1. The proposed solution must satisfy the recurrence, i.e.
kc
n

9
kc
n

1
+ 14
kc
n

2
=
c
n
kc
2

9
kc
+ 14
k
=
c
2
k
(
c
2

9
c
+ 14)
=
c
2
We can choose
k
=
c
2
c
2

9
c
+14
(and hence the solution of the required form
exists), unless
c
2

9
c
+ 14 = 0, i.e., unless
c
= 2 or
c
= 7.
2. 8.1.8: This is the same as the number of solutions to
y
1
+
y
2
+
y
3
+
y
4
= 39
with 0
≤
y
i
≤
15 for
i
= 1
,
2
,
3
,
4. Consider the set of all
(
42
3
)
solutions to
this equation for that the variables are nonnegative (but possibly greater
than 15), and let
p
i
be the property that
y
i
≥
16.
Then
N
(
p
k
) (for
k
= 1
,
2
,
3
,
4) is the same as the number of solutions to
z
1
+
z
2
+
z
3
+
z
4
= 39
with
y
k
≥
16, which is
(
26
3
)
.
Similarly,
N
(
p
k
∩
p
j
) =
(
10
3
)
for any
k
and
j
such that 1
≤
k < j
≤
4.
By principle of inclusion and exclusion, the
number of solutions to the original equation is
(
42
3
)

4
(
26
3
)
+ 6
(
10
3
)
.
8.1.13: Let
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This note was uploaded on 07/15/2010 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.
 Spring '09
 MarniMishna

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