MATH1081 Discrete Maths (1 Viewer)

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

lol i remember that q, its surprisingly hard just give to tutor
Lol fair enough
_______________________

How do I create my sets for incl./excl.?

"What is the probability that a hand of eight cards dealt from a shuffled pack (w/o jokers) contains:
b) exactly three cards in at least one of the suits
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: Discrete Maths Sem 2 2016

Not sure if this is just a really tediously long question that I should give to my tutor, or if there's shortcuts.

Q: How many eight letter words can be formed from the letters of PARRAMATTA? (10 letters)
No shortcuts, you must exhaust all the cases.
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Lol fair enough
_______________________

How do I create my sets for incl./excl.?

"What is the probability that a hand of eight cards dealt from a shuffled pack (w/o jokers) contains:
b) exactly three cards in at least one of the suits
You could try letting Sj be the set of all hands such that there are exactly three cards from suit j. We want to find |S1 U S2 U S3 U S4|.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

Same question

c) exactly three cards in exactly one suit

I can't interpret this hint they gave:
Find the number of ways in which five cards can be chosen from three suits, with none of the three represented by three cards.


Sent from my iPhone using Tapatalk
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Same question

c) exactly three cards in exactly one suit

I can't interpret this hint they gave:
Find the number of ways in which five cards can be chosen from three suits, with none of the three represented by three cards.


Sent from my iPhone using Tapatalk
Having the eight-card hand have exactly three cards in exactly one suit (say suit 1) means we have exactly three cards from 1, and so the remaining five cards in the hand do not contain anything from suit 1. So the remaining five cards come from the other suits (2,3,4). But they must come from here in such a way that none of these suits are represented by exactly three cards in these remaining five cards (if that happened, that'd violate the condition of having exactly one suit contain exactly three cards).

So the hint is referring to this (remaining five cards chosen in such a way that they come from cards from suits 2,3,4, and so that none of these three suits are represented by exactly three cards in this set of five cards).
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

I've mind blanked

What's the difference between "negation" and "converse"
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

Just want some checking on my reasoning.

Q: A hand of thirteen cards is dealt from a shuffled pack. Giving reasons for your answers, determine which of the following statements are definitely true and which are possibly false.

a) The hand has at least four cards in the same suit.
- Not sure if this meant all 4 suits. Cause if that were the case then well no but if it were 1 suit I could argue true.

c) The hand has at least five cards in some suit.
- Sounds more like any suit now, so i just said

 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Just want some checking on my reasoning.

Q: A hand of thirteen cards is dealt from a shuffled pack. Giving reasons for your answers, determine which of the following statements are definitely true and which are possibly false.

a) The hand has at least four cards in the same suit.
- Not sure if this meant all 4 suits. Cause if that were the case then well no but if it were 1 suit I could argue true.

c) The hand has at least five cards in some suit.
- Sounds more like any suit now, so i just said

a) It means at least one suit, i.e. the hand has a suit with at least four cards from that suit. By the pigeonhole principle, this must be the case.

c) There need not be a suit with 5+ cards from it. Note ceil(13/4) = ceil(3.25) = 4 (not 5). So there must be a suit with 4+ cards, but need not be one with 5+ (e.g. imagine if the first three suits each had four cards and the last suit had one).
 
Last edited:

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

a) I'm pretty sure it means at least one suit, i.e. the hand has a suit with at least four cards from that suit. By the pigeonhole principle, this must be the case.

c) There need not be a suit with 5+ cards from it. Note ceil(13/4) = ceil(3.25) = 4 (not 5). So there must be a suit with 4+ cards, but need not be one with 5+ (e.g. imagine if the first three suits each had four cards and the last suit had one).
Ah damn, lost track of my arithmetic. That was probably why I doubted so for a).
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

Too lost in the wording.


Prove that there were two people in Australia yesterday who met exactly the same number of other people in Australia yesterday.
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Sorry lol, the definitions make sense but I can't do the question now
You can try investigating what happens for small values of |S| (e.g. |S| = 2 or 3 etc.) before generalising to |S| = N for arbtitrary N.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

You can try investigating what happens for small values of |S| (e.g. |S| = 2 or 3 etc.) before generalising to |S| = N for arbtitrary N.
Oh I see. The fact that it's symmetric matters a lot here it seems.

How would I communicate it for a generalised n though? Because taking |S|=5 and a,b,c,d,e in S, I'm not sure how to describe it in a way such that (for example) the following case is included

a meets b,c,d
b also meets e.
so f(c)=f(d)
___________________________

Also, I know how to do this but is this type of question really just stars and bars (lecturer didn't actually refer to the terminology)

"For non-negative integers xj, find the number of solutions to x1+...+x6=40"
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Oh I see. The fact that it's symmetric matters a lot here it seems.

How would I communicate it for a generalised n though? Because taking |S|=5 and a,b,c,d,e in S, I'm not sure how to describe it in a way such that (for example) the following case is included

a meets b,c,d
b also meets e.
so f(c)=f(d)
___________________________

Also, I know how to do this but is this type of question really just stars and bars (lecturer didn't actually refer to the terminology)

"For non-negative integers xj, find the number of solutions to x1+...+x6=40"




------------------------------
And yes, last one is just stars and bars.
 
Last edited:

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016





------------------------------
And yes, last one is just stars and bars.
Because it makes no sense for someone to meet nobody and someone else to meet everybody. Makes sense - I saw the link with the injectivity of a function as well but couldn't puzzle everything together. Thanks again :)
__________________________________

Possibly dumb, and last one for now. I got a) out because it's trivially 6^4 but I don't see what I'm doing for part b). Feels also like stars and bars but I just can't see it.

How many outcomes are possible from the roll of four dice
a) if the dice are distinguishable (for example, they are of different colours)
b) if the dice are not distinguishable
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Discrete Maths Sem 2 2016

Because it makes no sense for someone to meet nobody and someone else to meet everybody. Makes sense - I saw the link with the injectivity of a function as well but couldn't puzzle everything together. Thanks again :)
__________________________________

Possibly dumb, and last one for now. I got a) out because it's trivially 6^4 but I don't see what I'm doing for part b). Feels also like stars and bars but I just can't see it.

How many outcomes are possible from the roll of four dice
a) if the dice are distinguishable (for example, they are of different colours)
b) if the dice are not distinguishable


| | • • | • | | • .

So the answer is just found by finding the no. of ways to arrange the picture above (i.e. five bars and four dots).
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: Discrete Maths Sem 2 2016

A course has seven elective topics, and students must complete exactly three of them in order to pass the course. Show that if 200 students passed the course, at least six of them must have completed the same electives.


Please also provide your inspiration as to your train of thought... I'm getting way too lost with how to apply PHP with these problems for some reason...
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: Discrete Maths Sem 2 2016

A course has seven elective topics, and students must complete exactly three of them in order to pass the course. Show that if 200 students passed the course, at least six of them must have completed the same electives.


Please also provide your inspiration as to your train of thought... I'm getting way too lost with how to apply PHP with these problems for some reason...
There are 7C3=35 elective combinations possible. There are 200 students.

200/35 > 5

=> At least one elective combination must be taken by more than 5 students. (I.e at least 6 students.)

Not much inspiration to give, just identify the pigeonholes based on what you are asked to prove (in this case elective combinations), count these pigeonholes, and then divide the total number of pigeons by the number of holes to obtain a rational number r. There must then exist a hole with at least ceil(r) pigeons.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top