challset3sol - Challenge Set 3 Solution Sketches Page 247...

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Unformatted text preview: Challenge Set 3 Solution Sketches Page 247 Lb. Formal products are of the form xel xe2 xe3 , where 0 S ei S 4. We are interested in those of the form xel+e2+e3 where e1 + e2 + e 3 = 4. There are 15 formal products in all. The ordered 3 — tuples (e1 ,e2 ,e3)which satisfy this are: (4,0 0),(0,4,0),(0,0,4),(2,2,0),(2,0,2), (O,2,2),(2,1,1),(1,2,1),(1,1,2),(3, (3,0,1),(0,3,1),(0,1,3),(1,0,3),and (1,3,0). d. In the expansion of (1 + x + x2 +A +)3, we have product of the form xe‘ x°2x°3 where ci 2 0, for i = 1,2,3. We need not consider any ei > 4 . If so, then e1 + e2 + e3 > 4. We only need to consider the ei's such that el + C2 + C3 = 4 where 0 S ei S 4, i = 1,2,3. There are 15 formal product in all, and they are the same as those in part b. 2. a. e1+e2 +e3 + e4 +e5 = r, 0 S ei S 4. Generating function required is g(x) = (1 +x +x2 + x3 + x4)5. b. e1 + e2 + e3 = r, 0 < ei < 4. Generating function required is g(x) = (x + x2 + x3)3. 0. e1 +e2 +e3 + e 4 = r, 2 S ei S 8. Generating function required is (x2 +x4 +x6 +x8)(x3 +x5 +x7)(x2 +x3 +x4 +x5 +x6 +x7 +x8)2. d. e1+ e2 + e3 + e4 = r, 0 S ei. Generating function required is (1 + x + x2 +A )4. e. e1 +e2 +e3 + e4 2 r, ei > 0, e2, e4 odd, and e4 S 3. Generating function required is (x+x2 +x3+A )(x+x3 +x5+A )2(x+x3). 3. a. This is the same as the number of integer solutions to e + e2 + e3 = r, 0 S e1 S 3, 0 S e2 S 4, and 0 S e3 S 4. To generate all formal products of the form xel xez x“ with the given constraints, we need the generating function, g(x) = (1 + x + x2 + x3)(1+ x + x2 + x3 + x4)2. b. This is the same as the number of integer solutions to e1 + 62 + 63 = r, where 1 S eI S 5, l S e2 S 3 , 1 S e3 S 8. To generate all formal products of the form xe1 x‘:2 X“ with the given constraints, we need the generating function g(x) = (x + x2 + x3 + x4 + x5)(x + x2 + x3 )(x + x2 + x3 + x4 + x5 + x6 + x7 + x8). c. This is the same as the number of integer solutions to el + e2 + 63 + 64 = r, 0 S ei , i = 1,2,3,4. To generate all formal products of the form xel xe2 xe3 , with the given constraints, we need the generating function g(x) = (1+ x + x2 +A )4. d. This is the same as the number ofinteger solutions to eI + C2 + e3 + e4 + C5 + C6 + C7 = r, 0 S el = 0d 0 S e2 : odd, 0 S ei , i = 3,4,5,6,7. To generate all formal products of the form xex x” x:3 with the given constraints, we need the generating function (x + x3 + x5 +A )2 (l + x + x2 +A )5. 6. This is the same as the number of integer solutions to e1 + 62 + e3 + e4 + e5 2 r, where e]. 2 0. To generate formal products of the form xe1x°2x°3 x“ X“ with the given constraints, we need the generating function, g(x) = (1+ x + x2 +A )5 . 8.a. This is like the number of ways to distribute 27 identical balls into four boxes which at most 27 ball in any one box. The balls represent votes, and since all votes for one candidate are identical, this mode will work. We could also look at this as the number of integer solutions to e1 + e2 + e3 + c4 = r where O S ei , i = l,2,3,4. The generating function required is g(x) = (1+ x + x2 +A )4 and we require the coefficient of x”. b. Since each candidate votes for him / herself, it follows that the boxes in our model have at least one ball in them. Considering our number of integer solutions model, we have e] + 62 + e3 + C4 = r, where ei 2 l, i = 1,2,3,4. This time, our generating function is g(x) = (x + x2 + x3 +A )4 and we still want the coefficient of x27 c. If no candidate can receive a majority of the vote (i.e. 14 votes), then no box can have more than 13 balls in it. Our number of integer solutions model will now be, e, + e2 + e 3 + C4 = r, 0 S ei S 13, i = 1,2,3,4. The generating function will be g(x) = (1+ x + x2 +A +x‘3)4 , and we still want the coefficient of X”. 9. You can think of this as the number of ways of distributing r - identical objects into n - distinct boxes (where each element of the n - set represents a box) such that any box can have as many objects as it likes, or as the number of integer solutions to C1 + 62 +A +el1 = r, 0 S ei , ' = 1,2,...,n. The required generating function is g(x) = (1+ x + x2 +A )“. Aside 1: Let n = 3 and consider the 3 - set S = {1,2,3}. Any r - combination with repetition will be of the form, 1,1,...,1 , 2,2,...,2 , 3,3,...,3, where we have e1 1's, e22's and e3 3's. (Since this is acombination, order doesn't matter). Suppose you have three distinct boxes. If you treat the 1's, 2‘ s and 3's as identical objects, then the number of ways you can put these "identical" objects into three distinct boxes equals the number of r - combinations with repetition. 2: If repetition was not allowed, (i.e. each element of the n - set can be used at most once) then this is the same as the number ofinteger solutions to e1 + e2 +A +en = r, 0 S ei S Li = 1,2,...,n and the required generating function is g(x) = (1+ x)“. 14. a.This is like finding the number of integer solutions to e1 + e2 + e3 + e4 + e5 + e6 = r, where el‘e2 ,e3 are even, and e 4 ,e5,e6 are odd. The generating function for this is g(x) = (x + x3 + x5)3(x2 + x4 + x6)3. b. This is like finding the number of integer solutions to eI + e2 + 63 + e4 + e5 + c6 = r, where ei i i. The generating function for this is 6 Z[(x+x2 +x3 +x4 +x5 +x6)—xi] = i=1 (x2 +x3 +x4 +x5 +x6)(x+x3 +x4 +x5 +x6)(x+x2 +x4 +x5 +x6)(x+x2 +x3 +x5 +x6) (x+x2 +x3 +x4 +x6)(x+x2 +x3 +x4 +x5). 15. To generate all formal products of the form xe' xe2 x63 x64 with the given constraints, we need the generating function (x'3 + x"2 + x’1 + x0 + x1 + x2 + x3)". 16. This is like the number of ways to distribute r identical balls into 6 boxes with at most 9 balls in each box. For example, 3 balls in box 1, 2 balls in box 2 and no balls in the rest of the boxes represents the number 320000. Any number between 0 and 999999 can be represented using balls in boxes. We can look at this as the number of integer solutions to el+e2+e3+e4+e5+e6= r with O S ei S 9. The generating function for this is g(x)=( 1 +x+x2+x3+x4+x5+x6+x7+x8+x9)6. 17. This is like the number of integer solutions to el+ez+e3+e4+e5+e6+e7+eg= r where ei=5k, k=0,1,2,.... The generating function for this is g(x) = (1+x5+xl°+...)8. 19. To model this problem, select r-integers such that no two are consecutive, i.e. n,,n2,..., nr . The n—r non-selected integers can be distributed into r+1 “boxes”, before, between and after the chosen integers. Once they are put in the box, they can ordered from least to greatest. The first and last box can have none or some numbers in them. The boxes in the middle must have at least one number in them ( if not then at least one of the two numbers n,, n2,... nr must be consecutive). The generating function required is (1+x+x2+...)(x+x2+x3+...)4(1+x+x2+...). For the case where n:20 we are interested in the coefficient of x15 {r=5, and we are distributing 15 numbers into 6 boxes}. For general n, we are interested in the coefficient of x”. Aside: To show you how this works, let us consider the case where n=5 and r=2. There are six ways to select 2 nonconsecutive from the integers {1,2,3,4,5}, i.e. {1,3}, {1,4}, {1,5},{2,4},{2,5},and {3,5}. As we discussed above, this process is equivalent to the number of ways of putting 3 balls into 3 boxes with the middle box having at least one ball in it. There are six ways to do this, they are: (0,3,0), (0,2,1),(0,1,2), (1 ,1,1),(1 ,2,0) and (2,1,0) where the coordinates represent the number of balls in box 1, box2 and box3 respectively. 24. a. This is the same as the number of integer solutions to e1+e2+...+e50 = r, where 0 S ei S 1 . The generating function required is (1+x)5°. b. Similiar to part a except that this time, 0 S ei S 3. The generating function is (l+x+x2+x3)5°. 25. This is the same as the number of integer solutions to el+e2+e3+e4+es=r (for the chocolate bars) where Ci 2 0 and f1+f2+f3+f4+f5 = s ( for the lollipops) where O S f j S 3. To generate the formal products of the form xeifj where ei 2 0, O S f j S 3 we need the two variable generating function g(x,y) = (1+x+x2+...)(1+y+y2+y3)5. Page 256 1. Find the coefficient of x8 in the expansion of (1 + x + x2 +...)". 1+x+x2+A L, so (1+x+x2+A )n = = 1 .The coefficient l-x 1—x (1—)()11 f 8_ [8+n-1] [7+n) o x 18 = . 8 8 2. (x5 +x6 +x7+...)8 = (x5(1+x+x2+...))8 = x4°(1+x+x2+A )8 = x40 1 (I—xf : 40 r _ 1.40 . . x 1+ 1 x+A x +... . We need the coefficlent of x 1n the expanswn 1‘ 1 of . (1 -X)8 r-40+8—1) The coefficient is [ r-40 3. (1 + x2 + x4)(1+ x)’". From the binomial theorem, m m m 2 m m 2 4 m (1+x) = 1+ 1x+ 2x+A+mx .Letf(x)=1+x +x andg(x)=(1+x). _ 9 , m m m The coeffi01ent of x in f(x)g(x) = 1 9 +1 7 +1 5 . 4 4 5. (x2+x3+x4 +x5+x6+x7) =(x2(1+x+x2 +x3+x4+x5)) = x8(1+x+x2 +x3 +x4+x5)4 =x The coefficient of x18 = 1(10104'13-4i4i‘9=(131-4Q- 7 2 _. _. b. x _3:(=(x2—3x) 1 4 =(x2—3x)[1+(1+4 1]x+A+[r+4 l]x’+A] (1—K) (1-K) 1 r , 12. 10+4-1_ 11+4—1 _ 13 _ 14 The coeffiment of x IS 3 — 3 . 10 11 10 11 c. (1+x)5 =1+5x+10x2 +10x3+5x4 +x5 1 1+5—1 2+5~1 2 r+5——1 =1+ X+ x +A+ x'+A (l-x)5 1 2 r 5 1+5—1 +5—1 8+"; =(1+5x+10x2+10x3+5x4+x5)[1+( 1 )xm (r Jim/x]. _X r The coefficient of x]2 is, (12+5—1] [11+5—1J [105—1] [9+5—1] [EMS—1] [7+5-1] 1 +5 +10 +10 +5 +1 . 12 11 10 9 8 7 5 5 9. (x2+x3+x4+x5) =(x2(1+x+x2+x3)) =x1°(1+x+x2+x3)5.Thecoefficient of x9 is 0 because the smallest power of x would be 10. 1 _ (l—ufl ' 1+7—1 2+7—1 2 r+7—1 1+7—1 3 r+7—-1 3 1+ 1 u+ 2 u +A+ u‘+A =1+ 1 x +A+ x'+A I‘ I' 6+7-1 12 coefficient ofx18 is when r = 6, Le. [ 6 j =( 6). 10 For(1+x3 +x6 +x9+A )7, letu=x3. Now(l+u+u2+A )7 = ll. 1 =(l+x)_1= 1 l+x 1+(—x) (.1)12 =1. b. = 1+ (—x) + (—x)2 +A +(—-x)'+A . Coefficient of x12 is _5 1 [HS—1] (2+5—1) 2 [r+5—1) d. (1-4x) = 5 = 1+ 4x+ (4x) +A+ (4x)’+A = (1—4x) 1 2 r 1+5-l 2+5—1 2 r+5—l . 12_ 1+ 1 4x+ 1 16x +A + 4rxr+A . The coeffic1ent ofx IS I 13 a. The generating function which models this problem is x6 (l-X)3' g(x)=(x2 +x3 +x4+A +x‘°+A )3 =(x2(l+x+x2+A ))3 = x6(1+x+x2+A )3 = We want the coefficient ofxw. x6 6 1 6 1+3—1 r+3—1 r _ 10 (1_x)3=x (1_x)3=x 1+ 1 x+A+ r x+A sothecoefficrentofx = (“s-Hz). b. The generating function which models this problem is g(x)=(1+x+x2)(1+x+x2+A)2 =(1+x+x2)(1_1_] =(1+x+x2) ~x 1 _ (I—xr _ 2 1+2—1 r+2—1 ‘ 10 (l+x+x ) 1+ 1 x+A+ x‘+A .We wantthe coeffiment ofx . (“til+r£9+:‘l+l“:“l=tiél+t§l+£§l The coefficient of X“) c. The generating function which models this problem is 1 g(x)=(l+x+x2+A )2(1+x2+x4+A)=(1+x2+x4+A )(1 )2 = —X I' . , 10+2-1 8+2—l 6+2—1 4+2—1 2+2—l coeffic1entis + + + + +1 = 10 8 6 4 2 11 9 7 5 3 + + + + +1. 10 8 6 4 2 18. This is like the number of integer solutions to 1+2—1 r+2—1 (1 +x2 +x4+A )[1+[ 1 jx+A +[ )x’+A]. We want the coefficient ofxm. This e1+e2 +e3+A + e10 = rwhere ls ei S 6 for i = 1,2,...,10. The generating function re uired is x = x + x2 + x3 + x4 + x5 + x6 10. We want the coefficient ofxzs. q g 6 10 g(x)=(x(l+x+x2+x3+x4+x5))1°=x10[1 x) =x10(l—x6)10 1 lo= l—x (l—x) 10 10 . M 1+10—1 r+10—l _ 25 Z _ (—1)JxJ 1+ 1 x+A+ x’+A .The coeffic1entofx = j=0 J 1' [10] {HMO—1] [10] l(9+10—1] (10] {um—1) [24] [18) (12) (—1) + (—1) + (—1) = —10 +45 . 0 15 1 9 2 3 15 9 3 21. The generating function which models this problem is, g(x) = (x2 + x3 + x“)(x2 + x3 + x4 + x5)(x2 + x3+A +x9 ). Once you give two books of type Ito each teacher, there are two books of type I left. So each teacher can only end up with at most 4 books of type I. Once you give two books of type II to each teacher, there are three books of type 11 left. So each teacher can only end up with at most five books of type 11. Once you give two books of type III to each teacher, there are seven books of type 111 left. So each teacher can only end up with at most nine books of type 111. We want the coefficient of x”. 61—x3l—x4 l—x8_6_3_4_81_ g(x)—x£1-X][1_xj(l_xj—x(l x)(l x)(1 x)(1_x)3- 1+3—1 r+3——1 x"’(1—x]5 +x12 +xl1 —x8 +x7 —x4 —x3)[1+( 1 JX+A +( r (“3 “1-1223 41131-131111- 24. Lay out the 8T's, and put boxes before, after, and in between the T's. There )x'+A]. The 6+3-1] coefficient of x12 is ( 6 are 9 boxes in all. We distribute the H's into these boxes. Since we don‘t want a run of seven or more heads, each box can have no more than 6 H's. The generating function for this is g(x) = (1+ x + x2 + x3 + x4 + X5 + x6)9. We want the coefficient of X”. gm = [1- x7] 2 ((11:39 : imknixfi (1+{1+19—1JX+A {r +9 — ljxrm} 1—K i=01 I‘ _ l7_ {17+9-1] (10+9—1X9) [14+9—1X9] _ The coeffiCIent of x IS — + _ 17 10 1 14 2 25 9 18 9 22 I . . I E ]‘[ )+( j.Thenumberofwaystofllpacom25tlmesw1th8T'sand17H's 17 1 10 2 14 1”] (9)1‘81+1911221 17 1 10 2 14 1”) 8 . 25! 25 . . . 1s = . Therefore, the probablhty IS 17! 8! 8 10 26. If a die is rolled 2 times, then the generating function which represents all the possible outcomes is g2(x) = (x+x2+x3+x4+x5+x6) (x+x2+x3+x4+x5+x6). / / First roll Second roll We can generalize this to 11 rolls with the generating function gn(x)=(x+x2+x3+x4+x5+x6)“. It is not hard to see that g0(x)=1 and g1(x)= x+x2+x3+x4+x5+x6. Since our die can be rolled any number of times, we have to add up all these generating functions, i.e. 1+ (x+x2+x3+x4+x5+x6)+ (x+x2+x3+x4+x5+x6)2+. . .+(x+x2+x3+x4+x5+x6)“+. .. .If we let u=(x+x2+x3+x4+x5+x6), we have g(x)= 1+u+u2+... = 1/(1-u) = 1/(l-x-x2-x3-x4-x5-x6)= (1 -x—x2—x3-x4-x5-X6)'l . 19. The key here is that any combination of 15 boxes will yield 300 chocolates. If we look at this way, then this is equivalent to the number of integer solutions to e1 + e2 +A +67 2 r where 1 S ei S 5. The generating function which models this is g(x) = (x + x2+A +x5)7 , we want the coefficient ofx”. 1_ 5 7 g(x)=x7(l+x+x2 +x3+x4)7 =x7fi=x(l—x5)7 (1—x>7 z °° '+7— x704):5 +21x‘° — 35x15 + 35x20 —21x25 +7x3° —x35)Z(J j=0 l5. [8+7-1) 7(3+7—1] X 15 8 —- 3 . 1 . J ij . The coefficient of ll 23. Collect $24 from 4 children and 6 adults. Each child gives at least one dollar and atmost four dollars. Therefore we need (x + x2 + x3 + x4 )4 as part of our generating function. Each adult gives at least one dollar and atmost seven dollars. Therefore we need (x + x2 +A +x7 )6 as part of our generating function. The generating function which models this is g(x) = (x+x2 +x3 + x4)4(x +x2+A +x7)6 = x4(1 +X+ x2 +x3)4x6(l + x+ x2+A +X6)6 = Xlo (1_x4)4 (1_x7)6 (1-K)4 (lrx)6 = X]0(1 * X4)4 (1 — x7)6 We need the coefficient ofx14 in (1 _x)10- (1—x4)4(1—x7)6 ifm‘ljxi. j:0 J °° '+10—1 . (1—x4)4(l—x7)6Z[J J, JXJ: j:0 °° '+10—1 . (l—4x4 +6x8 —4xI2 +X16)(l—6X7 +15x14 —20x21+15x28 *6x35 +x42)Z:(J j jxj. j=0 The coefficient of x14 2 [14+10—1] [7+10—1] (6) (10+10-1] 24(3+10—1] 6[6+10-1] {mm—1] 14 ‘6 7 +2'4 10 + 3 + 6 _ 2 ' 27. m + n [ r J is the coefficient of xr in the expansion of (1+ x)""" .Suppose m S n. m m . " n . (1+x)"‘+" = (1+x)"‘(1+x)n JXI Z[.]x’. The coefficient ofxr is, i=0 1 j=0 J (3)+(T‘Jt.‘iJ+A+(:‘Jt.fJ+ATill.)Hencewe (“1”)=0+tTJtrill+A+tEthJ+A43(39- 12 28. (1-X2)" ——= l “. 1 " , n . °° '+n—1 . LHS: (1-x2)“ n =Z(—1)‘[_jx2‘ Z[J , )XJ. The general term ofthis (l'x) '=0 1 j=0 J . . . n J'+ 11-1 . - . . . expansmn IS (-1)1 [i j jxw‘. For the coefficient of xm , we need 1 and j such that m = j + 2i. For m even, the coefficient of xm is the sum over all pairs of (i, j) such that m i+j=m, i.e.§(—1)k[:j[ n_1 n+m—2i—l n+m—2i—l n+m—2i—l . = = . Therefore we can conclude n + m — 2i — l n j. The coefficient of x‘“ in the RHS = m—21 n+m-2ielim+2i n—l 7 n n+m—2i—1 n Z(—1>k = . k=0 k n-l m 29. a. m m n I m m . Evaluate 2 where r 15 fixed. If we write k = k , by usmg the symmetry H k r + k m — . . . m m n m n m . . property of b1nom1a1 coeffic1ents, then ; [k J [r + k] = g (I + k] [m _ k]. Combmatonally, m + n this is [I + In] , because if we want to select r + m objects out of n + m objects, we can split the n + m object into two sets of m and n objects. We can select r + k objects out of the set with n objects and m - k objects out of the set with m objects . Summing up all the possible cases gives our desired result. n k n ' ' - n n c. ;2 k = the binomral expansmn of (1+2) = 3 . 13 33. x —-a +a x+a x2+...+a x’+...+a x". g 0 1 2 r n k! ifn=k dn (xk)= O ifn>k We need the following result about derivatives, dxn (k-1)(k-2)...(k-n+l)xk'“ ifn< k dr dr r+l ‘ Hence d r [g(x)]= arr! + a”1 d r x +A . The terms w1th powers larger thanr x x will have positive powers of x left. Hence dr 1 dr ! r[g(x)]atx=0, = at r! + 0 +0+A .Hence — [g(x)]= M = ,. dx r! dxr r! 39. How can we get a sum of r and count the number of ways to get it? Let i be the out put of the first die where i= 1,2,3,4,5,6. Once we know this, the number of ways to get a sum of r is the number of integer solutions to el+e2+. . .+ei = r where 1S e1 S 6, 1S e2 S 6,..., 1S ei S 6. The generating function for this is (x +x2 + x3 + x‘1 + x5 +x6)i. Since i can go from 1 to 6, the generating function we want is the sum of the generating functions, (x+x2 +x3 +x4 + x5 + x6) + (x+x2 +x3 + x4 +x5 +x°)2+...+ (x+x2 +x3 +X4 +x5 +x6)6 = f(f(x)) where f(x) = (x+x2 +x3 +x4 +x5 +x6). 40. Px (t) = ZZaiti where ai = probability that it takes i minutes to save a customer. i=0 Py (t) = z b J.tj where b J. = probability thatj people line up to be served each minute. j=0 Pz (t) = chtk where ck = probability that k people line up to be served. k=0 ck =aibj where ij=k. bib]. =P(Y=iandY=j) = 0unlessi=j. bf =bi becauseP(Y=iandY=i)=P(y=i). A discrete random variable can't have values i and j at the same time. 14 (b0 +b,t+b2t2+A )2 = b1th +b22t4+A = blt2 +b2t4+A (b0 +b,t+b2t2+A )3 = b13t3 +b23t6+A = blt3 +b2t6+A So Pz(t) = chtk = ZZaibjtk = Zai(b0 +blt+b2t2+A )i = 1103(0). k=0 k=0§jik i=0 an<b1t+b2t2+A ) + a2<b1t2 +b2t4+A >+a3<b1t3 +b2t6+A )+ A ai(b1ti+b2t2i+b3t3i+... +A ...
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This note was uploaded on 03/04/2012 for the course BIO 510 taught by Professor Miller during the Fall '06 term at Carnegie Mellon.

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