Solution: 2014 Winter Midterm - 4

Author: Michiel Smid

Question

Let $S$ be a set of size 37, and let $x$, $y$, and $z$ be three distinct elements of $S$. How many subsets of $S$ are there that contain $x$ or $y$, but do not contain $z$?

(a)

$2^{37} - 2^{35}$

(b)

$2^{36} - 2^{34}$

(c)

$2^{37} - 2^{34}$

(d)

$2^{36} - 2^{35}$

Solution

A = set of subsets that contain x but not z
We can take for granted that one of the element has a fixed possibility of being x: 1
Since $z$ can't be in this subset and $x$ is in this subset, we can create subsets from the remaining 35 elements.
$|A| = 2^{35} $
B = set of subsets that contain y but not z
We can take for granted that one of the element has a fixed possibility of being y: 1
Since $z$ can't be in this subset and $y$ is in this subset, we can create subsets from the remaining 35 elements.
$|B| = 2^{35} $
Now, let's try to determine $ A \cap B $, which is, \enquote{subsets that have x and don't have z and have y and don't have z}
We can take for granted that one of the element has a fixed possibility of being x: 1
We can take for granted that one of the element has a fixed possibility of being y: 1
Since $z$ can't be in this subset and $x$ and $y$ are in this subset already, we can create subsets from the remaining 34 elements.
$|A \cap B| = 2^{34} $

$|A \cup B| = |A| + |B| - |A \cap B|$

$|A \cup B| = 2^{35} + 2^{35} - 2^{34}$

$|A \cup B| = (1+1) \cdot 2^{35} - 2^{34}$

$|A \cup B| = 2 \cdot 2^{35} - 2^{34}$

$|A \cup B| = 2^{36} - 2^{34}$