Wednesday, April 29, 2015

[Pri20150429PPP] Pythagoras for Primary Pupils?

Question


Introduction
     This primary (elementary) school mathematics examination question created quite a stir among some parent support groups on Facebook.  The issue is that the height of the triangle seems to have been omitted.

Pythagoras’ Theorem
     Some participants who know secondary school mathematics were quick to suggest the use of Pythagoras’ Theorem to find the height of the triangle, which works out to be  12 cm.  This leads to the answer (2) 114 cm2,  which is correct.  The problem is that pupils are not taught Pythagoras’ Theorem until secondary school, and so it would seem an unfair test for the pupils.  So the discussion turned to thinking of various methods by which a primary school pupil may uncover the answer without resorting to advanced knowledge.

Elimination and Educated Guessing
     Mr Teo Kai Meng, a tutor who regularly participates in the support groups, offered some insightful observations.  Assuming that the height measurement is a whole number of centimetres, only options (2) and (4) need to be considered as they were divisible by  19, which the area had to be under the said assumption.  [ Another tutor, Melissa Song had a similar idea by observing that since the triangles DKLN and DKMN  have the same height, the ratio of their areas is the ratio of their bases LN : MN = 19 : 16. ]  We can ignore choices (1) and (3).  As we know,  area = ½ ´ base ´ height .  Since teachers like to catch students for being careless in forgetting to multiply by ½  (or dividing by 2), it is quite likely that option (4) was set up as a booby trap.  Thus one may intelligently surmise that option (2) should be the answer.

     Mr K L Chua, a tutor who calls himself “Mathematics Specialist”, used a similar reasoning.  He worked backwards from each of the four choices to get the heights and chose the most plausible answer.  Of the two whole-number answers, (4) was eliminated and (2) was chosen since from the diagram the height should roughly be near to  16 cm  even though the diagram was not drawn to scale.

Scale Drawing

     Another participant suggested doing a scale drawing to estimate the height.  Indeed this can be done, and is a good tactic too, since this could be done quickly with a ruler and pencil.

A Visual Solution
     Assuming no knowledge of Pythagoras’ Theorem, it is possible to construct a visual solution.  First we take four copies (indicated in orange/light-orange) of the right-angled triangle  DLNK  and arrange them to surround a square of side  37  cm (indicated in green),  which is the same as the longest side (hypotenuse) of the said triangle.     This green area is  1 369 cm2.  Now we rearrange the triangular pieces as indicated by the red arrows.  The areas of the orangey triangles do not change when you shift them.  Neither will the green area change, since the total area everything in the containing square (orangey plus green areas) remains the same.

With pairs of right-angled triangles joined together along their longest sides, we now obtain two green squares, the larger of which has side  35 cm.  This gives an area of  1 225 cm2.  The total area of the two green squares is the same as the area of the large green sqaure before the shifting, namely  1 369 cm2.  Hence the area of the small square is  (1 369 – 1 225) cm2 = 144 cm2.


From here we quickly deduce that the unknown height is  12 cm,  12 being the square root of  144.  Hence we conclude that the area of  DKLM  is  114 cm2.

Remarks
     I have shown that it is theoretically possible for primary school pupils without knowledge of Pythagoras’ Theorem to derive the answer in an exact manner.  By the way, the method of shifting triangular pieces as indicated above can be generalised to give a proof of Pythagoras’ Theorem.
     Some parents expressed fear that this is another one of those Cheryl-like or olympiad type of problems.  Is there a conspiracy by the school teachers to purposely set difficult questions and make life difficult for pupils, disadvantaging those who cannot afford private tutors?  In this case, could this just have been an oversight on the part of the teacher who set the question?
     Entrepreneur John Low Jiayong and tutor John Lim sourced for and managed to obtain faithful copies of the original question.  It turned out that some school-paper vendors had inadvertently erased the 12 cm measurement.  Thus in the original question, the height  KN  was given as 12 cm, and the measurement of  37 cm for the hypothenuse  LK  was purposely given as extraneous information to distract students.
     Some parents observed that this is after all just a multiple-choice question that carries a credit of only one mark.  If this were an exam situation with the  12 cm omitted, it would be best to either sacrifice the  1  mark and move on, or to use tactics like elimination, educated guessing or estimation with scale drawing.
     Anyway, it had been quite a fruitful discussion, as adults (parents and tutors) attacked this problem from many angles in a purposeful way.  Many parties put in concerted effort and contributed in an engaged matter.  It would be good if this type of rich discussion were enacted in our classrooms everyday among pupils and teachers, perhaps enabled by social technology.  For then, pupils would deeply learn and perhaps would not be so stressed out, nor be needing so much extra external help.  Private tutors could then move on to focus on value-added  mentorship  for 21st Century Learning, instead picking up the tab where school teachers have left off.


H01. Act it out (e.g. scale drawing)
         (as a class learning activity, pupils can use scissors to cut out paper triangles
          and physically move them around)
H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H07. Use guess and check
H08. Make suppositions
H09. Restate the problem in another way

Suitable Levels
Primary School Mathematics
Lower Secondary Mathematics
* any precocious / gifted pupil who wants to learn
* any person of any age interested in creative problem solving





Tuesday, April 14, 2015

[AM_20150413RSD] Rationalising Denominators for #Surds

Question 

Introduction
     Expressions with surds in their denominators are cumbersome.  The good news is that we can make the denominators into rational numbers, which are nicer.  Rational numbers those that can be expressed as a ratio of integers i.e. they are (proper or improper) fractions or can be converted to fractions.  Whole numbers are also part of rational numbers because you can always put them upon a denominator of  1;  e.g. 2 = 2/1,  so  2  is a rational number.
     The standard trick for simplifying expressions with surds in their denominators is to rationalise the denominator by mutiplying the numerator and the denominator with its conjugate surd.  For example, the conjugate surd of   Ö5 + Ö2   is   Ö5 – Ö2.   Just change the  +  to  –  or the  –  to  +.  Let us see how the magic works.

Solution
Remarks
     Note that in the first step, I pulled out 2 as the common factor of the denominator, so that I get a simpler surd to work with.  Always try to work with simpler expressions.  This not only shortens your working, it reduces your chances of making a careless mistake.
     In mathematics, “rationalising” does not mean you give some reason or excuse for something that you know you have done wrong.  It means “make it into a rational number”.  Why does rationalising the denominator work?  This is because on the bottom (denominator) we have a difference-of-squares expression of the form
                                           (a + b)(ab)   which is equal to   a2b2.
Since squaring “gets rid” of square roots,  a2  and  b2  will give you rational numbers (whole numbers or fractions), you will end up with a nice number downstairs (on the denominator).  Pupils should make sure they have this technique in their repertoire of skills.

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more

Monday, April 13, 2015

[OlymUSec_20150412BDP] Guessing Cheryl’s Birthday

Question

Introduction
     This “Primary 5 mathematics” (actually an upper secondary Olympiad) logic puzzle has gone viral.  It has been making its rounds in various forums in Singapore and overseas, stumping adults and children alike.  It is actually a parody of an old puzzle.  Can it even be solved?  It seems that there is no information given by each parties that we can exploit.  Actually there is!  In a subtle way ...

Solution
     In the beginning, everybody knows that Albert knows only the month and Bernard knows only the numerical day of the month.
     When Albert tells us “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” he is leaking out information (from his knowledge of the month) that the day of the month appears more than once and cannot be (June 18 or May 19).  Actually, the original phrasing is more like “If I don’t know when Cheryl’s birthday is, then Bernard does not know too.”.  The person who set this question merely changed the names of the people and the dates, without appreciating the subtle but crucial difference between a statement of fact and an implication (an “if ... then ... ” statement). 
     Ruling out June 18 and May 19, we also know that Albert knows that the birthday month is neither June nor May.  Otherwise, how would he have been so confident in saying that he knows Bernard would not know Cheryl’s exact birthday?  So we can eliminate those months.
     Bernard acknowledges the above state of affairs and the embedded hint.  With the choice narrowed down and with his knowledge of the numerical date, he now knows Cheryl’s birthday.  Since we know that Bernard knows Cheryl’s birthday, we know that it cannot be a numerical date that appears more than once (otherwise he would not have been able to know).  So we can cross out July 14 and August 14.

     Now Albert would telepathically thank Bernard for this helpful hint.  Because now he is able to deduce Cheryl’s birthday with his knowledge of the month.  That would mean that this cannot be a month with two candidate dates.  We blot out the August dates and see for ourselves the only remaining possibility.

Conclusion: Cheryl’s birthday is  July 16.

Remarks
     This puzzle was solved using the process of elimination and analysing our knowledge of what each party knows and can know.  Thus we successively narrow down the possibilities until the answer becomes obvious.  Here we learn that
     knowledge of other people’s knowledge can itself give us knowledge
This principle is actually employed in cryptology (the use of secret codes) which finds applications in fields like banking, the military (cf. interesting story of how the German Enigma code was broken in WorldWar II) and communications.  As an example, radio communication can tell the enemy of troop positions and warn of an impending attack, and that is why radio silence is imployed as a precaution.  Sensitive information in certain organisations is restricted on a “need to knowbasis.

Suitable Levels
Upper Secondary Olympiad
* other syllabuses that involve knowledge or epistemology
* application of mathematical principles in real life
* for all people interested in logic puzzles

[OlymPri20150412LGC] Your #logic a House of Cards?

Question


Introduction
     This is an olympiad type of question that tests one’s use of logic.  Sadly, logic is seldom taught in schools, except maybe in courses like knowledge inquiry, theory of knowledge, or in selected topics like geometry.  Mathematics is actually very much connected with logic, but it seems to be regarded as difficult to teach.  Being able to use logic and think critically is very important in our lives in various aspects.
     Let us study how we can analyse conditional statements using logic.

Analysing Conditional Statement with Logic
     A conditional statement is a statement of the form “If  H  then  C.”  An example would be: “If the blue litmus paper turns red, then the liquid is an acid”.  The front clause H (the part about the litmus paper turning red) is called the hypothesis (or premise).  The latter clause  C  (the part about the acid) is called the conclusion.  Every conditional statement has three other related forms: the contrapositive, the converse, and the inverse.

     For example, let say there is a NC16-rated movie and you can watch it only if you are at least 16 years old.  So  H := “watch NC16”  and  C := “16 years old and above”.  The conditional statement means “If you want to watch the movie, then you must be 16 years old and above”.  The contrapositive is “If you below 16 years old, then you cannot watch the movie”.  The contrapositive always has exactly the same meaning as the original conditional statement.  If one is true, so is the other.  If one is false, so is the other.  The contrapositive is merely phrased in a different way.
     The converse says “If you are 16 and above, then you will watch the movie.”  The inverse would say “If you don’t want to watch the movie, then you are not 16 and above.”  Note that the inverse is the contrapositive of the converse.  The inverse always has exactly the same meaning as the converse.  If one is true, so is the other.  If one is false, so is the other.
     The converse (and the inverse) has a different meaning from the original statement.  If the original statment is true, the converse may or may not be true.  In our example, “If you want to watch the movie, then you must be 16 years old and above” is true, but the converse “If you are 16 years old and above, then you want to watch the movie.”  may not be true, as some people who are 16 years old and above may want to do something else.  Like reading a book.  Or playing golf.  So you see, there are really two camps: the original conditional statement and its contrapositive in one camp, and the converse and the inverse in the other.
     Let us analyse the given problem using logic.


Analysis
     The statement  ‘any card with a letter A on one side always has the number “1” on the other side’ can be rephrased as ‘If letter A appears, then the other side is “1”’.  Let us analyse each position in order.  For (a), since the letter A appears, we definitely need to know what is behind the card to test whether the claim is true.  For (b), note that the claim does not say ‘If the letter is not A, then the other side is not “1”’ (inverse)  nor  ‘If the number is “1”, then the other side must be an A.’ (converse).  It is possible that A is on the other side and we are fine.  But it is also possible to have a non-A with the number  “1”.  The given condition does not prohibit that.  So, if the other side is not an A, it is still OK.  So the other side can be anything, and it does not matter.  For (c) we definitely have to flip to check the other side in case the other side is an A, then the claim would be proven false.  For (d), since the card is not an A, it is irrelevant.  The claim is talking about  A.  It is not talking about B.  We can summarise the analysis in the diagram below:-

Ans:  We need to flip (a)  and  (c)

Suitable Levels
Primary School Olympiad

* syllabuses that involve logic and epistemology

[S1_AFMLCM_20150412] Conquering Algebraic Fractions

Question

Introduction
     This is an equation involving algebraic fractions, usually for secondary 1 (approximately grade 7) pupils in Singapore.  Many students (and teachers?) like to use the “cross-multiplying” method, as shown in Solution 1.  A usually more efficient method is to multiply every term by the Lowest Common Multiple (LCM) of all the denominators appearing in the equation, as shown in Solution 2.


Discussion
     Note that division by zero is not allowed.  Furthermore, in algebra, it is dangerous to cancel or divide by an unknown quantity, because there is a possibility that you are dividing by zero.  So any division or cancellation by an unknown quantity must be justified beforehand.  Mathematics is not a game of blind senseless manipulations.  If you look at the second solution, which is short and sweet (only 4 steps), multiplying through by the LCM of denominators not only avoids this awkwardness, but it clears all the fractions in one fell swoop.  The solution takes only  4  steps, and it is in fact the recommended method.  All students, whether “good” or “poor” in maths, should use the second method.  Teachers who refuse to use/teach the LCM method (out of habit, or because their own teachers taught them otherwise, or because this makes them or their pupils “uncomfortable”) are really doing the weaker students a huge disservice.  You are widening the achievement gap.  The better students are better, precisely because they use better methods.  The longer one’s working is, the higher the chances of making mistakes and the more time is wasted.  If the “weaker” pupils have to jump through lots of hoops to achieve a certain standard before they are allowed to learn this “advanced” method (actually it’s just the normal method), they will have to unlearn their old method and may get confused as they learn this method.  A triple whammy!  All learners need to practice anyway, so one might as well practice the correct thing right from the beginning and learn good habits (striving for efficient, effective, elegant solutions).  So please, please, please everyone: use the LCM method!



Suitable Levels
* Secondary 1 Mathematics
* GCE ‘O’ Level (“Elementary”) Mathematics Revision
* other syllabuses that involve algebraic Fractions
* precocious children who want to learn algebra


Sunday, April 12, 2015

[EM_20150412SCA] Is your Airport Design career “taking off”?

Question

An Airbus 380 has constant acceleration of  1 m/s2.  Its takeoff velocity is 280 km/h.  How long must the runway be at a minimum to allow the plane to take off?

Introduction
     A practical question for airport design, perhaps?  I present two solutions.  The first uses a graphical method (speed-time graph) which is in the (“Elementary”) Mathematics syllabus and the other uses formulas for motion under constant acceleration taught in Physics.  Whichever method is used, remember to convert from km/h to m/s.  The target velocity is  700/9 m/s,  and the time to achieve this is  700/9 s  starting from rest,  since the acceleration is  1 m/s2.

Solution 1 [“Elementary” Mathematics, speed-time graph]
     The speed-time graph is very useful because it is able show the acceleration (as the gradient or slope of a straight line) and at the same time the area under the graph gives the numerical value of distance travelled.  If the acceleration is constant, we usually we get a trapezium.  But since the aeroplane starts from rest, we get a triangle (see diagram below).  All we need to do is to calculate the area under the graph and get the answer.
     It is interesting to observe that if we used the average speed  700/18 m/s,  we would also get the answer because the area under the graph (yellowish green rectangle) is the same as the area of the triangle.  This trick works for constant acceleration, but it may not work in other situations.


Solution 2 [Physics, constant acceleration]
     We use the important formulas  v = u + at  and  s = ut + ½at2.  In our example,  u = 0  because the initial velocity is zero (the airplane starts from rest).  This makes our calculations very easy.  If we compare the two methods, you find that the calculations are very similar, and we get the same answer.  Remember that speed = |velocity|  the magnitude of velocity.  In this relatively easy problem, the velocity means the same as speed because we are going in a straight line and in one direction only.  In other situations, this may not be so.

Heuristics Used
H02. Use a diagram / model
H05. Work backwards
H09. Restate the problem in another way
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level “Elementary” Mathematics
GCE ‘O’ Level Physics
* other syllabuses that acceleration, speed and distance
* precocious kids who always want to learn more

[AM_20150412SQS] Perfect Squares Lurking Absurdly

Question


Introduction
     Surds are expressions involving roots like square roots.  They are usually irrational numbers.  If you try to put them into a ratio of integers, you are absurd!  Many students (and teachers?) are not sure of how to put square roots of numbers in a simple form.  The trick is to use square numbers or perfect squares.  These are squares of whole numbers.  For example:-
     12 = 1 ´ 1 = 1                      Ö1 = 1
     22 = 2 ´ 2 = 4                      Ö4 = 2
     32 = 3 ´ 3 = 9                      Ö9 = 3
     42 = 4 ´ 4 = 16                  Ö16 = 4
     52 = 5 ´ 5 = 25                  Ö25 = 5
A number like  4  can be represented by a real square whose sides have length  2  units.  Note also that if you take the square root of a perfect square, you always get a nice whole number.
     How do you deal with numbers that are not perfect squares?  You factor out as many perfect squares as possible.  This would eventually lead to surds with small numbers, which are more manageable.  Here are some examples:-
With this weapon in our hands, let us kick some butt.

Solution
Moral of the Story
     Using perfect squares reduces your square roots to surds involving square roots of prime numbers, which are easier to combine or cancel.  This gives a short and sweet solution.  In mathematics, always try to do things by the cleanest way (if you can).


H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more



Saturday, April 11, 2015

[JCH2CNEFTG_20150410] Exponential Half-Power Trick

Question

Introduction
     This is a complex-number question that appeared for ‘A’ Level Mathematics in November 1998.  Remember that mathematics is never out-dated.  Many Singapore schools keep this sort of questions in their question banks (for tutorial exercises, tests and examinations), in the hope that it becomes part of students’ repertoire.
     Students are expected to know the famous Euler’s Formula, one of the most beautiful formulas discovered by this visually-challenged but prolific mathematician.  It links the exponential function with trigonometry via the idea of angle rotation.  Adding eif  with its reciprocal  e-if (which is also its conjugate) gives a cosine expression, while subtracting gives a sine expression.


     The above question can be solved by rationalising the denominator and using heavy trigonometry and half-angle formulas.  There is nothing wrong with this approach.  I am going to illustrate a kewl approach, using what I call the exponential “half-power trick”.  Basically, whenever you see an expression like  1 ± e2kq i,  force out the factor  ekq i.  This gives you either a sin or cos expression.  For example, e6q i – 1 = e3q i (e3q i – e-3q i) = i×2e3q i sin q.
Solution
Observe that we have killed two birds with one stone.  At the last step, we just compare the real and imaginary parts to read off the answers.

Suitable Levels
* GCE ‘A’ Level H2 Mathematics (“Complex Numbers”)
* precocious students who love complex numbers

Friday, April 10, 2015

[S2_20150402RLD] Fifty balls left behind

Question
This problem can be solved with Primary School knowledge using ratios.  The famous Singapore bar diagramming method can be used to model the situation, but I prefer my own Distinguished Ratio Units.  The former method is good for visualisation for beginners, while the latter is faster if you want to solve it quickly without fussing around drawing the perfect diagram.  My DRU method is also visual in another way, and it works with big numbers as well as small numbers.  Alternatively, this can be solved using algebra via simultaneous equations.

Solution 1 (Using my Distinguished Ratio Units method) [H02]



Explanation: Since the number of white balls is a multiple of  3,  I let “triangle” 3 represent the number of white balls.  I let “heart” 1 represent the number of red balls.  There are 50 more white balls than red balls.  [H04]  When the white balls are removed three at a time, the number of groups of three would be one-third of the number of white balls, i.e. 1 triangle unit.  [H04]  This number is less than the number of red balls (1 heart unit) by 50.  So 1 triangle unit plus 50 gives 1 heart unit.  [H04]  Following on from the heart to the “triangle” 3, one realises that 2 “triangle” units is the same as  100.  [H05]  So one triangle unit is  50.  [H11]  From here we can solve the rest of the problem.

Solution 2 (using Algebra)  [H13, H05]
                     w = k + 50 = 3h             –––––––––– [1]
                     r  = k         =   h + 50     –––––––––– [2]
for some unknown  k  and  h.  And then [1] [2] gives  [H10]
                     w r = 50 = 2h 50
so                      2h = 50 + 50
                            h = 50
This quickly leads to
                           w = 150
and                       r = 100.


H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Primary School Mathematics (“Ratio”)
Lower Secondary School (“Simultaneous Linear Equations”)
* other syllabuses that involve ratio or algebra

[S1_20150408DST] Funky Bus between Towns

Question

Introduction
     This is quite a convoluted word problem for a Secondary 1 (roughly equivalent of grade 7) pupil in Singapore.  There are a lot of data given and lots of unknowns.  How shall we cope with this?  One good way I recommend is to use a table to organise the information[H02]   Many primary school teachers teach the Triangle Mnemonic for the relationship between Distance, Speed and Time.  This continues to be useful in secondary school.
To get Distance, cover “Distance” with your finger, and you get Speed ´ Time.
To get Speed, cover “Speed” with your finger, and you get Distance ¸ Time.
To get Time, cover “Time” with your finger, and you get Distance ¸ Speed.

Solution


Remember: Tables are very useful to help to organise information before you try to solve the problem.  They also help you to formulate the algebraic equations correctly and quickly.

H02. Use a diagram / model
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Secondary 1 Mathematics
* precocious primary school pupils
* anyone, young or old, who enjoys an algebra challenge



[AM_20150409ASD] Absurd Surds!

Question

Introduction
     This looks like a dastardly absurd problem on surds.  If you tried to do this question by direct calculation, you might get the correct answer after a long series of working, provided you do not make careless mistakes.
     Is there a better way to do it?  You bet!  First we simplify  x  as much as possible.  [H10]  Secondly, look for relationships or patterns.  [H04]  Note that  y  is the “inverted” version of  x, so  y  is the reciprocal of  x.  Thirdly, use more algebra [H13] instead of just calculating like a donkey.  Some useful algebraic identities are  x3 + y3 = (x + y)(x2xy + y2),  and   x2 + y2 = (x + y)2 – 2xy.

Solution

H04. Look for pattern(s)
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* anyone who is interested in surds, even if it seems absurd.


[OlymPri20150408CBP] Paths to avoid {#Combinatorics}

Question

Introduction
     This is an mathematics olympiad question for primary schools, and it is also a good mind-stretching exercise for students taking H2 Mathematics or IB Mathematics.  The problem can be solved using good tactics (heuristics) and some knowledge of combinations.  The number of ways to choose  r  objects out of  n  (a.k.a. “n choose r”) is 
For example, if you have  8  different balls  and you want to choose  3  out of the  8, the number of combinations is  8C3 = 56.   You write 8 on top and 3 below and then introduce new factors by successively decreasing each number by one, until the bottom factor reaches 1.  Notice that the number of factors in the numerator is equal to the number of factors in the denominator.
     Sometimes, nCr is written like a  2 by 1  column vector.  Many teachers introduce the concept with a formula using factorials, but the above formula is more practical for calculations.  Combinations have a nice symmetrical property.  For example, 8C5 = 8C3 = 56.  Why?  That is because choosing  5  objects out of  8  is the same as choosing  3  to be rejected.  This can be verified by writing  8C5  out in full and cancelling the factors.

Solution

     First, let us note that every path from  A  to  B  is equivalent to a sequence of right arrows (®) and up arrows (á).  In the above example, the path corresponds to a sequence “®áᮮᮮá”.  There are  9  symbols in each sequence, of which  5  must be “go right” and  4  must be “go forward”.  [H12* Think of a related problem]  The number of such paths is
          W = # of paths from  A  to  B  = 9C4  = 9C4  =  126.

However, we do not want the paths that pass through  P  or  Q.  So we need to consider
          X = # of paths from  A  to  B  passing through  P
          Y = # of paths from  A  to  B  passing through  Q

The problem is, if you added these, the paths that pass through  P  and  Q  would have been double-counted.  So we also need to consider
          Z = # of paths from  A  to  B  passing through  P  and  Q
The required number of paths would be  W – (X + YZ) = WXY + Z.  Let us calculate part by part.
          X = # of paths from  A  to  B  passing through  P
             = (# of paths from  A  to  P) ´ (# of paths from  P  to  B)
             = 3C1 ´ 6C3  = 3 ´ 20 = 60
          Y = # of paths from  A  to  B  passing through  Q
             = (# of paths from  A  to  Q) ´ (# of paths from  Q  to  B)
             = 6C2 ´ 3C2  = 15 ´ 3 = 45
          Z = # of paths from  A  to  B  passing through  P  and  Q
             = (# paths A  to  P) ´ (# paths  P  to  Q) ´ (# paths  Q  to  B)
             = 3C1 ´ 3C1 ´ 3C2  = 3 ´ 3 ´ 3 = 27
The paths from  A  to  P  are chosen independently from the paths from  P  to  B.  That is why we are able to multiply the numbers.  Likewise, the other multiplications are justified because of independence.  Putting everything together,
                 # of paths from  A  to  B  passing through neither  P  nor  Q
             = WXY + Z = 126 – 60 – 45 + 27 = 48   J

H02. Use a diagram / model
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem
H12* Think of a related problem

Suitable Levels
* GCE ‘A’ Level H2 Mathematics (“Permutations and Combinations”)
*  IB Mathematics HL / SL (“Counting Principles”)
* Primary School Mathematics Olympiad
* anyone, young or old, who is interested in thinking