Mathematics: Post your doubts here!

[smartangel]

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!

DUDE. YOU'RE AMAZING. thanks a million!

 

[farrukh]

(a) Simply 9! = 362880
(b) Regard the two trees as a unit. There are 7 locations available to place this unit. (Any two neighbouring slots can form a valid location but not on different sides.) 7 locations for 1 unit to fit in, that's 7P1.
Next, there are 7 slots left (9 minus 2) for the remaining 7 trees to fit in. This is 7P7.
In addition, in that unit, the 2 trees can be placed in different orders, so it's an extra 2P2.
Total number of ways is then 7P1 × 7P7 × 2P2 = 70560
(c) You can first plant the 2 magnolias, 1 on either side. So that is 4P1 × 5P1.
Then plant the remaining 7 trees in the 7 slots left. 7P7
It is notable that among the 7, 4 are of one species and 3 are of another. So the number should be divided by 4P4 then by 3P3.
Final answer is 4P1 × 5P1 × 7P7 / (4P4 × 3P3) = 700

thankyou so much bro! u r a genius! GOD bless u

 

[Zebedee]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_63.pdf

guys, dont understand the answer for question 2. help please

 

[mominzahid]

I have a severe confusion in normal distribution and binomial distribution.. :'(
i mainly have problem identifying when to apply continuity in conversion from binomial to normal. and this is as i've figured out mainly because i cant seem to identify the questions in which the continuity correction is to be applied.. Can you Please PLEASE explain to me what type of question does the examiner give in which the continuity is to be applied and what type of question he gives when it doesnt has to be applied.. please explain the major differences between the two questions if possible with an example.. its maths p6 day after tomorrow and it would really mean alot if u could help me.. Thanks in advancee....

 

[mukki]

Come on anyone please? This is a really tough one!

(a) 5!*4! (for arranging the 5trees and 4 trees which are on opposite sides of the road)

(b) (4!*2! *3!*2!)/5!*4!
im not even sure if these answers are correct

 

[mominzahid]

(a) 5!*4! (for arranging the 5trees and 4 trees which are on opposite sides of the road)

(b) (4!*2! *3!*2!)/5!*4!
im not even sure if these answers are correct

(a) Simply 9! = 362880
(b) Regard the two trees as a unit. There are 7 locations available to place this unit. (Any two neighbouring slots can form a valid location but not on different sides.) 7 locations for 1 unit to fit in, that's 7P1.
Next, there are 7 slots left (9 minus 2) for the remaining 7 trees to fit in. This is 7P7.
In addition, in that unit, the 2 trees can be placed in different orders, so it's an extra 2P2.
Total number of ways is then 7P1 × 7P7 × 2P2 = 70560
(c) You can first plant the 2 magnolias, 1 on either side. So that is 4P1 × 5P1.
Then plant the remaining 7 trees in the 7 slots left. 7P7
It is notable that among the 7, 4 are of one species and 3 are of another. So the number should be divided by 4P4 then by 3P3.
Final answer is 4P1 × 5P1 × 7P7 / (4P4 × 3P3) = 700

I have a severe confusion in normal distribution and binomial distribution.. :'(
i mainly have problem identifying when to apply continuity in conversion from binomial to normal. and this is as i've figured out mainly because i cant seem to identify the questions in which the continuity correction is to be applied.. Can you Please PLEASE explain to me what type of question does the examiner give in which the continuity is to be applied and what type of question he gives when it doesnt has to be applied.. please explain the major differences between the two questions if possible with an example.. its maths p6 day after tomorrow and it would really mean alot if u could help me.. Thanks in advancee....​

 

[farrukh]

I have a severe confusion in normal distribution and binomial distribution.. :'(​

i mainly have problem identifying when to apply continuity in conversion from binomial to normal. and this is as i've figured out mainly because i cant seem to identify the questions in which the continuity correction is to be applied.. Can you Please PLEASE explain to me what type of question does the examiner give in which the continuity is to be applied and what type of question he gives when it doesnt has to be applied.. please explain the major differences between the two questions if possible with an example.. its maths p6 day after tomorrow and it would really mean alot if u could help me.. Thanks in advancee....​

it is applied when u are approximating binomial as a normal distribution.....the question will ask to use a binomial to normal approximation or if a very large number of trails are done.....then u should use normal distribution

 

[farrukh]

(a) 5!*4! (for arranging the 5trees and 4 trees which are on opposite sides of the road)

(b) (4!*2! *3!*2!)/5!*4!
im not even sure if these answers are correct

your answers are wrong...check DragonCub's solution...it is perfect except that he didnt calculate probabilities

 

[shan5674]

So what is the value of Z at probability of (1-0.1016)?

0.8984

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_62.pdf

4(ii) anyone?

 

[mominzahid]

it is applied when u are approximating binomial as a normal distribution.....the question will ask to use a binomial to normal approximation or if a very large number of trails are done.....then u should use normal distribution

Will the question always use the word 'approximation' where we are expected to apply this continuity correction?

 

[Zishi]

0.8984

That's the probability of the value of a number (z) in the table. Get get value from the table and use your normal distribution formula to get sigma in terms of mu. Now you'll have two simultaneous equations, so solve them to find the mean and standard deviation. If you're able to do that, tell me then - I'll guide you to how to do next part.

 

[shan5674]

That's the probability of the value of a number (z) in the table. Get get value from the table and use your normal distribution formula to get sigma in terms of mu. Now you'll have two simultaneous equations, so solve them to find the mean and standard deviation. If you're able to do that, tell me then - I'll guide you to how to do next part.

Okay got that. Thanks. What about part (ii)?

 

[Zishi]

Okay got that. Thanks. What about part (ii)?

Y ~ N(33, 21). If you're familiar with a normally distributed curve, then you'd know that P(33 − a < Y < 33 + a) = 0.5 is the area in the mid of the curve. As the max probability is 1, so the sum of areas left is also 0.5. Then what does the area greater than 33 + a and lesser than 33 − a equal to?

 

[shan5674]

Y ~ N(33, 21). If you're familiar with a normally distributed curve, then you'd know that P(33 − a < Y < 33 + a) = 0.5 is the area in the mid of the curve. As the max probability is 1, so the sum of areas left is also 0.5. Then what does the area greater than 33 + a and lesser than 33 − a equal to?

0.5? :S I don't get it :/

 

[Aarjit]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_62.pdf

4(ii) anyone?

[Q4] H A P P I N E S S

(ii) The 9 letters of the word HAPPINESS are arranged in random order in a line. Find the probability that the 3 vowels (A, E, I) are not all next to each other. [4 marks]

Solving the problem via reverse approach;

i.e. find the number of arrangements in which A, E and I are all next to each other:

7 x A E I _ _ _ _ _ _

You may be wondering why I used the 7 x above; it's because A, E, I ('stuck' together) can alternate in 7 empty spaces between the _ . Let me number them for you:

1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7......... [in (a) A E I is in number 1]

n (A,E & I together) = 7 x 3! x 6! ............[P and S are recurring twice]
............clever, eh?.............2! x 2!

................................= 7560
now,

n (A,E & I NOT together) = # total possible arrangements - n (A,E & I together)
..........................................= [9! / (2! x 2!)] - 7560 = 90720 - 7580 = 83160

Probability of A,E & I NOT together = n (A,E & I NOT together) = 83160 = 11 .......... Q.E.D
...................................................... ..# total possible arrangements....90720....12

 

[Zishi]

0.5? :S I don't get it :/

 

[DragonCub]

your answers are wrong...check DragonCub's solution...it is perfect except that he didnt calculate probabilities

Oops... I didn't notice the "probability" AGAIN. (I lost some marks in the Mock due to this. ) Really sorry about that dude.
For (b), the probability is 70560 / 362880 = 7/36.
And for (c)... Well I guess I skipped directly to (d) just now.
(c) doesn't need to concern about the magnolias, so answer is 9! / (4! × 3! × 2!) = 1260.
For (d) it's 700 /1260 = 5/9.
My answer was not that perfect anyway.

 

[Wanzi21]

Help me with this.. thanks

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_63.pdf

4(ii)?