Q16 (1 Viewer)

[Rafy]

Discuss your answers to Q16 here.

 

[OMGITzJustin]

whats the first answer to this question?

(m+n)! ?

 

[hscwav2012]

divided by m!n! since they are all identical

 

[brianphamm]

divided by m!n! since they are all identical

+1

 

[OMGITzJustin]

what about the second question?

 

[nucgaek]

ignore

 

[v1]

Part 2 was (4C1)^10 if I'm not wrong

i got 13!/(10!*3!) pretty sure thats right

 

[Riaa]

i wrote down 4^10 as well xD

 

[unLimitieDx]

divided by m!n! since they are all identical

yes one mark!

 

[Zeroes]

My thoughts on this section: fml

But hey at least I got a. (i) by the looks of it

 

[GoldyOrNugget]

An alternative expression for ai) is ^m+nC_n. You have n+m slots, and n of them will be of one colour, and the other m will be of the other colour.

 

[deswa1]

Q16 was pretty nice. The induction was too long for the mark value I thought. I didn't get a) ii), b was pretty straightforward (but the induction was long), I knew how to do c) i) but I miswrote a line in my working so I might get docked for that, c) ii) was easy, I sorta got, sorta fudged c) iii) and I didn't have time for iv)

 

[Nooblet94]

Didn't get a)ii). Got all of b, although that was way too easy for a Q16 IMO.

Had no idea how to do c)i), but c)ii) followed from that easily. I could see why c)iii) was true, but I struggled to actually write down why - I'll probably get 1 for that. I got iv) but I may get docked a mark for my reasoning which was sloppy.

 

[jenslekman]

was aii) 4^10/10!? - soudned like quite a dodgily worded question

 

[qrpw]

was aii) 4^10/10!? - soudned like quite a dodgily worded question

well, obviously there is 0.3 ways of putting coins in boxes, genius

 

[lolcakes52]

ai) (M+N)!/M!N!
aii) 13!/10!3! (as we put three dividers in amongst the 10 coins, thus creating 4 groups.)
If Carrot hadn't put his holomorphic question on the BOS trial I would have gotten that wrong.

bi) straight forward
bii) required manipulation, I was stretched for time so I didn't finish
biii) I believe it was pi/4

ci) p(k) = P(k different integers)*p(one of those k integers)
= (nPk/n^k)*(k/n)
= required expression
cii) simple inequality
ciii) didn't understand this
civ) didn't understand this.....

 

[seanieg89]

ai) (M+N)!/M!N!
aii) 13!/10!3! (as we put three dividers in amongst the 10 coins, thus creating 4 groups.)
If Carrot hadn't put his holomorphic question on the BOS trial I would have gotten that wrong.

bi) straight forward
bii) required manipulation, I was stretched for time so I didn't finish
biii) I believe it was pi/4

ci) p(k) = P(k different integers)*p(one of those k integers)
= (nPk/n^k)*(k/n)
= required expression
cii) simple inequality
ciii) didn't understand this
civ) didn't understand this.....

holomorphic question?

 

[Jashua_Long]

holomorphic question?

speak engrish

 

[lolcakes52]

The one where you had to do arrangements of the word holomorphic, where you had to have no vowels together.

 

[deswa1]

Can someone that knows how they mark explain this. For 16b) i), it gave that |x|, |y| <1. So obviously you're meant to deal with this restriction when you take the tan inverse etc. and get +kpi to show that k=0 and hence the result follows so I did that. My question is though, its only one mark- if I didn't deal with the restriction, would I get 0/1 or 1/1?