EVEN & ODD SUBSETS
ZACH NORWOOD
One of your homework problems asks you to prove, for n even:
n/2 n/2 X
X
n
n
n−1
=2
=
.
2k
2k
−
1
k=0
k=1
First let’s see what this means. Recall that nk is the number of k-subsets
of (‘k-selections from’)P{1, . . . ,n}, and 2n is the total number of subsets of
n
n
n−1
is half the number of subsets of
{1, . . . , n}. So 2n =
k=0 k , and 2
Pn/2 n {1, . . . , n}. Notice also that k=0 2k = 2n−1 is the number of subsets of
Pn/2 n {1, . . . , n} of even size, and k=1 2k−1
is the number of subsets of {1, . . . , n}
of odd size.
Solution 1. Recall Pascal’s identity:
n+1
n
n
=
+
.
k+1
k
k+1
Use Pascal’s identity to combine consecutive terms in the sum given by the
binomial theorem:
n−1 X
n−1
n−1
2
=
k
k=0
n−1
n−1
n−1
n−1
n−1
n−2
+··· +
+
+
=
+
+
n−3
n−2
n−1
0
1
2
{z
}
|
{z
}
|
apply Pascal
apply Pascal
n−1
n
n
n−1
=
+
+ ··· +
+
0
2
n−2
n−1
n
n
n
n
=
+
+ ··· +
+
.
0
2
n−2
n
(The last equality follows from the fact that n0 =
n−1
= 1.)
n−1
Date: 8 Feb 2014.
1
n−1
0
= 1 and
n
n
=
2
ZACH NORWOOD
Solution 2. Since we’re trying to show that the number of even subsets of
{1, . . . , n} is the same as the number of odd subsets of {1, . . . , n}, it suffices
to exhibit a bijection from the set of even subsets to the set of odd subsets.
That is, we want to define a function
f : {X ⊆ {1, . . . , n} : |X| is even} → {X ⊆ {1, . . . , n} : |X| is odd}
and prove that it’s a bijection. Define f as follows:
(
X r {1} if 1 ∈ X
f (X) =
.
X ∪ {1} if 1 6∈ X.
First notice that, if |X| = n, then |f (X)| = n ∓ 1 (according to whether
1 ∈ X or 1 ∈
/ X), so f in fact sends even subsets to odd subsets. We must
show that f is injective and surjective. To see that f is injective, suppose
that X 6= Y . This means that there is some element that is one of X, Y
that is not in the other one. Suppose that a ∈ X r Y . If a 6= 1, then
a ∈ f (X) r f (Y ), so f (X) 6= f (Y ). The alternative is that a = 1, in which
case a = 1 ∈
/ X r {1} = f (X) but a = 1 ∈ Y ∪ {1} = f (Y ), so f (X) 6= f (Y ).
Either way, f (X) 6= f (Y ), so f is injective.
To show that f is surjective, simply notice that f (f (Y )) = Y , so given
Y ⊆ {1, . . . , n} with |Y | odd, we can take X = f (Y ) to see that there is some
X such that f (X) = Y .
We have shown that f is injective and surjective, so f is a bijection.
Solution 3. The final solution is the slickest. It just uses two clever applications of the binomial theorem:
n
X
X n X n
k n
n
(−1)
0 = (1 − 1) =
=
−
.
k
k
k
k=0
k even
k odd
P
P
P
n
n
Add k odd k to each side to get k even k = k odd nk .