MA017 2010 HOMEWORK 5
(1)
One
can
prove
(
n
k
)
+
(
n
k
+1
)
=
(
n
+1
k
+1
)
by
“pure
thought”
as
follows.
The
(
k
+ 1)element
subsets
of
{
1
, . . . , n
+ 1
}
are
of
two
mutually
exclusive
types:
those
that
contain
n
+ 1,
and
those
that
do
not;
the
former
are
in
bijection
with
the
k
element
subsets
of
{
1
, . . . , n
}
,
the
latter
with
the
(
k
+ 1)element
subsets
of
{
1
, . . . , n
}
.
Give
similar
proofs
for
whichever
of
the
following
combinatorial
identities
strike
your
fancy.
Assume,
in
all
formulae
where
(
n
k
)
appears,
that
k
≤
n
.
(a)
(
n
k
)
=
(
n
n

k
)
. Choosing a
k
subset of an
n
set is equivalent to choosing its complement, which
has
n

k
elements.
(b)
∑
n
k
=0
(
n
k
)
= 2
n
.
Both sides count the number of subsets of an
n
set.
One can partition the
subsets of an
n
set according to their size
k
. Summing over the number of
k
subsets for each size
k
, one gets the number of subsets of an
n
set. Alternatively, a subset of an
n
set is determined
by the membership of each of its
n
elements, so there are 2
n
possibilities.
(c)
∑
n
k
=0
k
(
n
k
)
=
n
2
n

1
. Both sides count the number of ways to choose a subset of an
n
set with
one marked element.
For each subset we choose, of size say
k
, there are
k
elements we can
mark. Alternatively, we can choose the marked element first, and then there are 2
n

1
choices
for the unmarked part of the subset.
(d)
∑
n
k
=0
k
2
(
n
k
)
=
n
2
n

1
+
n
(
n

1)2
n

2
. Both sides count the number of ways to choose a subset
of an
n
set with two distinct markings, possibly of the same element. We can choose our subset
first, and then choose which of the two elements to mark, giving the LHS. Alternatively, we can
choose the markings first. In the
n
cases in which they are applied to the same element there
are 2
n

1
possibilities for the unmarked part of the subset, while in the
n
(
n

1) cases in which
they are applied to distinct elements there are 2
n

2
choices for the remaining unmarked part
of the subset.
(e)
∑
n
k
=0
k
3
(
n
k
)
=
n
2
n

1
+ 3
n
(
n

1)2
n

2
+
n
(
n

1)(
n

2)2
n

3
. Both sides count the number of
ways to choose a subset of an
n
set with three distinct markings, possibly of the same elements.
We can choose our subset first, and then choose which of the three elements to mark, giving the
LHS. Alternatively, we can choose the markings first. In the
n
cases in which they are applied
to the same element there are 2
n

1
possibilities for the unmarked part of the subset, in the
3
n
(
n

1) cases in which they are applied to two distinct (there are
n
(
n

1)
2
choices for the pair of
elements to mark, and 6 ways to mark them) there are 2
n

2
choices for the remaining unmarked
part of the subset, while in the
n
(
n

1)(
n

2) cases in which all three marked elements are
distinct, 2
n

3
possibilities remain.
(f)
∑
k
(
n
2
k
)
=
∑
k
(
n
2
k
+1
)
. The left hand side counts the number of even subsets of an
n
set, while
the right hand side counts the number of odd subsets of an
n
set.
One obtains a bijection
between these two collections by inverting the membership of some fixed element.