B S E A S A A = 3! S = 2! 9!/2!x3! = 30240
if n and k are integers then (n over k) is the amount of subsets that can be made using k elements in an n elements set.
(n over k ) = n choose k
= C(n,k) <--- Can be found on the calculator
= n!/ k!(n-k)!
-
Evaluate (9 over 4)
(9 over 4) = 9! / 4!(5!) = 126 -
Evaluate (6 over 3)
(6 over 3) = 6!/3!(6-3)! = 6!/3!x3! = 20 -
Given the word TALLAHASSE, how many arrangements can be made with no consecutive vowels?
T A L H S E A L S E A a) TLLHSS = 6! L = 2! S = 2! 6!/2!2! b) . T . L . L . H . S . S . . = 7 vowels = 5 (7 over 5) c) AAAEE A = 3! E = 2! 5! / 3!2!
Answer; (6!/2!2!) x (7 over 5) x (5!/3!2!)
Ø = { }
|Ø| = 0
|{Ø}| = 1
|{ }| = 0 , Because Ø = { }
{ form | rule }
{ m/n | m,n element in Z, n ≠ 0 }
A = {a,b,c}
B = {0,1}
then A x B = {(a,0),(a,1),(b,0),(b,1),(c,0),(c,1)}
|A| = 3
|B| = 2
if A and B are finite sets!
|A x B | = |A| * |B|
A1 x A2 x ... x An = {(A1, A2, ..., An)| Ai € Ai for all i}
B is a subset of A if every element in B is also in A.
A = {1,2,3,4,5,6}
B = {1,3,4,5}
B is a subset of A
is a set of all possible subsets
If |A| = n
|P(A)| = 2^n = 2^(|A|)
Each element can be in a subset or not.
P(Ø) = 0
A = {Ø}
P({Ø}) = {Ø, {Ø}} = 2
is A subset in P(A)?
Yes, {Ø} is in P({Ø})
is A element in P(A)?
Yes, Ø is in P({Ø})
each set A is in a universe U A is a subset of U
| p | ¬p |
|---|---|
| 0 | 1 |
| 1 | 0 |
| p | q | p ^ q |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
p ^ q = min(p,q)
| p | q | p v q |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
p ^ q = max(p,q)
| p | q | p --> q |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
p --> q == 1; iff p is less than or equal to q;
| p | q | p <-> q |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
p == q; then p <-> q = 1
note: ¬(p⊕q)= p <-> q and vice verca
| p | q | p ⊕ q |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
p is not equal to q; then p (+) q = 1
| p | q | ¬(p ^ q) |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
| p | ¬(p ^ p) |
|---|---|
| 1 | 0 |
| 0 | 1 |
p nand p <=> ¬p
nand is sometimes written as ↑: p ↑ q <=> ¬(p ^ p)
We can use logical equivalences to reduce complex formulas into simpler ones.
T0 = Tautology (Always 1, sometimes written as '1', instead of T)
F0 = Contradiction (Always 0, sometimes written as '0', instead of F)
P ^ T <=> P
P v F <=> P
P v T <=> T0
P ^ F <=> F0
(P v F ) ^ (q v T)
(P v F) == (Identity Laws) P
(q v T) == (Domination Laws) T0
P ^ T == (Identity Laws) P
Answer; P
¬¬p <=> p
¬(p ^ q) <=> ¬p v ¬q
¬(p v q) <=> ¬p ^ ¬q
¬(¬p ^ ¬q)
¬¬p v ¬¬q == (DeMorgans)
answer; p v q == (Double Negation)
p ^ (q v r) <=> (p ^ q) v (p ^ r)
p v (q ^ r) <=> (p v q) ^ (p v r)
p ^ (p ^ q) <=> P
p v (p ^ q) <=> P
¬(¬p) v ((p v F) ^ ¬(¬q))
p v ((p v F) ^ q) == (Double Negation)x2
p v (p ^ q ) == (Identity Law)
P == (Absorption Law)
p ^ q <=> q ^ p
p v q <=> q v p
p ^ (q ^ r) <=> (p ^q ) ^ r
p ^ ¬p <=> F
p v ¬p <=> T
p --> q <=> ¬p v q
| p | q | p --> q | ¬p | ¬p v q |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 |
¬(p ^ q) ^ q
( ¬p v ¬q ) ^ q == DeMorgan Law
(q ^ ¬p) v (q ^ ¬q) == Distributive Law
q ^ ¬p v False == Inverse Laws
q ^ ¬p == Identity Laws
¬p ^ q == Commutativity
p --> q <=> ¬q v p
¬q --> ¬p <=> q v ¬p
q --> q <=> q v ¬p
¬p --> ¬q <=> p v ¬q
https://datapor.no/loot/lVyftyrx298.png
p <-> <=> (p --> q) ^ (q --> p)
| p | q | p <-> q | p -> q | ^ | q --> p |
|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 1 |
∴ == Therefore
R --> W } Premise
R }
______
∴ W - Conclusion
This is the same as above, if seen as a truth table*
((R --> W) ^ R) --> W (Tautology)
-
A set of premises P1, P2,...,Pn prove some conclusion q in an argument
- (p1 ^ p2 ^ ... ^ pn ) --> q
-
An argument is valid if the premiseslogically entail the conlcusion
-
Modus Ponens (MPP)
P --> q P _______ ∴ q -
Modus Tollens (MTT)
P --> q ¬P _______ ∴ ¬p -
Hypothetical Syllogism (HS)
P --> q q --> r ________ ∴ p --> r -
Disjunctive Syllogism (DS)
P v q ¬P ________ ∴ q -
Addition (Or Introduction)
P _______ ∴ p v q -
Simplification (And Elimination)
P ^ q _______ ∴ p ∴ q -
Conjunction (And Introduction)
P q _______ ∴ p ^ q
- ∀ = Universial
- ∀x == for all x
- ∃ = Extensial
-
∃x == there exists at least one x/ fOr some x
∀x P(x): "For all x, x is P" ∃x P(x): "For some x, x is P"
-
For every real number n, there is a real number m such that m^2 = n.
\ / \ / \ /
∀n∈ℝ ∃m∈ℝ : m^2
-
Define ∀x, ∃x for a universe with elements {1,2,..., n}
∀xP(x) <=> p(1) ^ p(2) ^ ... ^ p(n)
∃xP(x) <=> p(1) v p(2) v ... v p(n)
Show that ¬∀xP(x) <=> ∃x[¬P(x)]
¬∀xP(x) = ¬(p(1) ^ p(2) ^ ... ^ p(n))
= ¬P(1) v P(2) v ... v P(n)
= ∃x[¬P(x)]
∀xP(x) <=> ¬∃x[¬P(x)]
∃xP(x) <=> ¬∀x[¬P(x)]
¬∀xP(x) <=> ∃x[¬P(x)]
¬∃xP(x) <=> ∀x[¬P(x)]
https://datapor.no/loot/g6oRHTAIddv.png wtf
Negate the following
[∀x[∃y[ P(x,y) ^ Q(y)]]]
= ¬[∀x[∃y[ P(x,y) ^ Q(y)]]]
= ∃x¬[∃y[P(x,y) ^ Q(y)]]
= ∃x∀y¬[P(x,y) ^ Q(y)] == DeMorgans Law
= ∃x∀y[¬P(x,y) ^ ¬Q(y)]