HSC 2016 MX2 Combinatorics Marathon (archive) (1 Viewer)

VBN2470 · Nov 5, 2015

This thread is to encourage you to post all your harder Probability + Permutations & Combinations questions here.

Some Things to Consider:

1. Please answer one question at a time.

2. When posting a new question, please make it clear that it is a new question (write in capital letters 'NEW QUESTION' or bold/italicise the text) so it clear to others that a new question has been posted (Credits to DrSoccerBall for this suggestion

)

3. Discuss your approach(es) to the questions, there are always many ways to answering a combinatorics problem and it would be good if everyone can benefit from this.

4. Since harder questions are being posted, it will naturally follow that a lot of the problems require a lot more reasoning & argument, so make sure you do your best to be clear in explaining your reasoning.

5. Respect each other at all times and have fun learning

To start off:
---------------------------------------------------------------------
Words of 12 letters are constructed from the English alphabet.
How many of these words
(a) contain the sub word 'MATHS' twice?
(b) do not contain the sub word 'MATHS' at all?
---------------------------------------------------------------------

calamebe · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

I may have got this wrong, Perms and Combs is my worst topic, but here's what I got: https://imgur.com/QFNuhjR

InteGrand · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

calamebe said:
I may have got this wrong, Perms and Combs is my worst topic, but here's what I got: https://imgur.com/QFNuhjR

$Note that for (a), the arrangements can be like this too:$

$$\text{MATHS}\, \text{MATHS}\, *\, *$$

$$* \,\text{MATHS} \, * \text{MATHS}$$

$and other arrangements, where * represents an arbitrary letter from the alphabet.$

VBN2470 · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

calamebe said:
I may have got this wrong, Perms and Combs is my worst topic, but here's what I got: https://imgur.com/QFNuhjR

For (a) you are close, but you forgot to consider all the possibilities in which 'MATHS' can go. Treat 'MATHS' as one letter, then you have 4 places to arrange your 'letters' including the sub word 'letter' 'MATHS'. Pick 2 places in which 'MATHS' can go, this can be in $C(4,2)=6$ ways then arrange the remaining two letters in $26^2$ ways, thus total number of ways is $26^2 \times C(4,2)$ ways.

T-R-O-L-O-L · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

This looks like it's going to be a great thread!

lita1000 · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

for b) should be something like 26^12-(8C1*26^7-answer from part a)

couldve made errors, literally walking home from the station with no pen or paper

Drsoccerball · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

VBN I think you should also put in the rules that questions are to be either bolded or surrounded in underscores/scores so we can see the question better as combinations don't require Latex. Also should writing the answer in white be something in this thread or no ?

VBN2470 · Nov 5, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Drsoccerball said:
VBN I think you should also put in the rules that questions are to be either bolded or surrounded in underscores/scores so we can see the question better as combinations don't require Latex. Also should writing the answer in white be something in this thread or no ?

Good point, I'll get that fixed. Umm, about the answers, putting in white coloured font (or in a spoiler box) is optional but because people will be explaining how they get to answers in the first place, I'd say it's not necessary. But still a good option.

leehuan · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Long-winded question (DO NOT CHEAT

):

a) 5 cards are drawn from a typical 52-card deck (i.e. without jokers). How many possible sets of cards can be drawn?
b) There are a variety of hands that can be dealt in a 5 card poker hand. Determine the probability of every type hand being drawn, given what the hands actually are, and that no types of hands overlap (i.e. if it is classified as a straight flush, it doesn't constitute as a straight anymore).
(i) Royal Flush - 10-J-Q-K-A of one suit
(ii) Straight Flush - All cards are of same suit; rank of all cards form a sequence (e.g. 7-8-9-10-J)
(iii) Four of a Kind - Four cards of identical rank
(iv) Full House - Three cards of identical rank, two cards of a different identical rank
(v) Flush - All cards are of same suit
(vi) Straight - Rank of all cards form a sequence
(vii) Three of a kind - Three cards of identical rank
(viii) Two pair - Two cards of identical rank, another two cards of identical rank
(ix) One pair - Two cards of identical rank
(x) High card - No link between cards' rank and suit
c) Max and Scarlett were both extremely lucky people, and both received a four of a kind. Nobody else got a straight flush or the royal flush. What is the probability that Max beat Scarlett?

Glossary:
Suit - Category of the card, determined using the symbol of the card. (Spades, Hearts, Clubs, Diamonds)
Rank - Value of the card, from lowest to highest - 2,3,4,5,6,7,8,9,10,J,Q,K,A

Italics - Skippable; they're archived in last year's 3U marathon.

Ambility · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

leehuan said:
Long-winded question (DO NOT CHEAT ):

a) 5 cards are drawn from a typical 52-card deck (i.e. without jokers). How many possible sets of cards can be drawn?
b) There are a variety of hands that can be dealt in a 5 card poker hand. Determine the probability of every type hand being drawn, given what the hands actually are, and that no types of hands overlap (i.e. if it is classified as a straight flush, it doesn't constitute as a straight anymore).
(i) Royal Flush - 10-J-Q-K-A of one suit
(ii) Straight Flush - All cards are of same suit; rank of all cards form a sequence (e.g. 7-8-9-10-J)
(iii) Four of a Kind - Four cards of identical rank
(iv) Full House - Three cards of identical rank, two cards of a different identical rank
(v) Flush - All cards are of same suit
(vi) Straight - Rank of all cards form a sequence
(vii) Three of a kind - Three cards of identical rank
(viii) Two pair - Two cards of identical rank, another two cards of identical rank
(ix) One pair - Two cards of identical rank
(x) High card - No link between cards' rank and suit
c) Max and Scarlett were both extremely lucky people, and both received a four of a kind. Nobody else got a straight flush or the royal flush. What is the probability that Max beat Scarlett?

Glossary:
Suit - Category of the card, determined using the symbol of the card. (Spades, Hearts, Clubs, Diamonds)
Rank - Value of the card, from lowest to highest - 2,3,4,5,6,7,8,9,10,J,Q,K,A

Italics - Skippable; they're archived in last year's 3U marathon.

$$For iv, is it P(Full House)$ =\frac{4C1\times 13C3 \times 3C1 \times 13C2}{2598960}=\frac{429}{4165}$ ?$

InteGrand · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Ambility said:
$$For iv, is it P(Full House)$ =\frac{52\times12\times11\times39\times12}{2598960}=\frac{5148}{4165}$ ?$

EDIT: I just realised that's more than one, lol.

You also apparently got a probability greater than 1 – a sign there's something wrong somewhere in the working.

Sy123 · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

$\\ $For coins$ \ c_1, c_2, \dots , c_n \ $we have probability$ \ c_i \ $returning heads is$ \ p_i \ $for all$ \ i = 1,2, \dots , n$

$\\ $Let the probability that the first$ \ m \ $coins return$ \ k \ $coins be$ \ s(m,k)$

$\\ $Find a recurrence relationship for$ \ s(m,k)$

Ambility · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

InteGrand said:
You also apparently got a probability greater than 1 – a sign there's something wrong somewhere in the working.

Yeah, I changed it.

leehuan · Nov 6, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Ambility said:
$$For iv, is it P(Full House)$ =\frac{4C1\times 13C3 \times 3C1 \times 13C2}{2598960}=\frac{429}{4165}$ ?$

You have 13 different ranks. You pick one to be the triple than choose 3 out of 4 from a suit. And similarly for the pair
So the favourable outcomes is actually 13C1*4C3 * 12C1*4C2

leehuan · Nov 13, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

$\\ $One vowel and any 10 other letters are chosen from the word 'SUPERCALIFRAGILISTICEXPIALIDOCIOUS' to make 'words'. How many possible arrangements can there be? $$

(No I don't have the answer)

Vowels: A, E, I, O, U

Drsoccerball · Nov 13, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

leehuan said:
$\\ $One vowel and any 10 other letters are chosen from the word '[' to make 'words'. How many possible arrangements can there be? $$

(No I don't have the answer)

Vowels: A, E, I, O, U

Very very very very very uselessly long...

Drsoccerball · Nov 13, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Drsoccerball said:
Very very very very very uselessly long...

The question can be reasonable if the words you're making is <6 letters.

braintic · Nov 14, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

leehuan said:
$\\ $One vowel and any 10 other letters are chosen from the word 'SUPERCALIFRAGILISTICEXPIALIDOCIOUS' to make 'words'. How many possible arrangements can there be? $$

(No I don't have the answer)

Vowels: A, E, I, O, U

My question - aren't you too young to have heard this word?

Sy123 · Nov 14, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

Sy123 said:
$\\ $For coins$ \ c_1, c_2, \dots , c_n \ $we have probability$ \ c_i \ $returning heads is$ \ p_i \ $for all$ \ i = 1,2, \dots , n$

$\\ $Let the probability that the first$ \ m \ $coins return$ \ k \ $coins be$ \ s(m,k)$

$\\ $Find a recurrence relationship for$ \ s(m,k)$

$s(m,k) = c_ms(m-1,k) + (1-c_m) s(m-1,k-1)$

Drsoccerball · Nov 14, 2015

Re: HSC 2016 MX2 Combinatorics Marathon

braintic said:
My question - aren't you too young to have heard this word?

I heard this word in like year 2

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