3

Let $A$ be a finite set. Prove or disprove: the number of subsets of $A$ whose size is even is equal to the number of subsets of $A$ whose size is odd.

Example: $A = {1,2}$. The subsets of $A$ are {},{1},{2}, and {1,2}. Since there are two subsets of odd size ({1} and {2}) and two subsets of even size ({} and {1,2}) the claim holds for this particular example.

Yiorgos S. Smyrlis
  • 83,933
user117911
  • 89
  • what did you get so far. – Yimin May 12 '15 at 04:12
  • 2
    Induction: Suppose any set $A_k$ of $k$ people has $2^{k-1}$ subsets of odd size, and $2^{k-1}$ of even size. Take a set $A_{k+1}$ of $k+1$ elements, $A_k$ plus Beti. How many subsets of odd size does the new set have? (i) The ones that do not contain Beti, so they must be an odd subset of $A_k$. There are $2^{k-1}$ of these; (ii) The ones that contain Beti, so the rest must be an even subset of $A_k$, $2^{k-1}$ possibilities. Total is $2^{k-1}+2^{k-1}=2^k$. – André Nicolas May 12 '15 at 04:26

3 Answers3

6

Choose an element $a\in A.$

The number of odd-sized subsets containing $a$ is equal to the number of even-sized subsets not containing $a.$

The number of odd-sized subsets not containing $a$ is equal to the number of even-sized subsets containing $a.$

Add these two equations: the number of odd-sized subsets is equal to the number of even-sized subsets.

bof
  • 78,265
3

The number of subsets of size $k$ in a set with $n$ elements is $\binom{n}{k}$.

The Binomial Theorem states that $$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^k$$ In particular, if we let $a = b = 1$, we obtain $$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$ Therefore, the number of subsets of a set with $n$ elements is $2^n$. Furthermore, if we set $a = 1$ and $b = -1$ in the Binomial Theorem, we obtain \begin{align*} 0 & = 0^n\\ & = [1 + (-1)]^n\\ & = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}(-1)^k\\ & = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots + (-1)^{n}\binom{n}{n}\\ & = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} - \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} \end{align*} where $\lfloor x \rfloor$ represents the largest integer less than or equal to $x$. Thus, $$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1}$$ Hence, the number of subsets of a set with $n$ elements that contain an even number of elements is equal to the number of subsets that contain an odd number of elements.

Moreover, since a set with $n$ elements has $2^n$ subsets and the number of subsets with an even number of elements is equal to the number of subsets with an odd numbers of elements, the set has $\frac{1}{2} \cdot 2^n = 2^{n - 1}$ subsets with an even number of elements and $2^{n - 1}$ subsets with an odd number of elements.

N. F. Taussig
  • 76,571
0

$\sum_{i=0}^{n/2} C_n^{2i}x^{2i + 1} = \frac{1}{2}((1 + x)^n - (1 - x)^n)$

$\sum_{i=0}^{n/2} C_n^{2i+1}x^{2i } = \frac{1}{2}((1 + x)^n + (1 - x)^n)$

take $x = 1$.

Yimin
  • 3,311
  • 17
  • 32