Math 55 Homework 13 solutions
Section 8.4.
2. 1 + 4x + 16x2 + 64x3 + 256x4 .
4. (a) −1 − x − x2 − x3 − x4 − x5 − x6 .
x2
(d) 1 + 2x + x2 + x3 + · · · = 1 + 2x + x2 (1 + x + · · · ) = 1 + 2x + 1−x
.
(e) 70 + 2 71 x + 22 72 x2 + 23 73 x3 + 24 74 x4 + 25 75 x5 + 26 76 x6 + 27 77 x7 .
1
(h) 1 + x2 + x4 + x6 + · · · = 1 + (x2 ) + (x2 )2 + (x2 )3 + · · · = 1−x
2.
1
.
6. (a) −1 − x − x2 − x3 − · · · = −(1 + x + x2 + x3 + · · · ) = − 1−x
2x
2 2
3 3
2
(b) 2x + 2 x + 2 x + · · · = 2x(1 + (2x) + (2x) + · · · ) = 1−2x
.
(c) −1 + x2 + 2x3 + · · · = −1 + x2 (1 + 2x + 3x2 + · · · ) = −1 +
x2
(1−x)2 .
x
1
1
1
1
1 2
1
(d) 1!
+ 2!
x + · · · = x1 (( 0!
+ 1!
x + 2!
x + · · · ) − 0!
) = e x−1 .
(e) Note that
1
2
3
4 2
=
+
x
+
x + ··· .
2
2
2
(1 − x)3
From this, we see that
0
1
2 2
3 3
2
3
x2
x+
x +
x + · · · = x2
+
x + ··· =
+
.
2
2
2
2
2
2
(1 − x)3
10
2 10 3 10 4 10 5 10 6 10 7 10 8
(f) 10
+ 2 x+ 10
x + 4 x + 5 x + 6 x + 7 x + 8 x + 9 x +
1
3
10 9
10 x .
8. (b) an = n3 3n (−1)3−n .
n
(c) an = −1
n (−2) .
n
(e) a0 = 0, a1 = 4, and an = −1
if n ≥ 2.
n (−3)
3m
(h) a0 = 0, a2m−1 = 0, and a2m = (2m)! if m ≥ 1.
x6
9
3
is −3
10. (b) The power series is (1−x)
3 , so the coefficient of x
3 (−1) .
(d) If suffices to consider (x + x4 + x7 )(x2 + x4 + x6 + x8 ) to compute the
coefficient of x9 in the power series. In the expansion, the coefficient of x9 is 2.
12. (d)
−3
12
(−4)12 .
14. We need to compute the coefficient of x12 in (1 + x + x2 + x3 )5 . That is 35.
18. We need to compute the coefficient of x14 in (1 + x + · · · + x100 )(x3 + · · · +
x10 )(1 + x + · · · + x100 ). We can ignore x15 , x16 , . . ., and in the expansion of
(1 + x + · · · + x14 )(x3 + · · · + x10 )(1 + x + · · · + x14 ), the coefficient of x14 is 68.
1
20. (1+x10 +x20 +· · · )(1+x20 +x40 +· · · )(1+x50 +x100 +· · · )(1+x100 +x200 +· · · ).
22. The interpretation is the number of solutions of x1 + · · · + xn = 6 when
x1 , . . . , xn are nonnegative integers. The coefficient of x6 is n+6−1
.
6
40. (a)
− 21
n
− 12 − 12 − 1 · · · − 12 − (n − 1)
n!
1 · 3 · · · · · (2n − 1)
=
(−2)n n!
(2n)!
=
2 · 4 · · · · · (2n) · (−2)n n!
2n
(2n)!
n
= n
=
.
2 (−2)n n!n!
(−4)n
=
43. The coefficient of xr in (1 + x)m+n is
coefficient of xr in (1 + x)m (1 + x)n is
m+n
r
. On the other hand, the
r X
m
n
r−k
k
k=0
by Theorem 8.4.1, so they are equal.
44. Consider first (x2 + x)(1 − x)−3 = (x2 + x)( 22 + 32 x + · · · ). The coefficient
of xn is n2 + (n+1)
= n2 . Now, consider (12 x + 22 x2 + 32 x3 + · · · )(1 − x)−1 =
2
2
2 2
2 3
2
(1 x+2 x +3 x +· · · )(1+x+x2 +· · · ). The coefficient of xn is 12 +22 +·
· ·+n
.
3
3
2
−4
2
On the other hand, the coefficient of (x +x)(1−x) = (x +x)( 3 + 3 x+
n(n+1)(2n+1)
· · · ) is n+1
+ n+2
=
. By comparing the two computations, we
3
3
6
n(n+1)(2n+1)
2
2
get 1 + · · · + n =
.
6
Section 8.5.
4. Let S, T be the set of owners that will buy a printer next year and will buy
at least one software package respectively. The problem is to compute |S ∩ T |,
which is |S| + |T | − |S ∪ T | = 650000 + 1250000 − 1450000 = 450000.
6. (a) In this case, A1 ∪ A2 ∪ A3 = A3 , so |A1 ∪ A2 ∪ A3 | = 10000.
(b) |A1 ∪ A2 ∪ A3 | = |A1 | + |A2 | + |A3 | = 11100.
(c) The condition says that |A1 ∩ A2 | = |A2 ∩ A3 | = |A3 ∩ A1 | = 2 and that
|A1 ∩ A2 ∩ A3 | = 1, so by the inclusion-exclusion principal, we have
|A1 ∪ A2 ∪ A3 | = 10000 + 1000 + 100 − 2 − 2 − 2 + 1 = 11095.
2
10. Let S, T be the set of positive integers not exceeding 100 that are divisible
100
by 5, 7 respectively. Then |S| = b 100
5 c = 20, |T | = b 7 c = 14, and |S ∩ T | =
100
b 35 c = 2, so the answer is
|S ∪ T | = 100 − (20 + 14 − 2) = 68.
16. By the inclusion-exclusion principal in Example 8.5.5, the answer is
100 + 100 + 100 + 100 − 50 − 50 − 50 − 50 − 50 − 50 + 25 + 25 + 25 + 25 − 5 = 195.
18. In Theorem 8.5.1, we see that there are
10
10
+ ··· +
= 1023
10
1
terms in the formula.
26. Let E1 , E2 , E3 , E4 be the events. Then by the condition, p(Ei ∩Ej ∩Ek ) = 0
for different i, j, k. Then we also have p(E1 ∩ E2 ∩ E3 ∩ E4 ) = 0, so by the
inclusion-exclusion principal,
p(E1 ∪ E2 ∪ E3 ∪ E4 ) =p(E1 ) + p(E2 ) + p(E3 ) + p(E4 )
− p(E1 ∩ E2 ) − p(E1 ∩ E3 ) − p(E1 ∩ E4 ).
− p(E2 ∩ E3 ) − p(E2 ∩ E4 ) − p(E3 ∩ E4 )
28. Let E1 , . . . , En be the events. Then by the condition, p(Ei ∩ Ej ) = 0 for
different i, j. Then we also have p(Ei1 ∩ · · · ∩ Eir ) = 0 for different i1 , . . . , ir if
r ≥ 2, so by the inclusion-exclusion principal,
p(E1 ∪ · · · ∪ En ) = p(E1 ) + · · · + p(En ).
Section 8.6.
2. By the inclusion-exclusion principal, the answer is
1000 − (450 + 622 + 30 − 111 − 14 − 18 + 9) = 32.
4. Let y1 = 3 − x1 , y2 = 4 − x2 , y3 = 5 − x3 , and y4 = 8 − x4 . Then the
problem is to find the number of solutions of the equation y1 + y2 + y3 + y4 = 3
where y1 , y2 , y3 , y4 are nonnegative integers such that y1 ≤ 3, y2 ≤ 4, y3 ≤ 5,
and y4 ≤ 8. Because y1 + y2 + y3 + y4 = 3, the inequality conditions are always
satisfied, so the answer is 63 = 20.
3
5. The primes whose square is less than 200 is 2, 3, 5, 7, 11, 13. Note that
any products of different these numbers are always greater than 200. By the
inclusion-exclusion principal, the number of positive integers less than 200 such
that is divisible by at least one of 2, 3, 5, 7, 11, 13 is
199
199
199
199
199
199
c+b
c+b
c+b
c+b
c+b
c
2
3
5
7
11
13
199
199
199
199
199
−b
c−b
c−b
c−b
c−b
c
6
10
14
22
26
199
199
199
199
199
199
c−b
c−b
c−b
c−b
c−b
c
−b
15
21
33
39
35
55
199
199
199
199
199
199
−b
c−b
c−b
c−b
c+b
c+b
c
65
77
91
143
30
42
199
199
199
199
199
199
+b
c+b
c+b
c+b
c+b
c+b
c
66
78
70
110
130
154
199
199
199
199
199
199
+b
c+b
c+b
c+b
c+b
c+b
c
182
286
105
165
195
231
199
199
199
199
199
199
+b
c+b
c+b
c+b
c+b
c+b
c
273
429
385
455
715
1001
= 99 + 66 + 39 + 28 + 18 + 15 − 33 − 19 − 14 − 9 − 7 − 13
b
−9−6−5−5−3−3−2−1+6+4+3+2+2+1+1+1+1
+ 0 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 158.
If p is a positive integer less than 200 and greater than 1, then p is a prime if
and only if p is one of 2, 3, 5, 7, 11, 13 or p is not divisible by 2, 3, 5, 7, 11, 13, so
the answer is 199 − 1 − 158 + 6 = 46.
8. The number of functions from a set with seven elements to one with n
elements
is n7 . Moreover, the number of subsets of {1, 2, 3, 4, 5} with n elements
5
is n , so by the inclusion-exclusion principal, the answer is
5 7
5 7
5 7
5 7
5 7
57 −
4 +
3 −
2 +
1 −
0 = 16829.
1
2
3
4
5
10. That question is to find the number of onto functions from a set with eight
elements to one with three elements. The number of functions from a set with
eight elements to one with n elements
is n8 . Moreover, the number of subsets
3
of {1, 2, 3} with n elements is n , so by the inclusion-exclusion principal, the
answer is
3 7
3 7
3 7
38 −
2 +
1 −
0 = 6180.
1
2
0
12. 2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321.
14. The probability is
D10
10!
=
16481
44800 .
4
16. That is Dn = n!
1
0!
−
1
1!
1
+ · · · + (−1)n n!
.
18. Let S be the set of functions f : {1, . . . , n} → {1, . . . , n} such that f (u) 6= u
for all u. Then |S| = Dn . Also, for 1 ≤ i ≤ n, |{f ∈ S | f (1) = i, f (i) = 1}| =
Dn−2 . Moreover, there is a bijection between {f ∈ S | f (1) = i, f (i) 6= 1} and
the set of functions f : {1, . . . , n} − {i} → {1, . . . , n} − {i} such that f (u) 6= u
for all u where f corresponds to
f (i)
if u = 1
g(u) =
f (u) otherwise.
This shows that |{f ∈ S | f (1) = i, f (i) 6= 1}| = Dn−1 . There are n − 1 possible
ways to choose i, so Dn = (n − 1)(Dn−1 + Dn−2 ).
19. We will prove Dn = nDn−1 + (−1)n by induction. If n = 1, it is true
because D1 = 0 and D0 = 1. If Dk = kDk−1 + (−1)k where k is a positive
integer, then by the previous problem,
Dk+1 = k(Dk +Dk−1 ) = kDk +kDk−1 = kDk +Dk −(−1)k = (k+1)Dk +(−1)k+1 ,
so by induction, we proved the formula.
20. By the previous problem, we get
Dn = (−1)n + nDn−1 = (−1)n + n (−1)n−1 + (n − 1)Dn−1
n
n−1
= · · · = (−1) + n(−1)
0
+ · · · + n(n − 1) · · · 1(−1) = n!
1
1
1
− + · · · + (−1)n
0! 1!
n!
26. The problem is to find the number of functions f : {1, 2, 3, 4, 5, 6} →
{1, 2, 3, 4, 5, 6} such that f (1), f (2), f (3) ∈ {4, 5, 6} and that f (4), f (5), f (6) ∈
{1, 2, 3}. That is 33 × 33 = 36 .
5
.