Math Discrete Math Combinatorics MULTIPLICATION PRINCIPLE:


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1 Math Discrete Math Combiatorics Notes MULTIPLICATION PRINCIPLE: If there m ways to do somethig ad ways to do aother thig the there are m ways to do both. I the laguage of set theory: Let S, T be sets with m, elemets respectively. If S is a set of possible outcomes for a first task ad T is the set of possible outcomes for a secod task the the set of outcomes for completig both tasks is S T ad S T S T. This geeralizes to tasks: If m i couts the umber of ways to complete task i for i 1,,..., the m i couts the ways to do all the tasks. i1 Example: How may positive divisors does 10, have? Each positive divisor correspods to i 5 j where i ad j are chose from {0, 1,, 3, 4}. So by the multiplicatio priciple there are 55 5 positive divisors of 10, 000. Theorem: A set of objects has subsets. Proof : A subset of a elemet set is equivalet to a uique labelig of each of the elemets with either a 0 or a 1 0 meas ot i the subset ad 1 meas i the subset. So the umber of subsets is equal to the umber of ways to label each of the elemets with either a 0 or a 1. That s by the multiplicatio priciple Let k, be oegative itegers. A kelemet permutatio of objects is a ordered list of k of the objects. The umber of kelemet permutatios of objects is deoted P, k. By the multiplicatio priciple, m! P, k 1 k+1 k!, where 0! 1 ad for m N we have m! i. Example: A traditioal startig lieup i basketball cosists of a poit guard, a shootig guard, a power forward, a small forward, ad a ceter. How may traditioal lieups ca be formed from a 1 perso roster if ay player ca play ay positio? It s a ordered list of 5 of the 1; there are P 1, , 040 lieups. i1
2 Let k, be oegative itegers ad k. A kelemet combiatio aka selectio from objects is a uordered collectio of k of the objects equivaletly a kelemet subset of the set of objects. So, the umber of kelemet subsets of a elemet set is equal to the umber of kelemet combiatios aka selectios take from objects. This umber is deoted C, k or. It s read choose k. k I prefer the C, k otatio most of the time. Let s derive a formula for C, k. A kelemet permutatio of objects ca be uiquely determied by the followig two step process: Select a kelemet combiatio from the objects, ad the put these k objects i a particular order. The umber of kelemet combiatios is C, k ad the umber of ways to put the k objects selected ito a order is kk 1 1 k!, so by the multiplicatio priciple, P, k C, k k!. Hece C, k P, k k!! k! k!. Example: Jack has a coupo for a 3toppig large pizza from a local pizzeria. How may choices does Jack have for his 3toppig large pizza, give that the pizzeria has 10 total toppig choices? The aswer is 10 choose 3 which is C10, by the formula above. How may 4 digit strigs have exactly 3 oes? Note: A digit strig of legth 4 is a 4elemet permutatio of the digits. There are C4, 3 4 locatios i which the oes ca occur, ad 9 choices for the digit ot equal to 1. So by the multiplicatio priciple we get such strigs. Theorem: Suppose k, are oegative itegers ad k. The C, k C, k. Proof : Choosig k elemets from objects is equivalet to choosig k elemets to leave behid. Therefore the umber of selectios of k elemets from is the same as the umber of selectios of k objects from
3 Theorem: For ay iteger, ad for ay positive iteger k < we have C, k C 1, k 1 + C 1, k. Proof : Let X be a set of objects. Let S be the set of kelemet combiatios hece subsets of the objects of X. Thus S C, k. Let x X. We ca partitio S ito two subsets A ad B, where A are the kelemet combiatios that cotai x ad B are the kelemet combiatios that do ot cotai x. Clearly A ad B are disjoit ad their uio is S. So S A + B. The cardiality of A is C 1, k 1 sice to form a kelemet combiatio already cotaiig x, k 1 more elemets must be selected from the 1 elemets left i X \ {x}. The cardiality of B is C 1, k sice to form a kelemet combiatios ot cotaiig x, all k elemets must be selected from the 1 elemets left i X \ {x}. That completes the proof Pascal s triagle: Begi with a triagle of 3 oes i two rows. The rows below exted the triagle ad are geerated by puttig 1 o each ed ad i other locatios addig the two etries directly above as follows: We call the top row the Oth row ad it s C0, 0 1. The ext row is the 1st row ad it s C1, 0 C1, 1. Now let be a iteger. The fact that C, 0 C, 1 ad the previous theorem tells us that the th row of the triagle is: C, 0 C, 1 C,... C, C, 1 C,
4 Biomial Theorem: If x, y R ad N the x + y C, k x k y k. k0 Proof : We shall retur to the method of iductio. Suppose 1. The x + y x + y ad C, k x k y k C1, 0xy 0 + C1, 1x 0 y x + y. This establishes the base case. k0 Let N. Assume x + y ad simplify: C, k x k y k. The we multiply both sides by x + y k0 x + y x + y C, k x k y k x + y, k0 x + y +1 C, k x k+1 y k + x k y k+1, k0 1 1 x + y +1 x +1 + x y + C, kx k+1 y k + C, kx k y k+1 + xy + y +1, k1 k1 1 x + y +1 x +1 + C, ix i+1 y i + C, jx j y j+1 + y +1, i1 j0 x + y +1 x +1 + C, ix i+1 y i + i1 C, i 1x i+1 y i + y +1, i1 x + y +1 x +1 + C, i + C, i 1 x +1 i y i + y +1, i1 x + y +1 x +1 + C + 1, ix +1 i y i + y +1, i1 +1 x + y +1 C + 1, ix +1 i y i, i0 The secodtolast step uses that C + 1, i C, i + C, i 1 for i 1,,..., by a previous theorem
5 Let s apply the biomial theorem to fid the coefficiet of a 4 b 3 i a b 7. Usig x a, y b ad selectig the term of the sum correspodig to k 3 we get C7, 3a 4 b a 4 b 3 560a 4 b 3. So the coefficiet is 560. Multiomial Theorem: Let m, N ad m. The x 1 + x + + x m x k 1, k,..., k k t t, m k 1 +k + +k m 1 t m! where is a multiomial coefficiet. The sum is take k 1, k,..., k m k 1!k! k m! over all combiatios of oegative iteger idices k 1 through k m that sum to. Proof : Let N. We shall use iductio o m ad the biomial theorem. Whe m the theorem holds as it s the Biomial theorem. Throughout assume that k, ad k j for ay j, all represet oegative itegers. Let m be a iteger. Assume that x 1 +x + +x m k 1 +k + +k m The x 1 + x + + x m + x m+1 x 1 + x + + x m 1 + x m + x m+1 k 1 +k + +k m 1 +k sice k 1, k,..., k m 1, k! k 1!k! k m+1! k 1 +k + +k m 1 +k k 1, k,..., k m 1, k k 1 +k + +k m+k m+1 k 1, k,..., k m 1, k 1 t m 1 x k t t k m +k m+1 k k 1, k,..., k m 1, k k 1 +k + +k m +k m+1 k k m, k m+1. k 1, k,..., k m, k m+1 1 t m 1 k 1, k,..., k m x k t t x m + x m+1 k k x k m, k k m m x k m+1 m+1 m+1 k k m, k m+1 k 1, k,..., k m, k m+1! k 1!k! k m 1!k! 1 t m+1 1 t m+1 x k t t x k t t, k! k m!k m+1! 1 t m x k t t. That completes the proof by iductio What s the coefficiet of x 3 y z i x + y + z 7? 7 7! It s ,, 3!!!
6 Theorem: Suppose a sequece of legth has i idetical terms of type i, for 1 i k. The the umber of distict orderigs of this sequece is the multiomial coefficiet. 1,,..., k Proof : Let a 1, a,..., a be a sequece of legth with i idetical terms of type i, for 1 i k. I particular, k. Cosider a sequece of empty slots. There are C, 1 ways to isert the 1 terms of type 1 ito these slots. Now there oly remai 1 empty slots. There are C 1, ways to isert terms of type ito these remaiig slots. Cotiuig like way we get C 1 l 1, l ways to isert the l terms of type l ito the remaiig slots for l 3, 4,..., k. The by the multiplicatio priciple, the umber of distict orderigs of a 1, a,..., a is C, 1 C 1, C 1, 3 C 1 k 1, k! 1! 1! 1!! 1! 1! 3! 1 3! 1 k 1! k!0!! 1!! k! 1,,..., k Example: How may distict alphabet strigs ca be made with the letters of the word BLUEBERRIES? There are 3 E s, B s, R s, ad 1 each of I,L,S,U. So there are 11 11! 3,,, 1, 1, 1, 1 3!! 11! , 663, 00 such strigs. 4!
7 Now let s cosider selectios where order is ot take ito accout, but repetitio is allowed. Let s start with a simple example. Suppose we have a bag that cotais 3 idistiguishable red balls, 3 idistiguishable white balls, ad 3 idistiguishable blue balls. How may distict drawigs igorig order of 3 balls are possible? This oe is easy eough that we ca fid the selectios by brute force meaig we list all the selectios: We will use R, W, ad B to deote a red, white, ad blue ball respectively. Sice order does t matter let s adopt the covetio of writig R before W ad W before B. Here are all the possibilities: Ivolvig oly oe color: RRR, W W W, BBB. Ivolvig exactly two colors: RRW, RRB, RW W, W W B, RBB, W BB. Ivolvig all three colors: RW B. So there are 10 distict drawigs. Icidetally, what s C5,? Theorem: If X is a set cotaiig t elemets, the the umber of uordered kterm selectios from X, where repetitios are allowed, is Ck + t 1, t 1 Ck + t 1, k. Proof : Let X {a 1, a,..., a t }. We will show that every uordered kterm selectio from X, where repetitios are allowed, correspods to a diagram where t 1 of the k + t 1 slots are filled with the bar symbol ad vice versa every diagram correspods to a selectio. After that it will be simple to complete the proof. Suppose we have a uordered kterm selectio from X where repetitios are allowed. Let i cout the umber of times a i appears i the selectio for i 1,,..., t. Clearly t k. Let s build a diagram by startig with 1 slots, followed by a bar, followed by slots, followed by a bar,..., followed by t 1 slots, followed by a bar, followed by t slots. What we get is a diagram where t 1 of the k + t 1 slots are filled with the bar symbol. Now suppose we have such a diagram. Let s build the selectio correspodig to the diagram. The umber of slots to the left of the first bar maybe 0 slots is the umber of times a 1 appears i the selectio. For i, 3,..., t 1, we have that the umber of slots betwee the i 1st bar ad the ith bar maybe 0 slots is the umber of times a i appears i the selectio. The umber of slots to the right of the last bar maybe 0 slots is the umber of times a t appears i the selectio. Sice there are a total of k slots amogst the t 1 bars, we ed up with a uordered kterm selectio from X where repetitios are allowed. So coutig the umber of uordered kterm selectios from X, where repetitios are allowed, is equivalet to coutig how may diagrams there are where t 1 of k + t 1 slots are filled with the bars. So we get Ck + t 1, t 1 Ck + t 1, k same as choosig k slots to ot be bars
8 Example: Suppose a dout shop is havig a special o a doze douts where customers must choose their doze from 5 differet types of douts. How may distict dozes ca be made from the 5 types? I this example customers select k 1 douts from a set of t 5 types repetitio allowed. So by the previous theorem, there are C1 + 4, 4 C16, , 80 4! distict dozes that ca be ordered. Now let s cosider how to fid how may oegative iteger solutios there are to the equatio x 1 + x + + x t. We ca bars ad slots idea sice a solutio to the equatio correspods to a diagram where t 1 of + t 1 slots are filled with bars. Let s see why. Cosider such a diagram. Let x 1 be the umber of slots to the left of the first bar. For i, 3,..., t 1, let x i be the umber of slots betwee the i 1st bar the the ith bar. Let x t be the umber of slots to the right of the last bar. The we have a solutio to the equatio. It is also easily see that a solutio ca be coded as such a diagram. Therefore, by the theorem, the umber of oegative iteger solutios to the equatio is C + t 1, t 1. Example: Cosider the equatio x + y + z 10. There are C1, 111/ 66 oegative iteger solutios. ELFY Solve the problem if we assume the itegers have to be positive. Hit: Substitute x i x i 1 ito the equatio ad cout oegative iteger solutios to the resultig equatio... Cosider the followig example: 15 idetical dolls are to be distributed ito bis marked 1,, 3, 4. I how may ways ca this happe? This is the same as askig how may oegative iteger solutios are there to the equatio x 1 + x + x 3 + x So the solutio is C18,
9 Recursio is helpful for solvig some coutig problems. We will begi with a example: How may digit strigs of legth 4 do ot cotai the patter 00? Let S be the umber of digit strigs of legth N that do ot cotai the patter 00. Clearly S 1 10 ad S 99. Let 3. We partitio digit strig of legth that do ot cotai the patter 00 ito two sets, those that begi with a 0 ad those that do t. If the first digit is t a 0 the there are 9 choices for that digit followed by S 1 ways to fill the rest of the strig. If the first digit is a 0 the there are 9 choices for the secod digit followed by S ways to fill the rest of the strig. So we get S 9S 1 + 9S 9S 1 + S. So S 3 9S + S ad S 4 9S 3 + S , 70. So there are 9, 70 such strigs. What if we wat a closed formula for digit strigs? For that purpose we prove the followig: Theorem: Suppose the recursio a c 1 a 1 + c a holds or 3 ad a 1, a are give. If r is a root of t c 1 t c 0, the the sequece a r solves the recursio. If the polyomial above has roots r 1 r, the there exists costats b, c such that a br 1 + cr. proof : If r is a root of t c 1 t c 0 the r c 1 r + c. The c 1 r 1 + c r r c 1 r + c r r r. So r i solves the recursio. It s easy to see that ay liear combiatio of r 1 ad r will satisfy the recursio. Sice r 1 r there exist costats b, c such that br 1 + cr a 1 ad br 1 + cr a. Thus a br 1 + cr
10 So let s retur the recursio S 9S 1 + S with S 1 10 ad S 99. Let s fid the roots of t 9t 9. By the quadratic formula the roots are t 9 ± We ca fid b, c such that b c ad b + c 99. ELFY Solve the system above to show that b Hece S ad c or 1,, Pluggig i 4 yields the same aswer, 9, 70. The advatage? We ca plug i 10 for example better tha goig through the recursio oe step at a time. Fu with the coutig umbers: Let s deote by d the umber of ways to write N as a sum of distict positive itegers. We ca quickly determie some of the first few values of this sequece: d 1 1 as 1 is the oly way to do it, d 1 as is it, d 3 as we ca either write 3 or + 1, d 4 as we have 4 ad 3 + 1, d 5 3 as we have 5, 4 + 1, ad 3 +, d 6 4 as we have 6, 5 + 1, 4 +, ad ELFY Explai why d 7 5. Let s deote by o the umber of ways to write N as a sum of odd positive itegers. We ca quickly determie some of the first few values of this sequece: o 1 1 as 1 is it, o 1 as is it, o 3 as we have 3 ad , o 4 as we have ad , o 5 3 as we have 5, , , o 6 4 as we have 5 + 1, 3 + 3, , ad of course ELFY Explai why o 7 5.
11 I 1740 Euler proved the followig i a very clever way: Theorem: For all N, d o. Proof : Let s fid a patter i expadig the ifiite product 1+x1+x 1+x 3 1+x x1 + x 1 + x x x + x + x 3 + x 4 + 3x 5 + 4x d x +, sice for ay sum of distict itegers addig to there is a path of choices of factors from the first factor cotiuig idefiitely evetually always choosig 1 that produces a cotributio of x i the expaded ifiite product. Recall the geometric sum rule: k0 r k 1 1 r. Now cosider expasio of the ifiite product show below: x 1 x 3 1 x 5 1 x x + x x 3 + x x 5 + x x 7 + x x + x x 3 + x x 5 + x x 7 + x x + x + x 3 + x 4 + 3x 5 + 4x o x +, sice for ay sum of odd itegers addig to there is a path of choices of factors from the first factor cotiuig idefiitely evetually always choosig 1 that produces a cotributio of x i the expaded ifiite product. To elaborate, the path cosists of pickig x x 3 umber of 3 s i the sum from the secod factor, ad so o... umber of 1 s i the sum from the first factor, ad It oly remais to show that these two ifiite products are actually equal: 1 1 x 1 1 x 1 x 1 1 x 1 1 x x x 4 1 x 1 x1 + x 1 x 1 1 x x 3 1 x x 5 1 x 1 + x 1 x 4 1 x 7 1 x 6 1 x x x 7 1 x x 3 1 x x x1 + x 1 + x 3
12 A stadard method for provig that a equatio ivolvig biomial coefficiets is true is to show that both sides cout the same thig. Such a argumet is called a combiatorial argumet. We have examples of this already, such as the proof of C, k C, k for oegative itegers k, where k, or the proof of C, k C 1, k 1 + C 1, k for positive itegers k, where k <. Here are more examples: Theorem Vadermode s Idetity: For oegative itegers m,, r such that r is the least, r Cm +, r Cm, kc, r k. k0 Proof : Cosider formig a committee of r 0 people from a group of m r me ad r wome. Let S be the set of all possible committees that ca be formed. The left side, Cm +, r, couts how may ways there are to select r people to from m + people. Hece the left side couts S. We ca partitio S ito S 0, S 1,..., S r, where S i are the possible committees formed with i me ad r i wome, for i 0, 1,..., r. By the multiplicatio priciple we get that r S i Cm, ic, r i for i 0, 1,..., r. Hece the right side, Cm, kc, r k, also couts r S S i i0 k0 Theorem: For oegative itegers, r such that r, C + 1, r + 1 Ck, r. kr Proof : Cosider bit strigs strigs of 1 s ad 0 s of legth + 1 that have exactly r s. We will call these + 1bit strigs with r + 1 oes. Let S be the set of all such strigs. Clearly the left side, C + 1, r + 1, couts S. We ca partitio S ito S r+1, S r+,..., S +1 where the set S i is the subset of S cosistig of + 1bit strigs with r + 1 oes such that the rightmost 1 is i the ith positio from left to right. For each i r + 1,..., + 1 we get S i Ci 1, r coutig combiatios of r oes placed amog the first i 1 positios, as the ith positio is the rightmost oe for the bit strigs i S i. The the right side is equal to Ck, r kr +1 ir+1 Ci 1, r +1 ir+1 S i S
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