Find $\left( {A \cup B} \right) \cap \left( {A \cup C} \right)$.

Verify that A \ C ≠ C \ A.

Let S be a set containing $n$ elements where $n \in \mathbb{N}$.

Show that S has ${2^n}$ subsets.

EITHER

$\left( {A \cup B} \right) \cap \left( {A \cup C} \right) = \left\{ {1,\,2,\,3,\,5,\,7,\,9,\,11} \right\} \cap \left\{ {1,\,3,\,5,\,7,\,9,\,15,\,31} \right\}$    M1A1

OR

$A \cup \left( {B \cap C} \right) = \left\{ {1,\,3,\,5,\,7,\,9,\,11} \right\} \cup \left\{ {3,\,7} \right\}$     M1A1

OR

${B \cap C}$ is contained within A     (M1)A1

THEN

= {1, 3, 5, 7, 9} (= A)     A1

Note: Accept a Venn diagram representation.

 [3 marks]

A \ C = {5, 9}     A1

C \ A = {15, 31}    A1

so A \ C ≠ C \ A     AG

Note: Accept a Venn diagram representation.

[2 marks]

METHOD 1

if $S = \emptyset$ then $n = 0$ and the number of subsets of S is given by 2⁰ = 1     A1

if $n > 0$

for every subset of S, there are 2 possibilities for each element $x \in S$ either $x$ will be in the subset or it will not     R1

so for all $n$ elements there are $\left( {2 \times 2 \times  \ldots  \times 2} \right){2^n}$ different choices in forming a subset of S      R1

so S has ${2^n}$ subsets      AG

Note: If candidates attempt induction, award A1 for case $n = 0$, R1 for setting up the induction method (assume $P\left( k \right)$ and consider $P\left( {k + 1} \right)$ and R1 for showing how the $P\left( k \right)$ true implies $P\left( {k + 1} \right)$ true).

METHOD 2

$\sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)}$ is the number of subsets of S (of all possible sizes from 0 to $n$)     R1

${\left( {1 + 1} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)} \left( {{1^k}} \right)\left( {{1^{n - k}}} \right)$      M1

${2^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered} n \hfill \\ k \hfill \\ \end{gathered} \right)}$ (= number of subsets of S)     A1

so S has ${2^n}$ subsets      AG

[3 marks]