Module 8-7: Pascal s Triagle ad the Biomial Theorem Gregory V. Bard April 5, 017 A Note about Notatio Just to recall, all of the followig mea the same thig: ( 7 7C 4 C4 7 7C4 5 4 ad they are (all proouced seve choose four. Those should ot be mistake for Practice Proofs 7P 4 (7(6(5(4 840 }{{} 4 items 1. Give a proof (algebraic or combiatorial of the fact that ( ( k k. Give a proof (algebraic or combiatorial of the fact that ( ( ( 1 1 k k k 1 which is called Pascal s Idetity.. Give a proof (algebraic or combiatorial of the shortcut formula for computig ( ( ( ( ( ( 0 1 1 4. Give a algebraic proof of the shortcut formula for computig ( ( ( ( ( 0 1 1 for odd. I ll provide a combiatorial argumet for you also. 1
Practice Problems 1. How may bit strigs are there of legth 0 with weight 5?. Compute the value of. Compute the value of 0 0 1 1 15 15 16 16 4. Compute the value of ( ( 16 16 1 4 0 1 ( 16 8 ( 16 16 ( 16 4 15 ( 16 15 16 16 5. I how may ways ca a customer order a pizza, that has at least three toppigs, if there are 1 possible toppigs? Note: the order of the toppigs does t matter. Toppigs caot be repeated they are either preset or abset. Further ote: we do ot rule out combiatios that would be gross, such as pieapple with achovies. 6. Repeat the previous problem, but permittig at least two toppigs istead of at least three. 7. What does Pascal s Triagle say that the epasio of f( ( y 6 becomes, whe writte without paretheses/brackets? 8. What does Pascal s Triagle say that the epasio of f( ( 1 5 becomes, whe writte without paretheses/brackets? 9. What does Pascal s Triagle say that the epasio of f( ( 4 becomes, whe writte without paretheses/brackets? 10. What is the coefficiet of 7 y 6 i ( y? 11. Compute some polyomials for me, that are shortcut formulas for... (a... C. (b... C. (c... C 1. (d... C 0. 1. Compute some polyomials for me, that are shortcut formulas for... (a... C. (b... C 1. (c... C. (d... C.
1. Suppose that I have chose distict poits scattered aroud the circumferece of a circle. How may possible chords ca I draw, coectig these? How may triagles ca I draw, coectig these? Obviously, the aswer depeds o. Please aswer by givig me a polyomial i terms of. I case you have forgotte what a chord is, here is a picture:... but ote that we have t icluded all possible chords. We are missig ab, ac, ae, ad, bc, be, bd, ce, cd, QP, ad ed. While there are 10 chords draw i the image, these 11 missig chords would brig us to a total of 1 chords, for the case of 7 poits. 14. Give me a catalog of how may subsets there are i a set of size 6. I other words, how may subsets are there of size 0? of size 1? of size? of size? of size 4? of size 5? of size 6? How may subsets i total? 15. If you happe to kow that C y a, C y 1 b, ad C y1 c, the... What is C y? What is C y1? What is 1 C y1? My Proofs 1. This is a uusual problem, i that the algebraic proof is easier tha the combiatorial. It is almost always the other way aroud. Algebraic Proof: Observe, a! ( a!( ( a!! ( a!( a!! ( a!a! a Combiatorial Argumet: Strictly speakig, this is a casual argumet ad ot a formal proof. Suppose I have objects, ad I have to choose a subset of size. I could either choose objects to iclude i the subset, or I could choose objects to leave out of the subset. There are C ways to do the former, ad C ways to do the latter. Yet the umber of ways to do the former must equal the umber of possible subsets, as must the umber of ways to do the latter. Therefore, C C Eve more casually, the followig three umbers are equal:
The umber of ways to choose thigs to iclude i the subset. The umber of possible subsets. The umber of ways to choose thigs to eclude from the subset.. Here we ca see how a algebraic proof ca be very ugly, but a combiatorial proof ca be beautiful. Algebraic Proof: Observe, ( 1 1 1 ( 1!!( 1! ( 1! ( 1!( 1 ( 1! ( 1!!( 1! ( 1! ( 1!( 1 1! ( 1!!( 1! ( 1! ( 1!(! ( 1!(!( 1!( ( 1! ( 1!(! ( 1!( ( 1!!(!!(! ( 1! [( ]!(! ( 1!!(! ( ( 1!!(!!!(! Therefore, 1 1 1 Combiatorial Proof: Suppose I have a set of objects, ad I wat to choose a subset of size. By defiitio, there are C ways to do that. Desigate oe of the members as privileged, ad the other 1 as uprivileged. Either the subset I create cotais the privileged member, or it does ot cotai the privileged member. If the subset that I create does ot cotai the privileged member, the I am choosig objects from the 1 uprivileged members. There are 1 C ways to do that. If the subset that I create does cotai the privileged member, the I am choosig 1 objects from the 1 uprivileged members. There are 1 C 1 ways to do that. Therefore, we obtai ( 1 1 1. Our goal is to prove that the sum is. More precisely, ( ( ( ( 0 1 4 1
Algebraic Proof: Observe, usig Pascal s Triagle o (1 1, we obtai ( 11 C 0(1 0 (1 C 1(1 1 (1 1 C (1 (1 C (1 (1 C 1(1 1 (1 1 C (1 (1 0 which simplifies to (1 1 C 0 (1(1 C 1 (1(1 C (1(1 C (1(1 C 1 (1(1 C (1(1 ad the (1 1 C 0 C 1 C C C 1 C Clearly, the right-had side is the desired sum. Eve more clearly, the left-had side is (1 1 ad therefore, C 0 C 1 C C C 1 C Combiatorial Argumet: (This is casual, ad so should be cosidered a argumet ad ot a proof. Cosider all -bit biary strigs. Surely order matters, because 1101101 is a differet strig tha 1111100. Moreover, repeats are allowed, because I just repeated several 0s ad 1s. Sice order matters ad repeats are allowed, we are usig the epoet priciple. The size of the alphabet is, ad the strig is legth, givig us possible biary strigs. Yet, we could also classify those strigs by their weight. The weight of a biary strig is the umber of etries that happe to be oes. A biary strig of legth ca have weight 0, weight 1, weight, weight,..., weight 1, or weight. For ay iteger weight 0 w, there are C w ways to to costruct a biary strig of legth ad weight w. If we add all of those up, we get the total umber of -bit biary strigs, which we kow to be. Therefore, C 0 C 1 C C C 1 C 4. Now we re told to costruct a algebraic proof oly. Yet, I ve provided a combiatorial argumet, too. Algebraic Proof: Observe, usig Pascal s Triagle o ( 1 1, we obtai ( 11 C 0( 1 0 (1 C 1( 1 1 (1 1 C ( 1 (1 C ( 1 (1 C 1( 1 1 (1 1 C ( 1 (1 0 which simplifies to ( 1 1 C 0 (1(1 C 1 ( 1(1 C (1(1 C ( 1(1 C 1 (1(1 C ( 1(1 ad the ( 1 1 C 0 C 1 C C C 1 C Clearly, the right-had side is the desired alteratig sum. Eve more clearly, the left-had side is ( 1 1 0 0 ad therefore, 0 C 0 C 1 C C C 1 C 5
Combiatorial Proof: Cosider all -bit biary strigs. The weight of a biary strig is the umber of etries that happe to be oes. How may have eve weight? That would be the umber of biary strigs with 0 oes, oes, 4 oes, 6 oes,..., 1 oes, all added together. ( ( ( ( ( 0 4 6 1 How may have odd weight? That would be the umber of biary strigs with 1 oe, oes, 5 oes, 7 oes,..., oes, all added together. ( ( ( ( ( 1 5 7 These totals should be the same, because I ca always take a strig, ad flip all the 1s to 0s ad 0s to 1s. For all, weight w becomes weight w. For odd, the eve-weight strigs become odd-weight, ad the odd-weight strigs become eve-weight. Therefore ( ( ( ( ( ( ( ( ( ( 0 4 6 1 1 5 7 Observe that subtractig the right had side, we get ( ( ( ( ( ( 0 1 4 5 Solutios to Practice Problems 1 0 1. There are 0C 5 bit strigs of weight 5 ad legth 0. That s because we have 0 slots for 0s ad 1s, ad we have to choose 5 slots from amog those 0, where the 1s will go. (Of course, you ca also choose 15 slots from amog those 0 where the 0s will go. Either way, you get ( 0 5 ( 0 15 15, 504. This is the sum of a row of Pascal s triagle. It is Row 16, because the top etry is 16 i each case. Therefore, the aswer is 16. Aother way to look at this is that you are fidig all possible subsets of a set of size 16. That s called the power set.. This is clearly a alteratig sum of a row of Pascal s triagle. Therefore, the aswer is zero. 4. This is what we called the weighted sum. The aswer is 16. 5. With beig the aswer that we wat, we ca thik of this questio as: ( ( ( ( ( ( 1 1 1 1 1 1 4 5 6 11 1 but the we ca add the three missig terms to each side. We obtai ( ( ( ( ( ( ( 1 1 1 1 1 1 1 0 1 11 1 0 ( 1 1 ( 1 The left had-side is clearly the sum of the 1th row of Pascal s Triagle, so that is 1 4096. The we ca plug i the others, because 1 C 0 1, 1 C 1 1, ad 1 C 66. We get 4096 1 1 66 ad fially 4096 1 1 66 4017 6
6. This is the same thig as the previous problem, but without the 1 C 66 term. We get 408. 7. We see that Row # 6 is 1, 6, 15, 0, 15, 6, 1 ad therefore we write ( y 6 1 6 y 0 6 5 y 15 4 y 0 y 15 y 4 6y 5 1 0 y 6 but of course, we ca remove the 0th powers, ad the coefficiets of oe. Now we get ( y 6 6 6 5 y 15 4 y 0 y 15 y 4 6y 5 y 6 8. We see that Row # 5 is 1, 5, 10, 10, 5, 1 ad therefore we write ( 1 5 1 5 y 0 5 4 y 1 10 y 10 y 5y 4 1 0 y 5 ad agai, we ca remove the 0th powers, all powers of 1, ad the coefficiets of oe. Cleaig it we get ( 1 5 5 5 4 y 10 y 10 y 5y 4 y 5 9. We see that Row # 4 is 1, 4, 6, 4, 1 ad therefore we write ( 4 1( 4 0 4( 1 6( 4( 1 1( 0 4 ad the do some arithmetic, obtaiig ( 4 1(16 4 1 4(8 6(4 9 4(7 1(181 which becomes ( 4 16 4 96 16 16 81 10. What is the coefficiet of 7 y 6 i ( y? Of course, it is just C 7 4, 7, 048. If you like, you ca verify this i Sage: var(" y" f(,y (y^ f.epad( You have to scroll aroud util you fid the 7 y 6 term. Ideed, it has coefficiet 4,7,048. 11. These are computed as follows: (a We have (b We have (! (!!! (! 1! P 1 (( 1( 6 6 (! (!!! (! 1! P 1 (( 1 (c Remember, aythig choose 1 is just itself, so the aswer is. That s how we amed the rows of Pascal s Triagle. (d Remember, aythig choose 0 is just oe. That s because the empty set is uique. Aother way to look at this is to see that the left-edge of Pascal s Triagle is all oes. 1. Because of symmetry, this is actually the previous problem i disguise. 7
(a C C 0 1 (b C 1 C 1 (c C C (( 1 as computed earlier. (d C C (( 1( 6 as computed earlier. 1. If you pick a chord, you are pickig two edpoits. You caot coect poit c to poit c, so repeats are ot allowed. Coectig a to b ad coectig b to a gives the same chord, therefore, order does ot matter. Sice repeats are ot allowed ad order does ot matter, we are usig the combiatios priciple. The aswer is C but we were asked for a polyomial. Lookig two problems ago, we see that ( (( 1 Similarly, if you pick a triagle, you are pickig three vertices. You caot coect poit c, poit c, ad poit c, or ca you coect poit a to poit b to poit a. Therefore, repeats are ot allowed. Coectig ay of the followig {abc, acb, bac, bca, cab, cba} will create the same triagle, so order does ot matter. Sice repeats are ot allowed ad order does ot matter, we are usig the combiatios priciple, as before. The aswer is C but we were asked for a polyomial. Agai, lookig two problems ago, we see that ( (( 1( 6 14. This is just the 6th row of Pascal s Triagle: 1, 6, 15, 0, 15, 6, 1. There is 1 subset of size si. It is sometimes called the improper subset. There are 6 subsets of size five. There are 15 subsets of size four. There are 0 subsets of size three. There are 15 subsets of size two. There are 6 subsets of size oe. There is 1 subset of size zero. That is the empty set, sometimes called the trivial subset. If we add these up, we get the total umber of subsets of a set of size 6. That s totalig the 6th row of Pascal s Triagle. Therefore, we kow the total will be 6 64 without addig it up. Naturally, if you do add them up, you get the right aswer ayway: 1 6 15 0 15 6 1 64 15. If you happe to kow that C y a, C y 1 b, ad C y1 c, the... What is C y? a What is C y1? b What is 1 C y1? a b Note: If the previous problem is mysterious, let s check with 17 ad y 5, two arbitrarily chose specific values. (Of course, this is ot a proof. C y 17 C 5 6188 a 8
C y 1 17 C 4 80 b C y1 17 C 6 1, 76 c C y 17 C 1 6188 a C y1 17 C 1 80 b 1 C y1 18 C 1 8568 6188 80 a b 17 C 1 17 C 1 9