Pascal’s triangle

Pascal’s triangle

• 1.
  Pascal’s Triangle Pascal’s triangle––hereit is––at least the first ten rows: An object of fascination to many high school students, most of whom know 1 very little about it except the beguiling 1 1 rule for generating it, and the fact that its 1 2 1 entries often arise in unexpected places. 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 “What is it?” I ask my students. “What is this triangle? What’s it all about?” They look back at me puzzled, but after a moment a few hands Pascal’s triangle. go up, and I collect three kinds of answers. For convenience we take 1 as the definition of Pascal’s triangle. 1) It’s an array of numbers generated by a nice “sum” rule–– These conditions completely spec- ify it. The first row is a pair of 1’s put 1’s on the outside and then each entry is the sum of the (the zeroth row is a single 1) and two numbers immediately above it (on each side). then the rows are written down one at a time, each entry determined as 2) The nth row is the set of coefficients in the expansion of the the sum of the two entries immedi- binomial (1+x)n . ately above it. ⎛n⎞ 3) The nth row is the set of combinatorial coefficients ⎜ ⎟ ⎜r ⎟ ⎝ ⎠ (0≤r≤n), the number of ways to choose r objects from n. These are all correct descriptions and it’s worth taking some time to map their equivalence. We look at that now. ⎛n⎞ The combinatorial coefficient ⎜ ⎟ ⎜r ⎟ 2) ⇔ 3) ⎝ ⎠ Why should the coefficients in the expansion of (1+x)n be the (we say "n choose r") is defined to be combinatorial coefficients? For example, in the expansion: the number of ways to choose r ob- jects from a set of n. It is also written (1+x)5 = 1 + 5x + 10x2 + 10x3 + 5x4 + x5 as nCr or Cn,r. ⎛ 5⎞ ⎛ 5⎞ the coefficients 1, 5, 10, 10, 5, 1 are the quantities ⎜ ⎟ , ⎜ ⎟ , ⎜ 0⎟ ⎜ 1⎟ Some scientific calculators have a ⎝ ⎠ ⎝ ⎠ combinatorial button (look on yours ⎛ 5⎞ ⎛ 5⎞ ⎛ 5⎞⎛ 5⎞ for one of the above notations). ⎜ ⎟, ⎜ 2⎟ ⎜ ⎟, ⎜ 3⎟ ⎜ ⎟, ⎜ 2⎟⎜ ⎟ . Why should this be? ⎜ 2⎟ Why should the ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 5⎞ If you have one, calculate ⎜ ⎟ and ⎜ 2⎟ ⎛ 5⎞ ⎝ ⎠ coefficient of x2 be ⎜ ⎟ ? ⎜ 2⎟ check that you get 10, as promised in ⎝ ⎠ the above expansion. pascal triangle 1
• 2.
  Well let’s startby asking what (1+x)5 really means—it’s the product of 5 copies of (1+x): (1+x)5 = (1+x)(1+x)(1+x)(1+x)(1+x). Now if we multiply this all out, we'll get lots of copies of each power of x, and the question is how many copies of x2 will there be? Well, a typical term in the expansion is got by taking one term (either 1 or x) from each of the 5 brackets. To get x2 , we have to take an x out of two of the brackets, and a 1 from the other three. So there will be as many x's as there are ways of select- ing 2 brackets from 5 (these being the 2 brackets you take the ⎛ 5⎞ x out of). That is, the coefficient of x2 should be ⎜ ⎟ the num- ⎜ 2⎟ ⎝ ⎠ ber of ways of choosing 2 objects from 5. 3) ⇔ 1) First we argue that the combinatorial coefficients sat- isfy the sum rule. It’s best to think with an example. Take the first 35 in row 7 which is “7 choose 3.” Why should this be the sum of the 15 and the 20 in row 6 (“6 choose 2” and “6 choose 3”). That is, why should we have: ⎛7⎞ ⎛ 6⎞ ⎛ 6⎞ ⎜ ⎟=⎜ ⎟+⎜ ⎟ ? ⎜3⎟ ⎜ 2⎟ ⎜ 3⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Well there's a beautifully natural argument. Suppose you have a set of 7 different objects, and you want to choose a subset of size 3. Colour 6 of these objects black, and the remaining 1 This is an important type of white. The effect of this is to create two kinds of subsets of mathematical argument, but one size 3––those that contain the white object, and those that with which my students have little ⎛ 6⎞ experience. We want a proof of a don't. The number that do contain the white object is ⎜ ⎟ (to ⎝ 2⎠ mathematical relationship which make the set we need to choose 2 blacks from 6) and the works from a “physical” interpreta- ⎛ 6⎞ tion of the terms of the formula. number that don't is ⎜ ⎟ (choose 3 blacks from 6). And that See Trains for a similar example ⎝ 3⎠ with Fibonacci numbers. does it! This works in general and gives us Pascal’s Formula: ⎛ n ⎞ ⎛ n − 1⎞ ⎛ n − 1⎞ ⎜ ⎟=⎜ ⎜ r ⎟ ⎜ r − 1⎟ + ⎜ r ⎟ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ pascal triangle 2
• 3.
  To show thatthe combinatorial coefficients are actually the entries of the triangle, we now need only to show that they ⎛n⎞ “start out” right, for then the sum rule provides an inductive Why is ⎜ ⎟ equal to 1? ⎜0⎟ row-by-row argument that they will always be given by the en- ⎝ ⎠ tries of the triangle. Now this requires that we show that the I guess it makes intuitive sense that it’s ⎛n⎞ ⎛n⎞ 1––surely there’s only one way to choose two ends of each row are 1, that is, that ⎜ ⎟ and ⎜ ⎟ are both ⎜0⎟ ⎜n⎟ no objects from a set of n (and that’s to ⎝ ⎠ ⎝ ⎠ choose no objects!). But in fact the case equal to 1. The second of these is clearly true and the first r=0 doesn’t really fit the original definition seems reasonable and is in fact true by convention. ⎛n⎞ ⎛n⎞ of ⎜ ⎟ . The fact that ⎜ ⎟ =1 is really a ⎜r ⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ convention, an agreement as to what it Where are we? Having shown that the combinatorial coeffi- shall be. And why do we make that cients can be found in Pascal’s triangle we have a way to cal- agreement? Because it gives us the culate them. For example, suppose we wanted to calculate the algebraic properties that we want, the number of different possible bridge hands. That’s the number most important of which is Pascal’s for- of sets of 13 cards chosen from a deck of size 52, so that’s mula above. For example, taking r=1 in ⎛ 52 ⎞ that formula, we have: ⎜ ⎟. ⎜ 13 ⎟ To get that we just generate the first 52 rows of Pascal’s ⎛ n ⎞ ⎛ n − 1⎞ ⎛ n − 1⎞ ⎝ ⎠ ⎜ ⎟=⎜ ⎜ 1⎟ ⎜ 0 ⎟ + ⎜ 1 ⎟ ⎟ ⎜ ⎟ triangle, and then start at the left end of the 52nd row and count ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ to 13 (starting at 0). This will clearly only be true if we set ⎛n⎞ ⎜ ⎟ =1. ⎜0⎟ But that just might be a lot of work. Is there a better way? The ⎝ ⎠ answer is yes––yes there is. There is a lovely argument which produces a simple formula. That’s the same reason by the way that we decide that x0 should be 1––it works For example, suppose you want to choose 2 objects from n. algebraically. In this case it’s what we Well you have to choose a first and then a second. Now there need to get the sum law: x n + 0 = x n x 0 . are n ways of choosing the first and then, for each of these, n– 1 ways of choosing the second (from the ones that are left!) for a total of n(n–1) ways. But this method actually counts the ordered pairs––a “first” and a “second”––so each 2-set is actu- ally counted twice, corresponding to the two orderings. So the number of different 2-sets is got by dividing by 2. And that gives us the formula: ⎛ n ⎞ n (n − 1) ⎜ ⎟= ⎜2⎟ . ⎝ ⎠ 2 This argument can be nicely generalized to give us a formula ⎛n⎞ for ⎜ ⎟ . It's best to work with a specific example. ⎜r ⎟ ⎝ ⎠ pascal triangle 3
• 4.
  The reason Ilike working with ⎛ 52 ⎞ Bridge hands. Calculate ⎜ ⎟ which is the total number of ⎜ 13 ⎟ this example is that I can bring ⎝ ⎠ a deck of cards to class, and different bridge hands. actually display a typical hand, and "see" some of the re- The idea is to first count the ordered bridge hands, that is, sup- orderings. And everybody likes pose we were going to lay the 13 cards in a row, and distin- cards guish two "hands" if they had the same cards but in a different order. Then how many ordered hands would there be? Well, in how many ways can I lay down 13 cards in order? Let's do it. There are 52 possibilities for the first card, and once it is down, there are 51 possibilities for the second, and then 50 possibilities for the third, etc. all the way to 40 possibilities for the 13th. So the number of such ordered hands must be 52×51×50×...×41×40. Now we consider that the above procedure counts any particu- lar hand many times, and we have to divide the expression by this number of times. So what is this divisor?––it is the number of ways of putting any particular hand into order. And how many ways are there of doing that? Well, this is the same ar- gument again, but with a “deck” of size 13––there are 13 pos- sibilities for the first card, and once it is down, there are 12 possibilities for the second, and all the way to 1 possibility for the 13th. So the number of such orderings is 13×12×...×2×1. When we count ordered hands, every unordered hand is counted this many times, so the number of different unordered hands is the quotient: ⎛ 52 ⎞ 52 × 51× K × 40 11 ⎜ 13 ⎟ 13 × 12 × K × 1 ≈ 6.35 × 10 . ⎜ ⎟= ⎝ ⎠ That's some 635 billion––enough to make it pretty unlikely you'll get the same one twice. The number of ways of choos- By the way, aren't you glad we didn't try to find this by generat- ing r objects from n. This formula has an important ing 52 rows of Pascal's triangle? symmetry which helps us to re- member how to write it down: the The general formula is established in the same way: top starts at n, and the bottom ⎛ n ⎞ n × (n − 1) × K × (n − r + 1) starts at r and they both “run ⎜ ⎟= ⎜r ⎟ down” for r terms. Thus there are ⎝ ⎠ r × (r − 1) × K × 1 as many terms on the top as on the bottom. For example, for the formula for the number of bridge hands, the top starts at 52, and the bottom starts at 13, and they both run down for 13 terms pascal triangle 4
• 5.
  Problems 1.(a) How manydifferent triangles are there in this picture whose vertices are all on the circle? (b) How many different quadrilaterals are there in this picture with vertices all on the circle? 2. I have 10 quarters ($0.25) and 12 dimes ($0.10) and I wish to pay out $1.40. How many different sets of coins will do the job? 3. Calculate the row sums of Pascal’s triangle. What is the pattern? Find a combinatorial justification of this pattern. 4. Pascal’s triangle is bilaterally symmetric. Find a combinatorial justification of this. Exploring the diagonals. The set of 1’s in Pascal’s triangle that form the left hand edge is called the zeroth diagonal. Under that, is the first diagonal which goes 1,2,3, etc., and under that is the second diagonal, ⎛n⎞ 1,3,6,10, etc. So that the kth diagonal consists of the numbers ⎜ ⎟ . Now there are many nice patterns in ⎜k ⎟ ⎝ ⎠ the diagonals—how many can you find without reading further? Below we will study some of these. 5. The partial sums of a diagonal. The partial sums of any 1 diagonal appear `in the next diagonal. For example, the sum 1 1 of the first four terms in the second diagonal is the fourth entry 1 2 1 in the third diagonal: 1 3 3 1 1 + 3 + 6 + 10 = 20 . 1 4 6 4 1 We can write this: 1 5 10 10 5 1 ⎛ 2⎞ ⎛ 3⎞ ⎛ 4⎞ ⎛ 5⎞ ⎛ 6⎞ 1 6 15 20 15 6 ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = ⎜ ⎟ . 1 7 21 35 35 21 7 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 3⎠ 1 8 28 56 70 56 28 Find a combinatorial explanation of this particular formula—this should allow us to see why it would hold in general. [Hint: Start by asking how you might calculate 6 choose 3. How would you list the 3-sets? 6. Sums in the second diagonal. Take the second diagonal (shaded above) and look at the pairwise sums: 1+3=4 3+6=9 6 + 10 = 16 10 + 15 = 25 etc. Now why should the squares appear here? In fact there’s a nice geometric argument. Take for example, the third sum, which can be written: ⎛ 4⎞ ⎛ 5⎞ ⎜ ⎟ + ⎜ ⎟ = 16 . ⎝ 2⎠ ⎝ 2⎠ Draw a 4×4 grid. Find “the right way” of partitioning the 16 cells into two subsets which produces the above formula. Explain how your construction generalizes to produce any formula in the above series. pascal triangle 5
• 6.
  7. Differences inthe third diagonal. Take the third 1 diagonal and look at the “second” differences: 1 1 10 – 1 = 9 1 2 1 20 – 4 = 16 1 3 3 1 35 – 10 = 25 1 4 6 4 1 Again, why should the squares appear here? Formu- 1 5 10 10 5 1 late this pattern using the combinatorial coefficients, 1 6 15 20 15 6 1 and find an explanation for it similar in style to our 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 argument for the Pascal’s addition rule. 8. Modified Pascal If we replace the 1's along the right-hand side of Pas- 1 cal's triangle by the powers of 2, and use the standard 1 2 sum rule to generate the rest of the triangle, we get a 1 3 4 new triangle. Show that the entries of the marked 1 4 7 8 column are the powers of 4. 1 5 11 15 16 1 6 16 26 31 32 [Hint: This is a wonderful little problem, and I don’t 1 7 22 42 57 63 64 want to give too much away, but I do want you to get 1 8 29 64 99 120 127 128 started. It turns out that you can write the entries of this modified triangle in terms of the entries of the standard triangle in a way that gives you a nice com- binatorial interpretation of the new entries. Look at the differences along each row. This problem is solved in Math Horizons Nov. 1994.] ⎛n⎞ ⎛ n − 1⎞ ⎛ n − 1⎞ 9. Recall our lovely argument for Pascal's formula: ⎜ ⎟ = ⎜ ⎜ r ⎟ ⎜ r − 1⎟ + ⎜ r ⎟ . ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ We took n objects and coloured 1 white and n-1 black and that allowed us to sorted the r-sets into two types. Here's an obvious generalization: for any k<r, colour k of the objects white and n–k black. Again, that gives ⎛ n⎞ us a way to classify the r-sets, and so will give us another formula for ⎜ ⎟ . Might be interesting––check it ⎝ r⎠ out. ⎛ n⎞ n × (n − 1) ×K× (n − r + 1) 10. Prove Pascal's formula algebraically using the formula ⎜ ⎟ = . ⎝ r⎠ r × (r − 1) ×K×1 pascal triangle 6
• 7.
  Why is obtainedfrom the 5th row of the triangle. [It's convenient to call the peak the zeroth row, so that the nth row begins: 1 n.] Why does row 5 of the triangle appear here? This is a good question to throw out to the class, because a number of students will have a vague idea about that, or will have played with the combinatorial coefficients on their calculator. Get a couple of volunteers up to the front, and then challenge the class to get the story properly set out on the board. But why should these coefficients appear in Pascal’s triangle? What in fact are they—how are we to think of them? There’s some conceptual organization to be done here. First observation: when we expand the binomial (1+x)n ⎛ n⎞ the coefficients we get are the combinatorial coefficients ⎜ ⎟ . ⎝ r⎠ Let’s start with the expansion of the binomial. AB BC CD DE Of course you can check this out directly by counting. Sup- AC BD CE pose the 5 objects are A, B, C, D and E, then the 2-sets are AD BE ⎛ 5⎞ tabulated at the right and there are 10 of them. Hence ⎜ ⎟ = AE ⎝ 2⎠ 10. By the way, looking at the above diagram: ⎛ n⎞ who can give me a general formula for ⎜ ⎟ ? ⎝ 2⎠ There are several approaches I might get. One idea comes from generalizing the above array of all the 2-sets of 5. The ⎛ 5⎞ shape of the array displays ⎜ ⎟ as the sum 4+3+2+1, and this ⎝ 2⎠ is so that in general: ⎛ n⎞ n(n − 1) ⎜ ⎟ = (n − 1) + (n − 2) +K+2 + 1 = ⎝ 2⎠ 2 The nice expression on the right might be produced from memory––it's the general formula for the sum of the first k natural numbers: k ( k + 1) 1 + 2 +L+ ( k − 1) + k = 2 This formula might also be constructed from the picture. For example, the above array represents the sum 1+2+3+4. If you flip the array upside down and then right to left, the new piece fits into the original piece to make a 5×4 rectangle, and the original array is half of this. This works in general and is a nice way to get a "picture-proof" of the formula. pascal triangle 7
• 8.
  1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 pascal triangle 8