MATH1081 Discrete Maths (2 Viewers)

RenegadeMx · Sep 8, 2016

Re: Discrete Maths Sem 2 2016

did u check for when n=1? first of all

Drsoccerball · Sep 8, 2016

Re: Discrete Maths Sem 2 2016

Do cases and with the odd part do another induction to prove it.

Edit: That is write the statement first assuming k is even and solve, and then another case where k is odd and solve.

leehuan · Sep 8, 2016

Re: Discrete Maths Sem 2 2016

RenegadeMx said:
did u check for when n=1? first of all

Obviously lol

Drsoccerball said:
Do cases and with the odd part do another induction to prove it.

Edit: That is write the statement first assuming k is even and solve, and then another case where k is odd and solve.

Yeah I realised that the sign on the last term could've been clearly stated after I read the textbook. But I managed to get it out using strong induction anyway so I'll leave it. (That being said, for strong induction I also separated odd and even :/ )

leehuan · Sep 9, 2016

Re: Discrete Maths Sem 2 2016

Translate the following compound statement into symbolic notation:

Either going to bed or going swimming is a sufficient condition for changing clothes; however, changing clothes does not mean going swimming.

So I let b = going to bed
s = going swimming
l = changing clothes

$\text{So far I got up to this...}\\ \left[ (b \vee s ) \to \ell \right] \wedge \left[ \right]$

And then I got stuck. I have no idea how to translate the second statement.

Orthos · Sep 10, 2016

Re: Discrete Maths Sem 2 2016

"Changing clothes does not mean going swimming" ≡ "changing clothes means not going swimming"

RealiseNothing · Sep 10, 2016

Re: Discrete Maths Sem 2 2016

Orthos said:
"Changing clothes does not mean going swimming" ≡ "changing clothes means not going swimming"

How is this right at all?

InteGrand · Sep 10, 2016

Re: Discrete Maths Sem 2 2016

Orthos said:
"Changing clothes does not mean going swimming" ≡ "changing clothes means not going swimming"

Those would mean two different things.

One of those (the first one you wrote) is "A does not imply B", the other is "A implies not B" (assuming 'means' refers to 'implies'. It may have instead referred to something like 'is equivalent to'.) .These are different.

Drsoccerball · Sep 10, 2016

Re: Discrete Maths Sem 2 2016

Would the second statement be: l -> (b V s) ?

leehuan · Sep 10, 2016

Re: Discrete Maths Sem 2 2016

Drsoccerball said:
Would the second statement be: l -> (b V s) ?

Thing is "bed" doesn't actually appear in the second statement

Paradoxica · Sep 13, 2016

Re: Discrete Maths Sem 2 2016

Find the solution to this recurrence relation WITHOUT induction.

Orthos · Sep 14, 2016

Re: Discrete Maths Sem 2 2016

RealiseNothing said:
How is this right at all?

Still feels right

RealiseNothing · Sep 14, 2016

Re: Discrete Maths Sem 2 2016

Orthos said:
Still feels right

They are two different things as integrand explained.

leehuan · Sep 15, 2016

Re: Discrete Maths Sem 2 2016

leehuan said:
Translate the following compound statement into symbolic notation:

Either going to bed or going swimming is a sufficient condition for changing clothes; however, changing clothes does not mean going swimming.

So I let b = going to bed
s = going swimming
l = changing clothes

$\text{So far I got up to this...}\\ \left[ (b \vee s ) \to \ell \right] \wedge \left[ \right]$

And then I got stuck. I have no idea how to translate the second statement.

$(b \vee s \to \ell)\wedge [\neg (\ell \to s)]\\ \iff (\neg(b\vee s)\vee \ell)\wedge [\neg(\neg \ell \vee s)]\\ \iff [(\neg b \wedge \neg s)\vee\ell]\wedge(\ell \wedge \neg s)$
Not bothered right now to continue because the brackets are messy. A bit too much commutative and associative required.

Orthos · Sep 15, 2016

Re: Discrete Maths Sem 2 2016

RealiseNothing said:
They are two different things as integrand explained.

Understood, now. It took me a very long time to understand... what's wrong with me?!

leehuan · Sep 17, 2016

Re: Discrete Maths Sem 2 2016

Three letters are chosen out of SATURDAY to make words. How many possible words?

InteGrand · Sep 17, 2016

Re: Discrete Maths Sem 2 2016

leehuan said:
Three letters are chosen out of SATURDAY to make words. How many possible words?

$\noindent Note there is a pair of A's and the other letters are all single letters. Total of $8$ letters, consisting of $2$ A's and $6$ single letters. If we take both A's, the no. of words we could form is $3\times 6 = 18$. If we take exactly one A, we can form $\binom{6}{2} \times 3! = 15 \times 6 = 90$ words. If we take no A's, we can form $^6P_3 = 6\times 5\times 4 = 120$ words. Hence the answer is $18 + 90 + 120 = 228$ words.$

leehuan · Sep 17, 2016

Re: Discrete Maths Sem 2 2016

InteGrand said:
$\noindent Note there is a pair of A's and the other letters are all single letters. Total of $8$ letters, consisting of $2$ A's and $6$ single letters. If we take both A's, the no. of words we could form is $3\times 6 = 18$. If we take exactly one A, we can form $\binom{6}{2} \times 3! = 15 \times 6 = 90$ words. If we take no A's, we can form $^6P_3 = 6\times 5\times 4 = 120$ words. Hence the answer is $18 + 90 + 120 = 228$ words.$

Ah I get it

for some reason I tried to pick instead of choose 2 out of 6 I think...
_______________________

Consider the following.
“Problem: How many eight-card hands chosen from a standard pack have at least one suit missing?

Solution: Throw out one entire suit (4 possibilities), then select 8 of the remaining 39 cards.

The number of hands is 4³⁹C₈.”

a) What is wrong with the given solution?

So I know that there's been double counting and one way of approaching it is through inclusion-exclusion (which is part b) - do it correctly). But I'm not sure how to properly explain the double counting as I can't quite tell what's been double counted

InteGrand · Sep 17, 2016

Re: Discrete Maths Sem 2 2016

leehuan said:
Ah I get it for some reason I tried to pick instead of choose 2 out of 6 I think...
_______________________

Consider the following.
“Problem: How many eight-card hands chosen from a standard pack have at least one suit missing?

Solution: Throw out one entire suit (4 possibilities), then select 8 of the remaining 39 cards.

The number of hands is 4³⁹C₈.”

a) What is wrong with the given solution?

So I know that there's been double counting and one way of approaching it is through inclusion-exclusion (which is part b) - do it correctly). But I'm not sure how to properly explain the double counting as I can't quite tell what's been double counted

The given solution counts |S1| + |S2| + |S3| + |S4|, where Sj is the set of eight-card hands so that there is no card from suit j in the hand (j = 1,2,3,4).

But what we want is: |S1 U S2 U S3 U S4| (which is the no. of eight-card hands with suit 1 missing or suit 2 missing or ... or suit 4 missing (as usual 'or' meaning inclusive or).

The given answer would only generally equal the size of this union if the sets Sj were disjoint sets, but they are clearly not disjoint here, hence the answer given is incorrect.

(To see they aren't disjoint, note S1 and S2 both include a hand where all eight cards are from suit 3 for example. This corresponds to the given answer counting more than once things like the eight cards being cards #1,2,...,8 from suit 3. So it's double counting a lot of things.)

leehuan · Sep 18, 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)

RenegadeMx · Sep 18, 2016

Re: Discrete Maths Sem 2 2016

leehuan said:
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)

lol i remember that q, its surprisingly hard just give to tutor

MATH1081 Discrete Maths (2 Viewers)

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