Mr Boh Tak Chek is a terminal stage LMBFH Syndrome sufferer.

His teachers have given up on him. His loved ones have been told he has at most “three more weeks”, and were advised to make the necessary arrangements to keep his spirits up in the last leg of his academic life.

As Mr Boh wheeled himself into an exam hall to take the Paper 1 of a Mathematics subject on a gloomy Monday afternoon, he thought about the countless lessons he pon-ed, the endless Maple Story hours, the hundreds of late-night football matches, the Joss Sticks sessions he didn’t sign up …

But it’s now too late. A sudden inability to recall yet another formula (just memorized an hour ago) reminded him of his status in the final throes of this terrifying disease.

Only a miracle can save Mr Boh now.

The Mathematics paper contains 12 different questions of which 3 are on trigonometry, 4 are on algebra and 5 are on calculus. Ravaged by LMBFH Syndrome, Mr Boh is too weak to answer all and only has enough time to do 8 questions.

Calculate the number of different ways in which he can select 8 questions if there is no restriction.

Unfortunately Mr Boh did not revise one of the topics. Calculate the number of these selections which contain questions on only 2 of the 3 topics.

On the following day, Mr Boh wheeled himself again into the exam hall to take Paper II of this Mathematics subject. This time, to his horror, all 12 questions look alien to him!

As such he is leaving everything to fate as to the order of the questions to attempt:

The 12 questions are labeled A–L. As per Paper 1, Mr Boh only has time to do 8 questions.

Calculate the number of possible arrangements of 8 questions chosen from the 12.

Mr Boh thinks Question A looks familiar. Calculate the number of these arrangements in which Question A is appears as the first question.

Calculate the number of these arrangements which contain Question A.

Students, do not be become another Mr Boh. The Last-Minute Buddha Foot Hugging Syndrome can be cured. If you’re taking your O-Levels in 2008 or beyond, please contact Miss Loi today to begin a healthy lifestyle of regular Joss Sticks sessions.

Remember, Early Detection Saves Your Academic Life.

This community message is brought to you by Jφss Sticks.

part1 1) 12C8 = 495 2) split into 2 parts one is without trigo the other is without alegbra

(9C8) [no trigo] + (8C8) [ no alegbra) =9

part 2

1)(12C8) x 8! =19958400 2)(11C7 x 1) x 7! = 1663200 3) (11C7) x 8! =13305600 (11C7 represents the 7 questions besides question A and 8! is so that A and the other questions can be jumbled up)

Welcome to Jφss Sticks s.h & Yunheng! And welcome back Kiroii!

For the first question, all of you have correctly identified this as a Combination question, as can be seen from the lack of any order or arrangement requirement in the selections.

Part 1 is simple since there's no restriction, so a straightforward ^{12}C_{8} (12 choose 8 ) will give you the correct answer of 495.

Part 2 is not so simple, and unfortunately none of you got it right. First of all you have to consider all the possible scenarios to satisfy the following requirements:

1. Mr Boh must answer a total of 8 questions. 2. These 8 questions must contain 2 out of 3 topics.

So the scenarios could be

a. Answer all 3 trigo qns + all 5 calculus qns (^{3}C_{3}^{5}C_{5}) b. Answer all 4 algebra qns + 4 out of 5 calculus qns (^{4}C_{4}^{5}C_{4}) c. Answer 3 out of 4 algebra qns + all 5 calculus qns (^{4}C_{3}^{5}C_{5})

No other combination of two topics will be able to give Mr Boh a total of 8 questions.

Hence the no. of selections = ^{3}C_{3}^{5}C_{5} + ^{4}C_{4}^{5}C_{4} + ^{4}C_{3}^{5}C_{5} = 10

P.S. Mr Boh would be delighted if he has 212058 combinations to choose from!

For the second question, yes when you see terms like arrangement or order, you'll recognize that this is a Permutation question.

PART 1 ...

... is straightforward i.e. ^{12}P_{8} or ^{12}C_{8} x 8! (as Kiroii puts it) or 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 (as Miss Loi puts it - she doesn't like to use ^{n}P_{r} coz sometimes she tends to mix up with ^{n}C_{r} in the midst of heavy calculations) = 19958400.

PART 2 ...

... s.h. you got it wrong. Let's look at it this way:

The first place is CONFIRMED taken up by Question A, which means that there're only 11 questions left to vie for 2nd place, and 10 questions to vie for 3rd place etc. ....

So the number of possible arrangements = 1 x 11 x 10 x 9 x 8 x 7 x 6 x 5 = 1 x ^{11}P_{7} = 1663200

PART 3 ...

... Holy Moly only dear Kiroii got it right!

Let's look at it this way:

From your part 2 answer, when Question A is placed first, the number of arrangement = 1 x ^{11}P_{7}.

Do you all agree that when Question A is placed second (or third, or fourth for that matter) the number of arrangement is also equal to 1 x 1?

So since there are 8 scenarios (the 8 places) where Question A appears, the number of arrangements = (1 x ^{11}P_{7}) x 8 = 13305600.

P.S. Thanks for trying out the questions here. Anyone game for the log and indices question in the previous post? 😉

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

Gratitudes

It is not an overstatement when I say that she has so far been the best tuition teacher I have ever had!! … [read more]

## 8 Comments

曜

日

1. 12C8=495

2.9C8.8C8=9 ??

B

1.12P8= 19958400

2. 8C1=8

3. 11C8=165

曜

日

A

1. 12C8 = 495

2. 212058 ( too many cases to list and i'm lazy)

B

1. 12P8 = 19958400

2. 1 * 11P7 = 1663200

3. 8C1 * 7! = 40320

First time trying here 🙂

曜

日

part1

1) 12C8 = 495

2) split into 2 parts one is without trigo the other is without alegbra

(9C8) [no trigo] + (8C8) [ no alegbra)

=9

part 2

1)(12C8) x 8! =19958400

2)(11C7 x 1) x 7! = 1663200

3) (11C7) x 8! =13305600 (11C7 represents the 7 questions besides question A and 8! is so that A and the other questions can be jumbled up)

曜

日

Welcome to Jφss Sticks s.h & Yunheng! And welcome back Kiroii!

For the first question, all of you have correctly identified this as a

Combinationquestion, as can be seen from the lack of anyorderorarrangementrequirement in the selections.Part 1 is simple since there's no restriction, so a straightforward

^{12}C_{8}(12 choose 8 ) will give you the correct answer of 495.Part 2 is not so simple, and unfortunately none of you got it right. First of all you have to consider all the possible scenarios to satisfy the following requirements:

1. Mr Boh must answer a total of 8 questions.

2. These 8 questions must contain 2 out of 3 topics.

So the scenarios could be

a. Answer all 3 trigo qns + all 5 calculus qns (

^{3}C_{3}^{5}C_{5})b. Answer all 4 algebra qns + 4 out of 5 calculus qns (

^{4}C_{4}^{5}C_{4})c. Answer 3 out of 4 algebra qns + all 5 calculus qns (

^{4}C_{3}^{5}C_{5})No other combination of two topics will be able to give Mr Boh a total of 8 questions.

Hence the no. of selections =

^{3}C_{3}^{5}C_{5}+^{4}C_{4}^{5}C_{4}+^{4}C_{3}^{5}C_{5}= 10P.S. Mr Boh would be delighted if he has 212058 combinations to choose from!

曜

日

For the second question, yes when you see terms like

arrangementororder, you'll recognize that this is aPermutationquestion.PART 1 ...... is straightforward i.e.

^{12}P_{8}or^{12}C_{8}x 8! (as Kiroii puts it) or 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 (as Miss Loi puts it - she doesn't like to use^{n}P_{r}coz sometimes she tends to mix up with^{n}C_{r}in the midst of heavy calculations) = 19958400.PART 2 ...... s.h. you got it wrong. Let's look at it this way:

The first place is CONFIRMED taken up by Question A, which means that there're only 11 questions left to vie for 2nd place, and 10 questions to vie for 3rd place etc. ....

So the number of possible arrangements = 1 x 11 x 10 x 9 x 8 x 7 x 6 x 5 = 1 x

^{11}P_{7}= 1663200PART 3 ...... Holy Moly only dear Kiroii got it right!

Let's look at it this way:

From your part 2 answer, when Question A is placed first, the number of arrangement = 1 x

^{11}P_{7}.Do you all agree that when Question A is placed second (or third, or fourth for that matter) the number of arrangement is also equal to 1 x 1?

So since there are 8 scenarios (the 8 places) where Question A appears, the number of arrangements = (1 x

^{11}P_{7}) x 8 = 13305600.P.S. Thanks for trying out the questions here. Anyone game for the log and indices question in the previous post? 😉

曜

日

w00t for question 3 i was rather uncertain actually but i w as like thinking of this new method my teacher eleborated>.>

曜

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Kiroii, so are you more certain now? 🙂

曜

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hmm yeah its undoubtly explicable for my admission on which i understood the question to a higher extend due to thy solution celine

-__-

albeit i think mine is shorter (11C7) x 8!

11C7 is to pick the 7 other questions and 8! to jumble em up - like rojak?