Answering any General Mathematics! (1 Viewer)

tomholman

New Member
Hey everyone. I'd love to give back to all those battling through HSC general mathematics this year! I came 12th in the State for General Mathematics in 2014 and I have a whole range of resources and tutoring experience.

Post away!

SudhiTheBoat

Member
Hey man, what was your raw mark? And approximate marks at school if you remember?

DiZ

New Member
Sup! Just a question for probability.
I have the answer, but I have no idea how to do the working out for it.

David chooses five cards from a normal pack of 52 playing cards.
a) How Many different selections are possible (no need to do this one, I get how to do this type of question.)
b) How many five-card hands will have exactly one ace?
c) How many five-card hands will have exactly two aces?

b and c are the ones I can't work out.

Many Thanks.

Xstudying

New Member
Need help in my assignments questions... Anyone? Doing my last assessment for general maths

davidgoes4wce

Well-Known Member
I tutor 5 students in General Maths in 2016. Happy to give tutoring sessions at your place if you need.

Zoinked

Beast
Sup! Just a question for probability.
I have the answer, but I have no idea how to do the working out for it.

David chooses five cards from a normal pack of 52 playing cards.
a) How Many different selections are possible (no need to do this one, I get how to do this type of question.)
b) How many five-card hands will have exactly one ace?
c) How many five-card hands will have exactly two aces?

b and c are the ones I can't work out.

Many Thanks.
Correct me if I'm wrong, its been a while since general lol.
b) You know that there are four aces in a deck and the hand can only have one, which means after choosing your ace, there are 48 cards remaining. 4x48x47x46x45 should give you the answer.
c) 4x3x48x47x46.

That should be how its done (work these out and check the answers).

InteGrand

Well-Known Member
Sup! Just a question for probability.
I have the answer, but I have no idea how to do the working out for it.

David chooses five cards from a normal pack of 52 playing cards.
a) How Many different selections are possible (no need to do this one, I get how to do this type of question.)
b) How many five-card hands will have exactly one ace?
c) How many five-card hands will have exactly two aces?

b and c are the ones I can't work out.

Many Thanks.
Correct me if I'm wrong, its been a while since general lol.
b) You know that there are four aces in a deck and the hand can only have one, which means after choosing your ace, there are 48 cards remaining. 4x48x47x46x45 should give you the answer.
c) 4x3x48x47x46.

That should be how its done (work these out and check the answers).
You've done your answers assuming there's an order that matters for the cards, but for a hand, the order doesn't matter, so we should take this into account.

$\bg_white \noindent b) Pick the Ace: 4 ways.$

$\bg_white \noindent Pick the remaining four cards: We need to pick four cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \binom{48}{4}=\frac{48\times 47\times 46\times 45}{4!} ways.$

$\bg_white \noindent Hence the answer is 4\times \binom{48}{4} = 4\times \frac{48\times 47\times 46\times 45}{4!}.$

$\bg_white \noindent c) Pick the two Aces (from a set of four Aces, with order irrelvant): \binom{4}{2}=\frac{4\times 3}{2!}=6 ways.$

$\bg_white \noindent Pick the remaining three cards: We need to pick three cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \binom{48}{3}=\frac{48\times 47\times 46}{3!} ways.$

$\bg_white \noindent So the answer is \binom{4}{2}\times \binom{48}{3} = 6\times \frac{48\times 47\times 46}{3!} = 48\times 47\times 46 ways.$

fluffchuck

Active Member
You've done your answers assuming there's an order that matters for the cards, but for a hand, the order doesn't matter, so we should take this into account.

$\bg_white \noindent b) Pick the Ace: 4 ways.$

$\bg_white \noindent Pick the remaining four cards: We need to pick four cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \binom{48}{4}=\frac{48\times 47\times 46\times 45}{4!} ways.$

$\bg_white \noindent Hence the answer is 4\times \binom{48}{4} = 4\times \frac{48\times 47\times 46\times 45}{4!}.$

$\bg_white \noindent c) Pick the two Aces (from a set of four Aces, with order irrelvant): \binom{4}{2}=\frac{4\times 3}{2!}=6 ways.$

$\bg_white \noindent Pick the remaining three cards: We need to pick three cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \binom{48}{3}=\frac{48\times 47\times 46}{3!} ways.$

$\bg_white \noindent So the answer is \binom{4}{2}\times \binom{48}{3} = 6\times \frac{48\times 47\times 46}{3!} = 48\times 47\times 46 ways.$
Ooh very nice, I think you should explain to them what the binomial expressions mean (general doesn't cover it).

InteGrand

Well-Known Member
$\bg_white \noindent OK. For non-negative integers n and k\leq n, we can define the \textsl{binomial coefficient} \binom{n}{k} (read n choose k'', with one alternative notation being ^n C_k, which is why you will find a calculator button like this on some calculators) as the number of ways to pick k objects from a set of n objects where order is irrelevant, with the convention \binom{n}{0} = 1. It can be shown using basic combinatorics that it follows that \binom{n}{k} =\frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2)\ldots (n-k+1)}{k!}. Binomial coefficients possess many interesting properties. You can find out more about them here:$

https://en.wikipedia.org/wiki/Binomial_coefficient .

$\bg_white \noindent (The exclamation mark symbol means \textsl{factorial}, which can be defined for positive integers n as n! = n\cdot (n-1)\cdot \cdots \cdot 3\cdot 2\cdot 1. E.g. 4! = 4\times 3\times 2\times 1 = 24. Also, n! is read n factorial'' and the convention is that 0! = 1. Note that n! is the number of ways to permute a set of n objects.)$