Mathematics: Post your doubts here!

[Aarjit]

Q7..part ii onwards.. PLEASE HELP! its permutations n combinations question!

[Q7] Nine cards, each of a different colour, are to be arranged in a line. The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card? [3]

It's easier to solve by reversing the question, here's how you do it:

The total number of arrangements which do have the pink (P) card next to the green card (G):

[G P] 8! ....................................................[Here we are treating G and P as a single card attached together]
But remember that this can be [P G] 8! too !
n(P next to G) = 2! x 8! = 80640

now, finding the arrangements do not have the pink card next to the green card;

n(P away from G) = total # arrangements - n(P next to G) = 9! - 80640 = 282240

Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.

(iii) How many different arrangements in total of 3 cards are possible? [2]

arranging 3 cards from of 9;
= 9P3 = 504

(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card? [2]

Consider the restriction here: there MUST be a pink card (P) in the 3 cards, the remaining two can be randomly selected from the remaining 8 cards and arranged in the _ spaces;

n (P _ _ ) = 3P1 x 8P2...............................[Remember that P can be placed in a total of 3 ways P _ _ or _ P _ or _ _ P]
..............= 168

(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card?

Again, It's easier to solve by reversing the question like we did in (ii);

The total number of arrangements which do have the pink (P) card next to the green card (G):

[P G] 7P1................but this can also be [GP] 7P1 or 7P1 [PorG]!

n (P next to G) = 2! x 2! x 7P1 = 28

or, n (P away from G) = total # arrangements [see (iii)] - n(P next to G) = 504 - 28 = 476

Q.E.D.

 

[Aarjit]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf
Q2..how to find upper n lower quartile.my ans doesnt match the ms..

Draw an ORDERED stem and leaf diagram from the discrete data [n = 21 (odd number)] and use the following formulae:

Q1 = [(n+1)/4]th item = 5.5th item
now move to the 5th leaf in your stem and leaf plot. Q1 is the average of the 5th and 6th leaf.

Q3 = [3(n+1)/4]th item = 16.5th item
now move to the 16th leaf in your stem and leaf plot. Q3 is the average of the 16th and 17th leaf.

The answers will definitely concur with the m.s.!

 

[Aarjit]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf
Q6 please help..

Q6. Permutations and Combinations

(i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy and Monica if each receives an odd number of mugs. [3]

The possible distributions between Lucy (L) and Monica (M) are:

...L9M1,...L7M3,...L5M5,..L3M7,..L1M9

= 10C9 + 10C7 + 10C5 + 10C3 +10C1
= 10 + 120 + 252 + 120 + 10 = 512

(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a different design. Find in how many ways these 9 mugs can be arranged in a row if the china mugs are all separated from each other. [3]

Alright, let's try this out in a slightly creative way;

Plastic Mugs ~ \_/
China Mugs ~ (_)3

If the 3 China mugs are all separated from each other, they must fit in any of the alternating 7 _ !

_ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _

The (_)3 have 7 _ to fit in and the \_/ can arrange in 6! ways between themselves;

n = 7P3 x 6! = 151200

(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs. These 14 mugs are placed in a row. Find how many different arrangements of the colors are possible if the red mugs are kept together. [3]

If the red mugs are kept together, the remaining 11 mugs can be arranged randomly;

\_/ \_/ \_/ x 11 !...................... [Here, we are considering the \_/ to be 'attached' so they're always kept together]

= 3! x 11 ! x 2!......................[since there are 2 possible arrangements: \_/ \_/ \_/ x 11 ! or 11! x \_/ \_/ \_/ and the \_/ can arrange in 3! ways between themselves!]

but wait! we still have to consider the 7 and 4 identical yellow and blue mugs!

= 3! x 11! x 2! = 3960......Q.E.D
........7! x 4!

 

[leosco1995]

Help please:

Find the number of different ways in which the 9 letters of the word GREENGAGE can be arranged if exactly two of the Gs are next to each other.

The answer is 5040.. but I'm confused because there's 3 Gs and I don't know how to find the # of ways where 2 Gs are separated from the other one.

 

[Aarjit]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_63.pdf
Q3..last part!

Variance (V) = n.p.q...... (Remember transforming X ~ B (n,p) to X ~ N (np, npq) where np = μ and npq = σ^2 or Variance?)

You have, n =30 , p =0.6, q = 1 - 0.6 = 0.4

V = 30 x 0.6 x 0.4 = 7.2.........

 

[Aarjit]

Help please:

Find the number of different ways in which the 9 letters of the word GREENGAGE can be arranged if exactly two of the Gs are next to each other.
The answer is 5040.. but I'm confused because there's 3 Gs and I don't know how to find the # of ways where 2 Gs are separated from the other one.

Simply 'choose' 2 Gs out of 3, and the remaining 7 letters can be completely random!

G G _ _ _ _ _ _ _...............[consider GG to be 'attached' together so that it can alternate in 2! ways]

n = 2! x 3c2 x 7!.................. [but don't forget that there are 3 recurring Es too!]
..............3!
...= 5040...... Q.E.D

 

[kirashinagami]

Hai there. Can somebody help me to explain all the questions? http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_ms_72.pdf Just a brief explanation. I don't have a really good basic s2. So please help me! I am still trying to understand the concepts. Any tips in doing the questions will be greatly appreciated! Thank you..

 

[Aarjit]

Hai there. Can somebody help me to explain all the questions? http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_72.pdf Just a brief explanation. I don't have a really good basic s2. So please help me! I am still trying to understand the concepts. Any tips in doing the questions will be greatly appreciated! Thank you..

I recommend you to go through the Examiner's Report: http://www.xtremepapers.com/papers/...d AS Level/Mathematics (9709)/9709_s11_er.pdf

Also, you can check out the KhanAcadamy Statistics' playlist for the core concepts: http://www.youtube.com/playlist?list=PL4C863861E3B2E380

 

[leosco1995]

Simply 'choose' 2 Gs out of 3, and the remaining 7 letters can be completely random!

G G _ _ _ _ _ _ _...............[consider GG to be 'attached' together so that it can alternate in 2! ways]

n = 2! x 3c2 x 7!.................. [but don't forget that there are 3 recurring Es too!]
..............3!
...= 5040...... Q.E.D

But there are also 3 Gs... so don't we have to divide by 3! twice?

 

[Aarjit]

But there are also 3 Gs... so don't we have to divide by 3! twice?

There is only 1 G that you can possibly alternate. The 2 Gs are fixed/'attached', so that they're always together.

The typical logic of dividing by 3! for the recurring 3Gs doesn't really serve well for this question. Albeit, there are various methods of doing it, I simply outlined mine.

Spend some time in it; you'll definitely get it.

 

[kirashinagami]

Ok thx for ur advices! I will ask you specific questions then... http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_72.pdf how to solve no 2? I still don't understand how to find variance of a sample mean

 

[kirashinagami]

Please part 3
3
The number of goals scored per match by Everly Rovers is represented by the random variable X which has mean 1.8.
(i) State two conditions for X to be modelled by a Poisson distribution.
[2]
Assume now that X ∼ Po(1.8).
(ii) Find P(2 < X < 6).
[2]
(iii) The manager promises the team a bonus if they score at least 1 goal in each of the next 10 matches. Find the probability that they win the bonus. [3]

 

[leosco1995]

There is only 1 G that you can possibly alternate. The 2 Gs are fixed/'attached', so that they're always together.

The typical logic of dividing by 3! for the recurring 3Gs doesn't really serve well for this question. Albeit, there are various methods of doing it, I simply outlined mine.

Spend some time in it; you'll definitely get it.

I think I get it now. Basically, the 2 Gs are separated as a separate component so there isn't a need to use 3! for them. I think this was a pretty tricky question though. P&C sucks.

 

[Aarjit]

Please part 3

I can see that you're looking for help, but your paper 72 is on 29th May. We'll revise it after the 25th. Till then, post you queries in an S2 thread.
You might want to check out this too: http://www.examsolutions.co.uk/maths-revision/index.php#Statistics

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_61.pdf

Q5 pleasee

 

[angelicsuccubus]

doesn't anyone here do mechanics 2?

cause I need notes/help anything!

 

[Wanzi21]

How do you do the part 3? permutation and combination question

 

[Zishi]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf

Q5 pleasee

Which part?

 

[Zishi]

View attachment 10386

How do you do the part 3? permutation and combination question

The best way to do questions of this type is to subtract the "No woman is included" from the total number of selections that can be made.

 

[shan5674]

Which part?

both :/