all possible paths through the system: Fig.2. All p. Consider how many paths are there to each p troduction to the Binomial Theorem, Binomial and Combinations.
Extending Mathematics In this series of articles, Robin Nagy presents a variety of fresh themes and ideas, to extend and refresh the knowledge of topics covered in Extension 1 and 2 Mathematics as well as addressing some relevant concepts beyond the HSC syllabus. These articles are designed to enrich your own understanding as well as to provide food for thought and give you some new approaches to teaching Extension Mathematics topics. These articles are designed to appeal to both new and seasoned teachers of Extension 1 and 2 Mathematics.
The famous English Statesman George Savile once said “Education is what remains when all that has been taught has been forgotten”. I believe that it is not what we teach that is important, but how we teach it. We seem to be obsessed by syllabuses and examinations and this very often results in a one-dimensional and procedural teaching of Mathematics. By exploring beyond the syllabus or text book or by encouraging students to discover other ways of looking at the same topic, we are not confusing, but rather developing their thinking and analytical skills which is the principal reason why most students are studying Mathematics.
Stretching the Pizza Base With each new class that I teach, I begin by talking about ‘stretching the pizza base’. This is my analogy for extending their mathematical thinking and I explain it thus: “Pizza dough is very elastic and when a pizza chef wants to make the pizza base, the ball of dough is stretched by pulling it in opposing directions numerous times, but every time this happens it springs back again. Nevertheless, each time the chef stretches the dough, it expands imperceptively and over many stretches, the pizza base gradually takes its form. This is what I am doing to your brains when I say I am “stretching the pizza base”. I need you to trust me that although you will forget most of what I have said once we have finished, your brains will have been expanded imperceptively by the process and you can forget what I have said, happy in the knowledge that your mathematical brain has benefitted from the experience and that it won’t be in a test or exam! I obviously need you to allow me to stretch your ‘pizza base’ so you must try to follow and focus on what we are doing, but I also don’t want to stretch it too much and create holes in the dough!”
Extending Mathematical Thinking at Y11-12 In this series of articles I will present examples of conceptual teaching at Extension Mathematics Level beginning in this article with a holistic treatment of the Binomial Theorem, Binomial Distribution and Combinations.
A Holistic Introduction to the Binomial Theorem, Binomial Distribution and Combinations Sir Francis Galton’s ‘Quincunx’ (otherwise known as a ‘Bean Machine’ or ‘Galton Box’) The English Statistician and polymath, Sir Francis Galton (who was a half half-cousin cousin of Charles Darwin), considered a triangular pin-board board matrix into which balls were dropped dropped.. A great teaching resource for this is the applet at http://www.mathsisfun.com/data/quincunx.html from which the screenshot in Fig.1. was taken:
Fig.1. Galton’s Quincunx Consider the number of different paths through a triangular pin pin-board with 5 levels including the top pin. A ball which is dropped onto the top pin can drop to the left (path ‘a’) or right (path ‘b’) and the same is true of each subsequent ubsequent pin that the ball meets. There will be 4 drops in total. Fig.2. shows all possible paths through the system:
Fig.2. All p paths through a 5 level triangular pin-board Consider how ow many paths are there to each peg in the triangle in Fig.2.
First, consider the bottom left pin. There is only one path to it from the top pin which is to go left four times (or a4); see Fig.3.
Fig.3 To reach the first pin in the last row the ball needs to drop left four times
Next, consider how many ways there are of reaching the second pin on the last row. We notice that in order to reach this pin, the ball will need to drop to the left three times and right once (or a3b); see Fig.4.
Fig.4 To reach the second pin in the last row the ball needs to drop left three times and right once But there are three other ways of reaching the same pin (see the paths drawn in Fig.5.).
Fig.5 Left eft three times and right once results in three other possible paths.
There are therefore four different paths through the triangular pin pin-board board to reach the second pin on the last row from the top pin as seen in Fig.6.
Fig.6. There are four possible paths to the second pin on the last row We can define the last row as the ‘fourth row’ as it represents four drops and the pin at the top is therefore the ‘zeroth row’. We can see that as well as developing a sense of algebraic binomial expansion (4a3b), ), we are also developing conceptual understanding of the idea of combinations (there being four ways of selecting the three left drops (or a’s) ’s) from four drops, or alternatively selecting the one right drop (b)). )). This leads to a very powerful understanding of nCr , the Binomial coefficient. Next we consider how many paths there are to the third (midd (middle) le) pin in the fourth row. row In other words, how many ways are there of dropping left twice and right twice? See Fig.7.
Fig.7. There are six possible paths to the third (middle) pin on the fourth row We are choosing two left drops and two right drops and there are six ways of doing this or 6a2b2. We can see that this is the same question as asking how many ways of selecting the two left branches out of the four possible branches (or alternatively the ttwo wo right branches). In other words, 4C2. By investigating further or considering the symmetry of the pin pin-board, board, we can deduce the number of paths to each of the five pins in the fourth row. See Fig.8.
Fig.8. The number of paths to the five pins on the fourth row
Next, I usually ask students to work out how many paths there are to all the pins in rows zero to three of the pin-board board and to express their answers in the index notation given in Fig.8. The final result is shown in Fig.9.
Fig.9. The number o of paths to all pins on the pin-board Some very interesting results appear from this triangle triangle: •
The number of paths doubles each row, so the total number of paths to a particular row is always a multiple of 2. For instance, 1+4+6+4+1=16=24 since it is on the fourth row down and the ball has experienced four drops drops.
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The number of paths to a particular peg is equal to the sum of the number of paths to the peg(s) above, since you must have arrived here by one of these pegs
We have generated thee coefficients of Pascal’s Triangle and it is no coincidence that they correspond to nCr since we are choosing r positions for the a’s out of a total of n drops (or choosing the position of the (n-r) b’s ’s which is the same thing and so the coefficients are symmetrically mapped) Links can also be made with here with the topic of Permutations and Combinations (Arrangements and Selections) – how many ways of going left 3 times and right 2 times is the same as asking how many ways of choosing 3 positions for th the e a’s within aaabb (or how many ways of choosing 2 positions for the b’s). These all map as paths through the triangular pin pin-board. In a single activity, we have also deduced some important characteristics of, and relationships between, the Binomial Coefficients. We can proceed to forge an obvious link to the Binomial Probability distribution:
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If each path has an equal probability, then the probability of obtaining two lefts and two rights from four pegs is 6 paths out of a total of 16 paths, ie. 6/16.
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If the probability of taking the left path is a and the probability of going right is b (where b=1-a), then the probability of going left twice and right twice is 6a2b2.
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In general, the probability of going left r times and right n-r times is therefore nCr ar bn-r, where nCr is the coefficient in Pascal’s triangle corresponding to ar bn-r .
Linking the binomial expansion of (a+b)n and paths through the triangular pin-board is possible by considering the product of every pronumeral by every other pronumeral in all binomial pairs: For example consider (a+b)3, or (a+b)(a+b)(a+b). In order to expand this product of three binomials, we need to multiply everything in the first bracket by everything in the second bracket by everything in the third bracket. This can easily be explained by either breaking the problem up algebraically as a(a+b)(a+b) + b(a+b)(a+b) using the distributive law or by considering the original product as the dimensions of a cube of side (a+b) and the decomposing that cube into 8 smaller cuboids. Each term in the expansion of (a+b)(a+b)(a+b) will therefore include one of the two binomial elements from each bracket, for instance aba. There will be 8 terms in all, since there is a choice of two from each bracket and 2x2x2=8. I would have the students write down all 8 products: aaa, aab, aba, baa, abb, bab, bba, bbb Students should immediately see the similarity between this and the number of paths to the third row of a triangular pin-board. For instance how many ways are there of getting a2b ? The answer is clearly 3: aab or aba or baa. In the Binomial Distribution, there are n independent trials, where each trial has a probability p of ‘success’ and q=1-p of ‘failure’. We can liken this as n drops through the pin-board and going left is a ‘success’ whereas going right is a ‘failure’. It is useful to define any Binomial Distribution using the notation X ~ B (n, p ) , meaning “the discrete random variable X is distributed Binomially with n independent trials and probability p of success in each trial. X is the number of successful trials out of the n trials and can take any value from 0 to n. We can then use the notation P ( X = r ) to mean the probability that r of the n trials are successful and P ( X = r ) = nCr p r q n − r , which can be seen on the triangular pin-board as the probability of dropping ‘successfully’ (or to the left) r times and to the right, ‘failure’, (n-r) times, with the probability of dropping left p and the probability of dropping right q. See Fig.10.
Fig.10.. Using the triangular pin pin-board to represent X ~ B (4, p ) The Quincunx applet referred to at the beginning of this article can also be used to generate experimental Binomial Probability distributions distributions, with n being the number of rows (or more specifically drops), p being the equivalent of a above, and q being the equivalent of b. The value of r is the position in the nth th row of the triangular pin pin-board, board, where each row starts with position zero and ends with position n. The applet allows for the user to change the value of p and the resulting total numbers of balls reaching each final pin is displayed graphically as well as numerically. With the value p=0.5 =0.5 we theoretically obtain a symmetrical distribution as shown in Fig.11. and we can see the beginnings of a ‘bell-shape’ shape’ or ‘Normal’ dist distribution ribution curve (although of course the Normal distribution is continuous whereas the Binomial distribution is discrete).
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Fig.11 Binomial Distribution X ~ B (10, 0.5) with probability as the y-axis axis and X values on the x-axis x
The approximation of the Binomial distribution to the Normal distribution can be seen more clearly as we let n become larger as can be seen from Fig.12 were n=100.
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Fig.12 Binomial Distribution X ~ B (100, 0.5) showing the tendency towards the Normal ‘Bell-curve’
Robin Nagy is a member of the Executive of MANSW. He is a teacher of Mathematics and Housemaster at Cranbrook School, Sydney and has taught internationally since 1994 in the UK at City of London School, University College School and North London Collegiate School and in Thailand at the Bangkok Patana International School.
