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EECS 203: Homework 8 Solutions
Section 5.4
1. (E) 8
By binomial theorem, the coeﬃcient of the term
x
8
y
9
in the expansion of
(
x
+
y
)
17
is
(
17
8
)
= 24310
.
When considering the term
x
8
y
9
in the expansion of
(3
x
+ 2
y
)
17
, let
x
0
= 3
x
and
y
0
= 2
y
, meaning that
(3
x
+ 2
y
)
17
= (
x
0
+
y
0
)
17
. If we substitute
x
0
= 3
x
and
y
0
= 2
y
, then this term becomes
(
17
8
)
(3
x
)
8
(2
y
)
9
=
(
(
17
8
)
3
8
2
9
)
x
8
y
9
. As a conclusion, we get the coeﬃcient of the term
x
8
y
9
in the expansion of
(3
x
+ 2
y
)
17
to be
(
17
8
)
3
8
2
9
= 81
,
662
,
929
,
920
2. (E) 24
Because
(
p
k
)
=
p
!
k
!(
p

k
)!
, the numerator of the fraction is clearly divisible by
p
. On the contrary, none of the
factors in the denominator are divisible by
p
because all of them must be less than
p
and
p
is prime. As
a result, since we know that any combination
(
p
k
)
are integers, we know that
(
p

1)!
must be divisible by
k
!(
p

k
)!
. Therefore,
(
p
k
)
/p
=
(
p

1)!
k
!(
p

k
)!
must be an integer.
3. (M) 28
a) Both the lefthanded side and the righthanded side are counting the number of ways to choose 2
objects out of the total of
2
n
objects.
Left side:
Choose
2
objects from a pile of
2
n
objects
→
(
2
n
2
)
Right side:
Suppose that there are two piles each with
n
objects. There are three cases of choosing
two objects out of the two piles.
(1) The two objects are drawn from diﬀerent piles
→
n
2
(2) The two objects are drawn from the ﬁrst pile
→
(
n
2
)
(3) The two objects are drawn from the second pile
→
(
n
2
)
Therefore, the lefthanded side and the righthanded side are counting the same set of objects.
b) The below algebraic derivation from the righthanded side to the lefthanded side proves that both
sides of the equation are equivalent.
±
2
n
2
²
=
(2
n
)!
2!(2
n

2)!
=
(2
n
)(2
n

1)(2
n

2)!
(2
n

2)!2!
=
(4
n
2

2
n
)(2
n

2)!
(2
n

2)!2!
=
(4
n
2

4
n
+ 2
n
)(2
n

2)!
(2
n

2)!2!
=
4(
n
)(
n

1)(2
n

2)! + 2
n
(2
n

2)!
(2
n

2)!2!
=
4(
n
!) + 2
n
(
n

2)!
(
n

2)!2!
×
(2
n

2)(2
n

3)
...
(
n

1)
(2
n

2)(2
n

3)
...
(
n

1)
=
2(
n
!) + (2(
n
!) + 2
n
(
n

2)!
(
n

2)!2!
=
2(
n
!) + 2
n
2
(
n

2)!
(
n

2)!2!
=
2(
n
!)
(
n

2)!2!
+
n
2
=
2
±
n
2
²
+
n
2
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View Full Document4. (C) 32
The binomial theorem is as followed:
(
x
+
y
)
n
=
n
X
j
=0
±
n
j
²
x
n

j
y
j
In order to prove the binomial theorem by induction, the base case and the inductive case must be proven.
Let
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 Spring '07
 YaoyunShi

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