Official BOS Trial 2016 Thread (5 Viewers)

porcupinetree · Sep 28, 2016

did someone say hsp

RCbest · Sep 28, 2016

i wanna sit the 4u exam btw

jamespaek · Sep 28, 2016

I'm in ext 2

DatAtarLyfe · Sep 28, 2016

porcupinetree said:
did someone say hsp

depends, were you haram dingo that gets tomato sauce?

RCbest · Sep 28, 2016

Yeeeee boiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

porcupinetree · Sep 28, 2016

DatAtarLyfe said:
depends, were you haram dingo that gets tomato sauce?

I think you're confusing me with @ekman

holy trinity for lyf

DatAtarLyfe · Sep 28, 2016

porcupinetree said:
I think you're confusing me with @ekman

holy trinity for lyf

well in that case you're all g.

Drsoccerball · Sep 28, 2016

For those who can't do the question here is what i did (may be wrong) :

a) First we choose k numbers out of the n and the out of the remaining n - k numbers we chose m - k of them since total selected is m.

$\therefore \binom{n}{k} \binom{n-k}{m-k}$

b) Consider all possible combinations. Firstly if bob selects 0 cards, $\binom{n}{0} \binom{n}{m}$ then when bob selects 1 card, $\binom{n}{1} \binom{n-1}{m}$ , and so on up to when he picks all m.

$\therefore $ Total possibilities : $ \binom{n}{0} \binom{n}{m} + \binom{n}{1} \binom{n-1}{m} +... + \binom{n}{m} \binom{n-m}{0}$

$However you can also count in a different way$

$\binom{n}{m} (2 \cdot 2 \cdot ... \cdot 2) = 2^m \binom{n}{m} $ Since there is two possibilities to put each card and there are m cards. $$

$EDIT: So first we pick the m cards and then decide whether or not to put them in one of two places.$
[]

Carrotsticks · Sep 28, 2016

Drsoccerball said:
For those who can't do the question here is what i did (may be wrong) :

a) First we choose k numbers out of the n and the out of the remaining n - k numbers we chose m - k of them since total selected is m.

$\therefore \binom{n}{k} \binom{n-k}{m-k}$

b) Consider all possible combinations. Firstly if bob selects 0 cards, $\binom{n}{0} \binom{n}{m}$ then when bob selects 1 card, $\binom{n}{1} \binom{n-1}{m}$ , and so on up to when he picks all m.

$\therefore $ Total possibilities : $ \binom{n}{0} \binom{n}{m} + \binom{n}{1} \binom{n-1}{m} +... + \binom{n}{m} \binom{n-m}{0}$

$However you can also count in a different way$

$\binom{n}{m} *(2+2+... + 2) = 2^m \binom{n}{m} $ Since there is two possibilities to put each card and there are m cards. $$
EDIT: So first we pick the m cards and then decide whether or not to put them in one of two places.
[]

Mostly correct, except the last line should have the 2's multiplied instead of added. Well done.

leehuan · Sep 28, 2016

Drsoccerball said:
$\binom{n}{m} *(2+2+... + 2) = 2^m \binom{n}{m} $ Since there is two possibilities to put each card and there are m cards. $$
EDIT: So first we pick the m cards and then decide whether or not to put them in one of two places.
[]

Ahh that was basically what I was thinking

I just wasn't so sure about it, especially how to word it

sjm46 · Sep 28, 2016

in for extension 2

study0214 · Sep 28, 2016

im in extension 2

Yunsar · Sep 29, 2016

Alternative method for part b:

$\begin{align*} \text {RHS} &= 2^m $${n \choose m} \\ &=\Bigg[ $${m \choose 0}+$${m \choose 1}+$${m \choose 2}+...+$${m \choose m}\Bigg] $${n \choose m} \\ &=$${n \choose m}+$${m \choose 1}$${n \choose m}+$${m \choose 2}$${n \choose m}+...+$${n \choose m} \\ &=$${n \choose m}+\frac{m!}{(m-1)!}\frac{n!}{m!(n-m)!}+\frac{m!}{2!(m-2)!}\frac{n!}{m!(n-m)!}+...+$${n \choose m} \\ \text {Note: } $${n-r \choose m-r} &= \frac{(n-r)!}{(m-r)!(n-m)!} \text { for all } r \in \mathbb N, n\geq m\geq r \\ &= $${n \choose m}+n\frac{(n-1)!}{(m-1)!(n-m)!}+\frac{n(n-1)!}{2!}\frac{(n-2)!}{(m-2)!(n-m)!}+...+$${n \choose m} \\ &=$${n \choose m}+$${n \choose 1}$${m-1 \choose n-1}+$${n \choose 2}$${n-2 \choose m-2}+...+$${n \choose m} \\ &=\text{LHS}\end{align*}$

Carrotsticks · Sep 29, 2016

Yunsar said:
Alternative method for part b:

$\begin{align*} \text {RHS} &= 2^m $${n \choose m} \\ &=\Bigg[ $${m \choose 0}+$${m \choose 1}+$${m \choose 2}+...+$${m \choose m}\Bigg] $${n \choose m} \\ &=$${n \choose m}+$${m \choose 1}$${n \choose m}+$${m \choose 2}$${n \choose m}+...+$${n \choose m} \\ &=$${n \choose m}+\frac{m!}{(m-1)!}\frac{n!}{m!(n-m)!}+\frac{m!}{2!(m-2)!}\frac{n!}{m!(n-m)!}+...+$${n \choose m} \\ \text {Note: } $${n-r \choose m-r} &= \frac{(n-r)!}{(m-r)!(n-m)!} \text { for all } r \in \mathbb N, n\geq m\geq r \\ &= $${n \choose m}+n\frac{(n-1)!}{(m-1)!(n-m)!}+\frac{n(n-1)!}{2!}\frac{(n-2)!}{(m-2)!(n-m)!}+...+$${n \choose m} \\ &=$${n \choose m}+$${n \choose 1}$${m-1 \choose n-1}+$${n \choose 2}$${n-2 \choose m-2}+...+$${n \choose m} \\ &=\text{LHS}\end{align*}$

Well done, this was the "or otherwise" method. Though my expectation was that students would do it the other way around.

Yunsar · Sep 29, 2016

A bit heavy for two marks though.

Carrotsticks · Sep 29, 2016

Yunsar said:
A bit heavy for two marks though.

The two marks was aimed at the combinatoric proof. If students could not see how to prove it combinatorically, I gave them the option of proving it from scratch algebraically (as you did). A bit more work than for 2 marks, yes. But better than receiving no marks because they are forced to use only one technique.

Trebla · Sep 29, 2016

Yunsar said:
A bit heavy for two marks though.

Not really if you argue how the result holds in terms of the arrangements in the question.

Even then, algebraically you just need to show is that

$\binom{n}{k}\binom{n-k}{m-k}\\\\=\dfrac{n!}{k!(n-k)!}\times\dfrac{(n-k)!}{(n-m)!(m-k)!}\times\dfrac{m!}{m!}\\\\=\dfrac{n!}{m!(n-m)!}\times\dfrac{m!}{k!(m-k)!}\\\\=\binom{n}{k}\binom{n}{m}\\\\ \sum_{k=0}^m \binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m} \sum_{k=0}^m\binom{m}{k}=2^m\binom{n}{m}$

Yunsar · Sep 29, 2016

Trebla said:
Not really if you argue how the result holds in terms of the arrangements in the question.

Even then, algebraically you just need to show is that

$\binom{n}{k}\binom{n-k}{m-k}\\\\=\dfrac{n!}{k!(n-k)!}\times\dfrac{(n-k)!}{(n-m)!(m-k)!}\times\dfrac{m!}{m!}\\\\=\dfrac{n!}{m!(n-m)!}\times\dfrac{m!}{k!(m-k)!}\\\\=\binom{n}{k}\binom{n}{m}\\\\ \sum_{k=0}^m \binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m} \sum_{k=0}^m\binom{m}{k}=2^m\binom{n}{m}$

Nice work. Certainly quicker than the approach I showed previously.

victorsim · Sep 29, 2016

Is coming to the exam. Keen

sharky564 · Oct 1, 2016

I'll be able to make it for MX1.

Official BOS Trial 2016 Thread (5 Viewers)

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