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Stat v. 62

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The question clearly mentioned different. Even from a biological point of view, there's significant variation even within a specie, so the trees could easily have been different. :)
little cloud u know the same question last part permutaion i wrote (8! *9*8*7*6) howmuch can i get :(
 
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But the question was regarding the arrangements of trees on a road for aesthetic purposes not a biological experiment. No one cares if trees that look almost the same are arranged in different ways. It was more of a trick question than a mathematics question.Many people got confused like me. Shame on CIE ......
 
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But the question was regarding the arrangements of trees on a road for aesthetic purposes not a biological experiment. No one cares if trees that look almost the same are arranged in different ways. It was more of a trick question than a mathematics question.Many people got confused like me. Shame on CIE ......
u know none of us got confused :/
 
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little cloud u know the same question last part permutaion i wrote (8! *9*8*7*6) howmuch can i get :(

I'm not too sure but from what I've seen in the mark schemes. 1 mark would have been for the answer, 1 for (9!*4!) for all the hibiscus being together, and the other for them not being together so 12!- (9!*4!). That's how they would have split the 3 marks.
So...
 
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I'm not too sure but from what I've seen in the mark schemes. 1 mark would have been for the answer, 1 for (9!*4!) for all the hibiscus being together, and the other for them not being together so 12!- (9!*4!). That's how they would have split the 3 marks.
So...
yeah but all of my friends did what i did :?
so it has to be one of the exception cases :/
a similar pastpaper gavetwo marks for it :/
 
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But the question was regarding the arrangements of trees on a road for aesthetic purposes not a biological experiment. No one cares if trees that look almost the same are arranged in different ways. It was more of a trick question than a mathematics question.Many people got confused like me. Shame on CIE ......

It shouldn't have been confusing since the question clearly mentioned different. You'd only get confused if you tried to read too much into it instead of just considering what was right there on paper.
 
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yeah but all of my friends did what i did :?
so it has to be one of the exception cases :/
a similar pastpaper gavetwo marks for it :/

Well, for your sake I hope they do consider that, since doing it as 8!*9P4 also means that the hibiscus plants are separated. So yeah, maybe.
 
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Well, for your sake I hope they do consider that, since doing it as 8!*9P4 also means that the hibiscus plants are separated. So yeah, maybe.
cause if i do then i can get 85 + percent and the uni of toronto asks for an A*s in either math or other subject of your major . and no more then one B so . i am abit worried as As result bases they accept provisionally :(
 
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cause if i do then i can get 85 + percent and the uni of toronto asks for an A*s in either math or other subject of your major . and no more then one B so . i am abit worried as As result bases they accept provisionally :(

You planning to go to UofT?
Strange, I got a conditional offer too but they said I only had to 'maintain academic standing', that basically translates to - don't drop more than 2 grade boundaries.
 
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I can't remember the questions and answers at all. I do remember I got 210 and 70 for upper and lower quartile though..
can anyone post the question and answer together?
 
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You planning to go to UofT?
Strange, I got a conditional offer too but they said I only had to 'maintain academic standing', that basically translates to - don't drop more than 2 grade boundaries.
yes but i am planning botha a major and a minor and scholarship too
so the demands are a bit higher
 
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I'm not too sure but from what I've seen in the mark schemes. 1 mark would have been for the answer, 1 for (9!*4!) for all the hibiscus being together, and the other for them not being together so 12!- (9!*4!). That's how they would have split the 3 marks.
So...


you remember the question?
 
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Total trees 15.
Hibiscus:4, Oleander: 2 J-something: 9

(i) selection of 12 trees where there has to be a minimum of two of each kind.
A: 2O, 2H, 8J = 2C2*4C2*9C8 = 54
2O,3H, 7J = 2C2*4C3*9C7 = 144
2O, 4H, 6J =2C2* 4C4*9C6= 84
total = 54+144+ 84 =282

(ii) The company picks 6J, 4H and 2O. How can they be arranged if the trees of the same type are are kept together.
A: 3!*4!*6!*2!= 207360

(iii) Arrange the 12 trees so that the hibiscus plants are not together.
A 12! -(9!*4!)

syed1995
saifookhan
 
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Total trees 15.
Hibiscus:4, Oleander: 2 J-something: 9

(i) selection of 12 trees where there has to be a minimum of two of each kind.
A: 2O, 2H, 8J = 2C2*4C2*9C8 = 54
2O,3H, 7J = 2C2*4C3*9C7 = 144
2O, 4H, 6J =2C2* 4C4*9C6= 84
total = 54+144+ 84 =282

(ii) The company picks 6J, 4H and 2O. How can they be arranged if the trees of the same type are are kept together.
A: 3!*4!*6!*2!= 207360

(iii) Arrange the 12 trees so that the hibiscus plants are not together.
A 12! -(9!*4!)

syed1995
saifookhan



Do u remember how much marks were allocated for each question?
 
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