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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 rintegers such that no two are consecutive, i.e. n,,n2,..., nr . The n—r nonselected 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
lx 1—x (1—)()11 f 8_ [8+n1] [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 r40+8—1) The coefﬁcient is [
r40 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(10104'134i4i‘9=(1314Q 7 2 _. _.
b. x _3:(=(x2—3x) 1 4 =(x2—3x)[1+(1+4 1]x+A+[r+4 l]x’+A]
(1—K) (1K) 1 r , 12. 10+41_ 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
(lx)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+51]
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+71 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. (14x) = 5 = 1+ 4x+ (4x) +A+ (4x)’+A =
(1—4x) 1 2 r 1+5l 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 (lX)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 = (“sHz). 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+21 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£1X][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 “11223 41131131111 24. Lay out the 8T's, and put boxes before, after, and in between the T's. There )x'+A]. The 6+31] 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+91] (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/(1u) = 1/(lxx2x3x4x5x6)= (1 x—x2—x3x4x5X6)'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+71) 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
(1K)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+101] 24(3+10—1] 6[6+101] {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.
(1X2)"
——= l “.
1 " , n . °° '+n—1 .
LHS: (1x2)“ n =Z(—1)‘[_jx2‘ Z[J , )XJ. The general term ofthis
(l'x) '=0 1 j=0 J . . . n J'+ 111 .  . . .
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+m2ielim+2i n—l
7 n n+m—2i—1 n
Z(—1>k = .
k=0 k nl 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
(k1)(k2)...(kn+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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