well.. fairly simple paper, think can get my B for overall due to screwing up Paper 1. Statistics was straight forward and simple, Pure Maths was so expected.

Happiest paper…lol. Finally got one paper I can do every question and got time to check through more than once. Anyway YAY Math over.

if everybody is going to do so well, i m going to get bell-curved down to a lousy grade.
so what is the point of studying?

It was pretty easy.

Lost 4 marks due to the Permutation and Combination question.

1 more mark cos i only included half the range for range of values for x in expansion.

Other than that, should be a rather okay thingy.

anyone still rmb e ans? care to share?

Exact area of R = pi - 2

P&C qn:
1)12!
2)6! x 2!
3)11!
4)5! x 6!
5)5! x 2!

Originally posted by teraexa:

Exact area of R = pi - 2

P&C qn:
1)12!
2)6! x 2!
3)11!
4)5! x 6!
5)5! x 2!

my ans for P and C..
1) 12!
2) 6! x 6(2!)
3) 11!
4) forgot
5) 5! x 2!

Originally posted by club18:

my ans for P and C..
1) 12!
2) 6! x 6(2!)
3) 11!
4) forgot
5) 5! x 2!

for 2), i tink it's supposed to be 2^6?

Originally posted by club18:

my ans for P and C..
1) 12!
2) 6! x 6(2!)
3) 11!
4) forgot
5) 5! x 2!

i tink for both question 2 n 5 nid to x 2^6...cos deres 2 arrangement for each couple..cos this qn is in the tjc prelims 2007 which i tried..so quite sure of it..
so for 2 shld be 6! x 2^6, 5.) 5! x 2^6..

Originally posted by xxxshadowxxx:

i tink for both question 2 n 5 nid to x 2^6...cos deres 2 arrangement for each couple..cos this qn is in the tjc prelims 2007 which i tried..so quite sure of it..
so for 2 shld be 6! x 2^6, 5.) 5! x 2^6..

Part 2 should be 6! x 2!, not 6! x 2^6. Here's the explanation:

Ok lets imagine 6 distinct groups first.

For the first slot, there are 6 possible combinations of groups to put into that slot. And then, within that group, you have 2! ways to arrange the people in the group.

For the 2nd slot, you have 5 possible groups to fit in now and again you have 2! ways to arrange the people in that group. Repeating this for the 4 remaining slots, the total combinations is 6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2! = 6! x 2!.

It's the same case for part 5 also.

part 2 should be 6! x 2^6? becoz you permutate 6 groups. then within each group the man and the woman can swop position.

part 5 should be 5! x 2 becoz you permutate 6 groups in a circle and times 2 only becoz the qn say man and woman must alternate. so it can only be MWMWMWMWMWMW or WMWMWMWMWMWM?

anybody else with me?

Originally posted by teraexa:

Part 2 should be 6! x 2!, not 6! x 2^6. Here's the explanation:

Ok lets imagine 6 distinct groups first.

For the first slot, there are 6 possible combinations of groups to put into that slot. And then, within that group, you have 2! ways to arrange the people in the group.

For the 2nd slot, you have 5 possible groups to fit in now and again you have 2! ways to arrange the people in that group. Repeating this for the 4 remaining slots, the total combinations is 6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2! = 6! x 2!.

It's the same case for part 5 also.

eh..no offence bt 6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2!= 6! x (2!)^6=6!x2^6

bt for part five i forget if dere is a nid for dem to alternate..i noe part 4 ask for alternate male n female if part 5 did ask for alternate den shld be 5!x2..if nt its 5! x 2^6

PCC ... I mean my P&C answers

(a)
1. 12!
2. 6! x 2^6

(b)
1. 11!
2. 6! x 6! / 12
3. 5! x 2

Originally posted by teraexa:

For the 2nd slot, you have 5 possible groups to fit in now and again you have 2! ways to arrange the people in that group. Repeating this for the 4 remaining slots, the total combinations is 6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2! = 6! x 2!.

Isn't ...

6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2!
= (6 x 5 x 4 x 3 x 2 x 1) x 2! x 2! x 2! x 2! x 2! x 2!
= 6! x (2!)^6 ?

here are e ans i rmb
1. $7.65

2(i) prove by mi
(ii)1-1/(N+1)^2
(iii)
sum to infinity=1
(iv) 1- 1/N^2

3(i) malcurin's series jux need to differentiate n show
(ii) 8-3x+(24/3/16)x^2-(8/127/12x^3
(iii) range: -1/(sq rt 2) < 1/(sq rt 2)

4(i) 5pi/6 +(sq rt 3)/8
5pi/6 - (sq rt 3)/8
(ii) pi-2

5. sampling qn

6.0.933
(i) 0.717
(ii) 0.616

7. z-test, p=0.0382(

8. chicken n turkey qn
(i) 0.395
(ii) 0.160
(iii) 0.392
(iv) cuz ans in (ii) is a subset of e ans in (iii)

9. (i) single line
(a) 12!
(b) 6!x2!x2!x2!x2!x2!x2!
(ii) circle
(a)11!
(b) 5!x6!
(c) 5!x2

**quite sure of (i)(b) n (ii)(c)

10. (i) 1/64
(ii) 21/256
(iii) 13/17

11. r=-.994
t= 155min[/list]

so sorry didnt know a part will appear as smiley face but there is 8 )
think when e paper is very easy its no longer a test of how gd u are but how carreful u are in ur work=)

anyone can explain why the answer to 9(b) is 5!x6!?
btw, 9(b) is the one alternating men and women around a circle

my answer is 6!x6!

this is how i get it:

first, arrange the men in a circle
then, there are 6 slots in between the men for the women
since there are 6 women and 6 slots, no. of ways for women = 6!
lastly, the men can switch places with each other so no. of ways for men = 6!
therefore, no. of ways = 6!x6!

Originally posted by xxxshadowxxx:

eh..no offence bt 6 x 2! x 5 x 2! x 4 x 2! x 3 x 2! x 2 x 2! x 1 x 2!= 6! x (2!)^6=6!x2^6

bt for part five i forget if dere is a nid for dem to alternate..i noe part 4 ask for alternate male n female if part 5 did ask for alternate den shld be 5!x2..if nt its 5! x 2^6

thanks for the reminder. went to check with my friends. seem like i got my part iib wrong. that's 7 marks gone already.

but for part 5, my stand on 5! x 2! stands.

Originally posted by teraexa:

thanks for the reminder. went to check with my friends. seem like i got my part iib wrong. that's 7 marks gone already.

but for part 5, my stand on 5! x 2! stands.

yup..seems like i got the 5! x 2! part wrong..didnt read the qn thoroughly..haha