Counting Elements in a List - Notes | CMSC 250, Study notes of Discrete Structures and Graph Theory

Download Counting Elements in a List - Notes | CMSC 250 and more Study notes Discrete Structures and Graph Theory in PDF only on Docsity!

Counting Elements in a List

• How many integers in the list from 1 to 10?• How many integers in the list from m to n?

(assuming m <= n)

How many in a list divisible by something:• How many positive three digit integers are there?

• (this means only the ones that require 3 digits)– 999 – 100 + 1 = 900
  • How many three digit integers are divisible by 5?
    • think about the definition of divisible by

x | y

↔ ∃

k

∈

Z, y = kx

and then count the k’s that work

100, 101, 102, 103, 104, 105, 106,… 994, 995, 996, 997, 998, 99920*

21*

…

199*

• count the integers between 20 and 199– 199 – 20 + 1 = 180

Flipping Two Coins

• Sample Space = {(H,H), (H,T), (T,H), (T,T)}• Probability of no heads• Probability of at least one head• Probability of same sides on the two coins• Note: probability & actual outcomes often differ

Standard Playing Cards

• values: 2,3,4,5,6,7,8,9,10,J,Q,K,A• suits: D(

), H(

), S(

), C(

• probability of drawing the Ace of Spades• probability of drawing a Spade• probability of drawing a face card• probability of drawing a red face card

Multi-level Probability

• If I toss a coin once – the probability of Head = ½• If I toss that coin 5 times
  • the probability of all heads– the probability of exactly 4 heads

5 (^12)

1 2

1 2

1 2

1 2

1 2

=

5 (^52)

Multiplication Rule

st

step can be performed n

1

ways

nd

step can be performed n

2

ways

• …• K

th

step can be performed n

k^

ways

• operation can be performed n

*n 1

2

…n

k^

ways

• Cartesian product

n(A)=3, n(B)=2, n(C)=

• n(AxBxC) = 24– n(AxB) = 6

n((AxB)xC) = 24

What if A wins 2 of every 3 games?•

Each line for A must have a 2/

-^

Each line for B must have a 1/

-^

How likely is A to win the tournament?

-^

How likely is B to win the tournament?

Using the Multiplication Rule for

Selecting a PIN

• Number of 4 digit PINs of (0,1,2,.)
  • with repetition allowed = 4444=256– with no repetition allowed = 4321=
    • Extra Rules : -. (the period) can’t be first or last– 0 can’t be first - with repetition allowed = 2443• without repetition allowed = 222

Difference Rule Formally

• If A is a finite set and B

A, then

n(A-B) = n(A) – n(B)

• One Application:probability of the complement of an event

P(E’) = P(E

c ) = 1-P(E)

PINs with less specified length

Addition Rule

• Assume it can be a 2,3 or 4 length PINPartition the problem

number of 2 length PINs w/rep allowed: 44 = 16number of 3 length PINs w/rep allowed: 444 = 64number of 4 length PINs w/rep allowed: 4444 = 256 Number PINs if allowing length of 2,3 or 4 = 336

Another example for

Multiplication Rule and Addition Rule• How many 3 digit integers are divisible by 5?

• How many end in a 0?

9101 = 90

• How many end in a 5?

9101 = 90

• These form a partition with the set of numbers divisible

by 5 so

• 90 + 90 = 180

Where Multiplication Rule Doesn’t Work• People= {Angel, Bob, Carol, Dan}• need to be appointed as

• president, vice-president, and treasurer– nobody can hold more than one office– Angel doesn’t want to be president– Only Bob and Dan want to be vice-president

Permutations

-^

Different ways of arranging objects– in a line or circle– without duplication/ all items distinguishable– note: order is taken into account

-^

Number of linear permutations of N objects = N!N possible for 1

st^

position * (N-1) for 2

nd

• …* (1) for last

-^

Number of circular permutations of N objects = (N-1)!Fix one person,then (N-1) possible for next position * (N-2) for 2

nd

• …* (1) for last

r-Permutations

If there are n things in the set,and you want to line-up only r of them.•^

Example: Class = {Alice, Bob, Carol, Dan}

• select a president and a vice president to represent the class

r

n

n

P

r

n

P

n

r