MATH 681
Exam #1
1.
(8 of 12 students attempted this)
Prove the combinatorial identity
∑
n
k
=1
k
2
(
n
k
)
=
n
2
n

1
+
n
(
n

1)2
n

2
. You may use any method you like.
A simple approach is purely combinatorial: the left side of the equation represents
the number of ways to choose a subset of size
k
from an
n
element set (which can be
done in
(
n
k
)
ways)and then select, with order, an element and then another (possibly
identical) element of the subset (which can be done in
k
2
ways). These numbers are
added up over all values of
k
, so that the left side represents the number of ways to
choose a subset of any size, and then choose two (not necessarily distinct) elements to
distinguish individually. In a less formal setting, the objects enumerated by this left
side might be considered to be the number of ways to choose a
k
person committee
from
n
people, and then to select a chair and secretary (who might be the same person).
In this context, the right side can be seen to enumerate the same thing through a
casewise division: if the chair and secretary are the same, then we could preemptively
choose the chair/secretary any of
n
ways, and then assign committee membership to
the remaining
n

1 prospectives in any of 2
n

1
ways. On the other hand, if the chair
and secretary were different, we would choose any of the
n
people to be the chair, any
of the remaining
n

1 to be the secretary, and then committ3ee membership for the
remaining
n

2 people could be resolved in any of 2
n

2
ways.
There are other approaches possible, using either generating functions or direct algebra,
but this is easily the simplest approach.
2.
(12 of 12 students attempted this)
Answer the following questions relating to paths
through the following grid (note the excluded portions):
(a)
(5 points)
How many walks are there from the lower left corner to the upper
right corner taking upwards and rightwards steps only?
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 Fall '09
 WILDSTROM
 Math, Recurrence relation, ways, Generating function, OGF

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