ξ = {integer $x$ : $0 < x \le 20$} $P = \{\text{perfect square}\}$ $Q = \{\text{even number which solves }3x>11\}$ $R = {\text{multiple of 4}}

1. List all perfect squares in the universal set

The universal set is
$\xi = \{1,2,3,\ldots,20\}$.

The perfect squares in this set are
$1^2=1,\quad 2^2=4,\quad 3^2=9,\quad 4^2=16.$

Since $5^2=25>20$, there are no more perfect squares in $\xi$.

Therefore,
$P=\{1,4,9,16\}.$

1. Identify the elements of $Q$ and exclude those in $P$

From the definitions,

$P=\{1,4,9,16\}$ and $Q=\{4,6,8,10,12,14,16,18,20\}$.

$P' \cap Q$ means the elements in $Q$ that are not in $P$.

So,

$P' \cap Q=\{6,8,10,12,14,18,20\}$.

2. Count the number of elements

There are $7$ elements in $\{6,8,10,12,14,18,20\}$.

Hence,

$n(P' \cap Q)=7$.

1. Identify the relationship between the three sets

From the definitions,

$P=\{1,4,9,16\}$,

$Q=\{4,6,8,10,12,14,16,18,20\}$,

$R=\{4,8,12,16,20\}$.

Every element of $R$ is also in $Q$, so

$R \subseteq Q.$

2. Place all elements in the correct Venn diagram regions

Draw $R$ completely inside $Q$, and let $P$ overlap $Q$.

Place:

$P\cap Q\cap R=\{4,16\},$

$P\text{ only}=\{1,9\},$

$Q\text{ only}=\{6,10,14,18\},$

$Q\cap R\text{ but not }P=\{8,12,20\},$

and outside all three sets:

$\{2,3,5,7,11,13,15,17,19\}.$

1. Describe the shaded region as elements outside both sets

The shaded region is outside circle $A$ and outside circle $B$.

So it represents elements which are not in $A$ or $B$, i.e. the complement of $A \cup B$.