ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: FallonXay on April 21, 2017, 12:53:27 pm

Title: Discrete Maths
Post by: FallonXay on April 21, 2017, 12:53:27 pm: Hello!
I'm having a ton of trouble with proofs in discrete maths ( I can barely do any questions :( )
I've posted a few questions below (Q13, 15a, 16 b/c) any help would be greatly appreciated! :)
Thanks.
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 01:40:02 pm: Quote from: FallonXay on April 21, 2017, 12:53:27 pm
Hello!
I'm having a ton of trouble with proofs in discrete maths ( I can barely do any questions :( )
I've posted a few questions below (Q13, 15a, 16 b/c) any help would be greatly appreciated! :)
Thanks.
Remember, in discrete mathematics they are EXTREMELY fussy about how your proofs are set out. For these proofs, I'll reduce my explanation but at the reward of structuring my answer the exact same way I would've structured it in the exam. If you post any more from here onwards, I will focus more on explanation, but my proof will be all over the place and it'll be up to you to reconstruct it logically (keeping any necessary and omitting any extraneous details)

Proofs are long so I'm also split posting.

For Q13, the trick was to identify that the LHS took the form (2n)!/n!
$\text{Let }n \in \mathbb{N}\text{. Then}$
$\begin{align*}(n+1)(n+2)\dots(2n-1)(2n)&=\frac{1.2.3.4\dots (2n-1)(2n)}{1.2.3.4\dots (n-1)(n)}\\ &= \frac{(2n)!}{1.2.3\dots(n-1)(n)}\\ &= \frac{2^n (2n)!}{2^n(1.2.3 \dots (n-1)(n)}\\ &= \frac{2^n [1.2.3.4.5.6\dots(2n-2)(2n-1)(2n)]}{2.4.6\dots (2n-2)(2n)}\\ &=2^n.1.3.5 \dots (2n-1) \end{align*}$
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 02:17:42 pm: $\text{Let }a,n\in \mathbb{Z}^+ \backslash \{1\}$
This proof involves prime numbers. So somewhere in here we will use proof by contrapositive.
$\text{Suppose }a^n-1\text{ is prime}\\ \text{Consider }a^n-1 = (a-1)(a^{n-1}+a^{n-2}+\dots+a+1)$
Note: It is assumed that you know the formula for the sum of a G.P. Note also that the above factorisation is irreducible over Z.
$\text{Since }a^n-1\text{ is prime, one of these factors must be 1}\\ \text{If }a-1 = 1\text{, then }a=2$
$\text{If } a^{n-1}+a^{n-2}+\dots+a+1=1\text{ we have no solution}\\\text{as }n>1\text{ and }a>1 \implies 1+a+a^2+\dots+a^n > 1$
$\text{Hence }a=2.\text{ It falls to prove that }n\text{ is prime.}$
$\text{Suppose the contrapositive: }n\text{ is composite.}\\ \text{Then, }n = km\text{ where }1 < k,m < n$
$a^n-1 = (a^k-1)\left(\left(a^k\right)^{m-1}+\left(a^k\right)^{m-2}+\dots+a^k + 1\right)$
$\text{Since }a=2\text{ we have}\\ 1 = a^1-1 < a^k-1 < a^m -1\\ \text{Hence }a^k-1\text{ is a factor of }a^n-1\text{ that's not }1\text{ or }a^n-1\\ \text{Hence }a^n-1\text{ is composite}$
$\text{As the contrapositive is true}\\ \text{If }a^n-1\text{ is prime, }n\text{ is prime}$
If you just rearrange the formula by multiplying that denominator to the left, you get the factorisation trick I used.
$\text{The converse is that if }a=2\text{ and }n\text{ is prime, then }a^n-1\text{ is prime.}\\ \text{False: Let }a=2\text{ and }n=11\text{ which is prime}\\<br />\text{Then }a^n-1 = 2^{11}-1 = 2047 = 23\times 89\\ \text{So }a^n-1\text{ is composite.}$
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 21, 2017, 02:38:58 pm: Quote from: RuiAce on April 21, 2017, 01:40:02 pm
Remember, in discrete mathematics they are EXTREMELY fussy about how your proofs are set out. For these proofs, I'll reduce my explanation but at the reward of structuring my answer the exact same way I would've structured it in the exam. If you post any more from here onwards, I will focus more on explanation, but my proof will be all over the place and it'll be up to you to reconstruct it logically (keeping any necessary and omitting any extraneous details)

Thanks Rui you life saver!

Also, would you have any tips on knowing when to use contrapositive, direct proofs or contradictions? (Besides a lot of practice :p )
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 02:41:36 pm: $\text{If }n=30\text{, then take }a=5\text{ and }b=6\\ \text{Clearly, }30 \mid 30\\ \text{But }30 \nmid 5\text{ and }30\nmid 6\\ \text{Hence the statement is false}$
________________________________
$\text{In general, we are trying to prove that}\\ \text{There exists }a,b\in \mathbb{Z}\text{ such that }n\mid ab\text{ and }n\nmid a, n\nmid b$
$\text{Suppose }n\text{ is composite. Then we may write}\\ n=km\\ \text{ where }1 < k,m < n$
It may help to stare really hard at the proof in part b) where we had a numeric example.
$\text{Take }a=k\text{ and }b=m\\ \text{Then }n=km = ab = 1\times ab \text{ so }n \mid ab\\ \text{but }n \nmid k \text{ as }k < n\\ \text{and }n \nmid m\text{ as }m < n\\ \text{Hence the result is false.}$
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 02:42:40 pm: Quote from: FallonXay on April 21, 2017, 02:38:58 pm
Thanks Rui you life saver!

Also, would you have any tips on knowing when to use contrapositive, direct proofs or contradictions? (Besides a lot of practice :p )
Hmm.

"Contra-" proofs appear mostly in irrationality and prime/composite numbers. It is advised that you attempt proof by contradiction if you can't figure out why it must be right, but you do know why the negation is just blatantly wrong.
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 21, 2017, 02:57:11 pm: Quote from: RuiAce on April 21, 2017, 01:40:02 pm
$\begin{align*}(n+1)(n+2)\dots(2n-1)(2n)&=\frac{1.2.3.4\dots (2n-1)(2n)}{1.2.3.4\dots (n-1)(n)}\\ &= \frac{(2n)!}{1.2.3\dots(n-1)(n)}\\ &= \frac{2^n (2n)!}{2^n(1.2.3 \dots (n-1)(n)}\\ &= \frac{2^n [1.2.3.4.5.6\dots(2n-2)(2n-1)(2n)]}{2.4.6\dots (2n-2)(2n)}\\ &=2^n.1.3.5 \dots (2n-1) \end{align*}$

In the 4th line, what happened to the 2^n in the denominator?
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 02:58:08 pm: Quote from: FallonXay on April 21, 2017, 02:57:11 pm
In the 4th line, what happened to the 2^n in the denominator?

Note that there are n terms.

I multiplied 2 to EACH of the n terms. This requires n lots of 2s to multiply, i.e. 2^n.

Essentially I expanded it in. Note how the denominator goes from $1.2.3\dots (n-1)(n)$ to $2.4.6\dots (2n-2)(2n)$
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 21, 2017, 03:03:49 pm: Quote from: RuiAce on April 21, 2017, 02:58:08 pm
Note that there are n terms.

I multiplied 2 to EACH of the n terms. This requires n lots of 2s to multiply, i.e. 2^n.

Essentially I expanded it in. Note how the denominator goes from $1.2.3\dots (n-1)(n)$ to $2.4.6\dots (2n-2)(2n)$

Sorry, I don't quite understand how you expanded it? wouldn't you have to multiply it out by 2^n to get (2^n) x (2 x 2^n) x ... x (n x 2^n)?
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 03:11:53 pm: Quote from: FallonXay on April 21, 2017, 03:03:49 pm
Sorry, I don't quite understand how you expanded it? wouldn't you have to multiply it out by 2^n to get (2^n) x (2 x 2^n) x ... x (n x 2^n)?
$\begin{align*}2^n.1.3.5\dots (n-1)(n) &= \underbrace{2.2.2. \dots .2.2}_{n\text{ terms}}.1.3.5\dots (n-1)(n)\\ &= (2\times 1)(2\times 3)(2\times 5)\dots (2\times (n-1))(2\times n)\tag{rearranging}\\ &= 2.4.6\dots (2n-2)(2n)\end{align*}$
I only multiplied 2 to every term. Not the bulky 2^n to every term.
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 21, 2017, 03:13:47 pm: Quote from: RuiAce on April 21, 2017, 03:11:53 pm
$\begin{align*}2^n.1.3.5\dots (n-1)(n) &= \underbrace{2.2.2. \dots .2.2}_{n\text{ terms}}.1.3.5\dots (n-1)(n)\\ &= (2\times 1)(2\times 3)(2\times 5)\dots (2\times (n-1))(2\times n)\tag{rearranging}\\ &= 2.4.6\dots (2n-2)(2n)\end{align*}$

Ohhhhhh, very nice. Thanks :D
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 21, 2017, 04:01:46 pm: Hello,

how would you prove the following question attached?

(Thanks in advance!)
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 21, 2017, 04:15:04 pm: Quote from: FallonXay on April 21, 2017, 04:01:46 pm
Hello,

how would you prove the following question attached?

(Thanks in advance!)
This one requires a bit of skillful thinking and preferably an abuse of notation to go with it.

(The chapter and question reference is given in your screenshot. If there's any line in the proof they gave that troubles you feel free to bring it up.)
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 22, 2017, 10:17:06 am: Quote from: RuiAce on April 21, 2017, 04:15:04 pm
This one requires a bit of skillful thinking and preferably an abuse of notation to go with it.

(The chapter and question reference is given in your screenshot. If there's any line in the proof they gave that troubles you feel free to bring it up.)

I'm assuming the '10' in the answers are factorising out the decimal places? But how do they know what powers to put it to? (i.e 10^m, 10^(r-1), etc)?
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 22, 2017, 10:40:54 am: Quote from: FallonXay on April 22, 2017, 10:17:06 am
I'm assuming the '10' in the answers are factorising out the decimal places? But how do they know what powers to put it to? (i.e 10^m, 10^(r-1), etc)?
The idea is that in general (i.e. not always), a decimal number that is rational has one part that is terminating, and a part after it that's recurring.

E.g. $\frac1{12}=0.083333333\dots=0.08\overline{3}$
The $0.08$ part is the terminating part
The $0.00\overline{3}$ is the recurring part

m and r are integers made to distinguish the purpose. Note that this is a proof where you would say let m and n be blah, not 'suppose' m and n be blah.

I believe that the question wants you to take the first r decimal places (or r-1, you can figure that little thing out) to be the terminating part, and the NEXT m-r decimal places to be the recurring part.
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 22, 2017, 10:49:20 am: Quote from: RuiAce on April 22, 2017, 10:40:54 am
The idea is that in general (i.e. not always), a decimal number that is rational has one part that is terminating, and a part after it that's recurring.

E.g. $\frac1{12}=0.083333333\dots=0.08\overline{3}$
The $0.08$ part is the terminating part
The $0.00\overline{3}$ is the recurring part

m and r are integers made to distinguish the purpose. Note that this is a proof where you would say let m and n be blah, not 'suppose' m and n be blah.

I believe that the question wants you to take the first r decimal places (or r-1, you can figure that little thing out) to be the terminating part, and the NEXT m-r decimal places to be the recurring part.

Ahh, I see ~ thanks.

Also, how do you do a) and d) of this question? (For d, I'm not really suer about how to use the idea of the if then in the second part - i.e the relevance of 2^(4n+7) )
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 22, 2017, 10:53:06 am: Quote from: FallonXay on April 22, 2017, 10:49:20 am
Ahh, I see ~ thanks.

Also, how do you do a) and d) of this question? (For d, I'm not really suer about how to use the idea of the if then in the second part - i.e the relevance of 2^(4n+7) )

Hint: Induction.

Edit: Oh sorry missed part a)

Part a) hint is proof by exhaustion
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 22, 2017, 11:02:13 am: Quote from: RuiAce on April 22, 2017, 10:53:06 am
Hint: Induction.

Edit: Oh sorry missed part a)

Part a) hint is proof by exhaustion

Oooooh! okay, thanks :D
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 22, 2017, 09:10:22 pm: Hello :)

How would you do this question?

(Thanks again!)
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 22, 2017, 10:09:47 pm: Quote from: FallonXay on April 22, 2017, 09:10:22 pm
Hello :)

How would you do this question?

(Thanks again!)
$\text{Part a) is just substitution}\\ \text{If }x\equiv 3\mod 4$
$\begin{align*}x^3+2x-1 &\equiv 3^3+2(3)-1 \mod 4\\ &=32\\ &\equiv 0 \mod 4\end{align*}$
________________
$\text{The contrapositive is}\\ \text{If }x^3+2x-1\not\equiv 0\mod 4\\ \text{then }x\not\equiv 3 \mod 4$
$\text{The contrapositive of any known true statement is always true as well}$
________________
$\text{The converse is}\\ \text{If }x^3+2x-1 \equiv 0\mod 4\\ \text{then }x\equiv 3\mod 4$
$\text{This is true - We may check that if }x\equiv 0,1,2\mod 4\\ \text{Then }x^3-2x+1\not\equiv 0\mod 4$
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 23, 2017, 07:32:16 am: Quote from: RuiAce on April 22, 2017, 10:09:47 pm
$\text{Part a) is just substitution}\\ \text{If }x\equiv 3\mod 4$
$\begin{align*}x^3+2x-1 &\equiv 3^3+2(3)-1 \mod 4\\ &=20\\ &\equiv 0 \mod 4\end{align*}$
________________
$\text{The contrapositive is}\\ \text{If }x^3+2x-1\not\equiv 0\mod 4\\ \text{then }x\not\equiv 3 \mod 4$
$\text{The contrapositive of any known true statement is always true as well}$
________________
$\text{The converse is}\\ \text{If }x^3+2x-1 \equiv 0\mod 4\\ \text{then }x\equiv 3\mod 4$
$\text{This is false, as if we take }x=1\\ \text{We have }x^3+2x-1\equiv 0\mod 4\\ \text{but clearly }x\equiv 1\not\equiv 3 \mod 4$

x = 1 doesn't give 0 though? So would you just do a proof by cases? (Test, 0, 1, 2, 3 mod4)?

Also, for part a, how come you can just substitute it in? Is that just a property of congruency?

Thanks!
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 23, 2017, 07:35:06 am: Quote from: FallonXay on April 23, 2017, 07:32:16 am
x = 1 doesn't give 0 though? So would you just do a proof by cases? (Test, 0, 1, 2, 3 mod4)?

Also, for part a, how come you can just substitute it in? Is that just a property of congruency?

Thanks!
My bad, I think I messed up some of my mental arithmetic because it would be 2 mod 4.
In that case, upon exhausting all the cases we see that the converse is true.

Yes, you were taught several properties of congruence back in topic 2.
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on April 23, 2017, 12:24:08 pm: Quote from: RuiAce on April 21, 2017, 02:17:42 pm
$\text{Suppose the contrapositive: }n\text{ is composite.}\\ \text{Then, }n = km\text{ where }1 < k,m < n$
$a^n-1 = (a^k-1)\left(\left(a^k\right)^{m-1}+\left(a^k\right)^{m-2}+\dots+a^k + 1\right)$
$\text{Since }a=2\text{ we have}\\ 1 = a^k-1 < a^k-1 < a^k -1\\ \text{Hence }a^k-1\text{ is a factor of }a^n-1\text{ that's not }1\text{ or }a^n-1\\ \text{Hence }a^n-1\text{ is composite}$

Heyy, just looking over this proof. In this line:
Quote from: RuiAce on April 21, 2017, 02:17:42 pm
$a^n-1 = (a^k-1)\left(\left(a^k\right)^{m-1}+\left(a^k\right)^{m-2}+\dots+a^k + 1\right)$

How come it's 'a' to the power of 'k' to the power of '(m-1)'. Don't we just substitute 'km' into the place where there was originally an 'n' to give k to the power of (km - 1)?

Also, could you clarify this section? I'm a little confused as to what's happening here:

Quote from: RuiAce on April 21, 2017, 02:17:42 pm
$\text{Since }a=2\text{ we have}\\ 1 = a^k-1 < a^k-1 < a^k -1\\ \text{Hence }a^k-1\text{ is a factor of }a^n-1\text{ that's not }1\text{ or }a^n-1\\ \text{Hence }a^n-1\text{ is composite}$

Thanks! :)
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on April 23, 2017, 01:00:20 pm: Quote from: FallonXay on April 23, 2017, 12:24:08 pm
Heyy, just looking over this proof. In this line:
How come it's 'a' to the power of 'k' to the power of '(m-1)'. Don't we just substitute 'km' into the place where there was originally an 'n' to give k to the power of (km - 1)?

Also, could you clarify this section? I'm a little confused as to what's happening here:

Thanks! :)

That last one had typo's in it. One of those k's should've been a 1 and the other an m. Fixed in the original post.
___________________
$a^n-1=a^{km}-1=\left(a^k\right)^m-1 = (a^k-1)\left((a^k)^{m-1}+(a^k)^{m-2}+\dots+(a^k)+1\right)$
$\text{Of course, we could've done}\\ a^{km}-1 = (a-1)(a^{km-1}+a^{km-2}+\dots+a+1)\\ \text{But this is useless. So we don't}$
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on May 01, 2017, 07:23:26 pm: Question 2 of Discrete Exam Working Out

(http://i.imgur.com/zFFQOAe.jpg)
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on May 01, 2017, 07:43:15 pm: Quote from: FallonXay on May 01, 2017, 07:23:26 pm
Question 2 of Discrete Exam Working Out

(http://i.imgur.com/zFFQOAe.jpg)
"Rationality" is just a colloquial adjective with no meaning. The set of rational numbers is what is closed under the four standard operations. "Taking integer powers" is more appropriate than "powers" by itself (note the consequences of raising to the power of pi, for example).

The flow of the proof is fine.
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on May 01, 2017, 07:47:53 pm: Quote from: RuiAce on May 01, 2017, 07:43:15 pm
"Rationality" is just a colloquial adjective with no meaning. The set of rational numbers is what is closed under the four standard operations. "Taking integer powers" is more appropriate than "powers" by itself (note the consequences of raising to the power of pi, for example).

The flow of the proof is fine.

Ahhk, cheers! Will fix that up :)
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on May 01, 2017, 07:52:02 pm: It may also be worth noting that closure under + and x is sufficient. This is because if we want to minus, we simply add the negative. Closure under division is also dangerous in that this may imply we may divide by 0, so closure under multiplication is better (just multiply by a multiplicative inverse, e.g. 1/2).
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on May 03, 2017, 08:36:49 am: Quote from: RuiAce on May 01, 2017, 07:52:02 pm
It may also be worth noting that closure under + and x is sufficient. This is because if we want to minus, we simply add the negative. Closure under division is also dangerous in that this may imply we may divide by 0, so closure under multiplication is better (just multiply by a multiplicative inverse, e.g. 1/2).

Oooh, very true. But if I only state closure under + and x, should I explicitly state in my proof multiplication by 1/2 instead of multiplication by 2 and addition of -a and - b instead of subtraction of a and b?
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on May 03, 2017, 08:45:16 am: Quote from: FallonXay on May 03, 2017, 08:36:49 am
Oooh, very true. But if I only state closure under + and x, should I explicitly state in my proof multiplication by 1/2 instead of multiplication by 2 and addition of -a and - b instead of subtraction of a and b?
Yes
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on May 21, 2017, 12:26:33 pm: Counting & combinatorics
Q25 h?
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on May 21, 2017, 12:54:59 pm: Quote from: FallonXay on May 21, 2017, 12:26:33 pm
Counting & combinatorics
Q25 h?
$\text{Taking }y_2 = x_2-5,\, y_3 = x_3-10\text{ and }y_j=x_j\text{ otherwise}\\ \text{We are now interested in the number of integral solutions to}\\ y_1+\dots+y_6=25$
$\text{Subject to }0\le y_1,y_2,y_3\le 9\\ \text{ and }y_4,y_5,y_6\ge 0$
This last part of this bundle was definitely the hardest of them all and my tutor didn't finish it off. I don't fully remember everything either so there's no guarantee that my final answer will be correct.
$\text{Consistency cannot be kept with any substitution, so we should consider the complementary event}\\ \text{Note that for the sake of stars and bars we will always hold }y_i \ge 0\\ \text{Hence, the complement event is when at least one of }y_1,y_2,y_3\text{ exceeds 9}$
________________
$\text{Note from above that the total number of outcomes is }\binom{30}5$
________________
$\text{For a specific case where one exceeds 9, take }y_1 = z_1 + 9\\ \text{Then we have }z_1+y_2+\dots+y_6 = 16\\ \text{This has }\binom{21}{5}\text{ solutions, and we have }\binom{3}{1}\text{ choices for the }y\text{'s to 'raise' by 9}$
________________
$\text{For a specific case where two exceeds 9, take the above and furthermore }y_2=z_2+9\\\text{Then we have }z_1+z_2+y_3+\dots+y_6=7\\ \text{This has }\binom{12}{5}\text{ solutions, and we have }\binom{3}{2}\text{ choices for the 'raise'}$
________________
$\text{Clearly we can't have them all exceeding 9 as then }\\z_1+z_2+z_3+y_4+y_5+y_6=-2\\ \text{which is a contradiction}$
________________
$\text{Hence, upon regrouping and applying the inclusion-exclusion principle}\\ \text{Favourable outcomes} = \binom{30}{5} - \binom{21}{5}\binom{3}{1}+\binom{12}{5}\binom{3}2$

Also I suggest renaming your thread to just discrete maths if you're gonna put questions from all topics here
Title: Re: Discrete Maths - Proofs Questions
Post by: FallonXay on May 21, 2017, 02:08:42 pm: Quote from: RuiAce on May 21, 2017, 12:54:59 pm
$\text{Subject to }0\le y_1,y_2,y_3\le 9\\ \text{ and }y_4,y_5,y_6\ge 0$

How did you come to this conclusion for the part quoted?

Also, I'm not sure how to rename a thread, sorry :/
Title: Re: Discrete Maths - Proofs Questions
Post by: RuiAce on May 21, 2017, 02:13:26 pm: Quote from: FallonXay on May 21, 2017, 02:08:42 pm
How did you come to this conclusion for the part quoted?

Also, I'm not sure how to rename a thread, sorry :/
$x_1\le 9\text{ was given}\\ \text{So if }y_1=x_1\text{ then }y_1\le 9$
$5\le x_2\le 14\implies 0\le x_2-5\le 9\\ \text{so if }y_2=x_2-5\\ 0\le y_2\le 9$
$\text{etc.}$
$\text{Basically, we just modify the restrictions originally given}\\ \text{to suit our new variables}$
Just rename it in the opening post
Title: Re: Discrete Maths
Post by: FallonXay on May 21, 2017, 02:56:11 pm: Quote from: RuiAce on May 21, 2017, 02:13:26 pm
$x_1\le 9\text{ was given}\\ \text{So if }y_1=x_1\text{ then }y_1\le 9$
$5\le x_2\le 14\implies 0\le x_2-5\le 9\\ \text{so if }y_2=x_2-5\\ 0\le y_2\le 9$
$\text{etc.}$
$\text{Basically, we just modify the restrictions originally given}\\ \text{to suit our new variables}$
Just rename it in the opening post

Ohhhhh! That makes a lot of sense!!! Thanks!

(Btw, the total combinations are C(30,5) because there are 25 'dots' and 5 'bars' (addition signs) which equals 30 and you choose 5 places to put the bars. Also for the specific case where one of the y's exceed 9 it should be 3 x C(20,5) because you need a minimum of 10 dots in the y to exceed 9 i.e y₂ = z₂ + 10 and similarly for the intersection of two y's should be 3 x C(10,5), etc.)

And I changed the thread name, thanks :)
Title: Re: Discrete Maths
Post by: RuiAce on May 21, 2017, 03:13:25 pm: Quote from: FallonXay on May 21, 2017, 02:56:11 pm
Ohhhhh! That makes a lot of sense!!! Thanks!

(Btw, the total combinations are C(30,5) because there are 25 'dots' and 5 'bars' (addition signs) which equals 30 and you choose 5 places to put the bars. Also for the specific case where one of the y's exceed 9 it should be 3 x C(20,5) because you need a minimum of 10 dots in the y to exceed 9 i.e y₂ = z₂ + 10 and similarly for the intersection of two y's should be 3 x C(10,5), etc.)

And I changed the thread name, thanks :)
Yep I agree with those corrections. I'll probably fix them another time though