Question

If $A = \left\{ {1,2,3} \right\},\,B = \left\{ {4,5,6} \right\},\,C = \left\{ {1,2} \right\}$, then the number of elements in the set $\left( {A - B} \right) \times \left( {A \cap C} \right)$ is

Answer 1

Hint: First calculate $A - B$ and then calculate $A \cap C$. Multiply the previously obtained two sets. Count the number of elements in the multiplied set to get the answer. Alternatively multiply the number of elements in the two sets, $A - B$ and $A \cap C$.

Complete step-by-step solution:
\[A = \left\{ {1,2,3} \right\}\] and $B = \left\{ {4,5,6} \right\}$
$A - B = \left\{ {1,2,3} \right\}$
\[A = \left\{ {1,2,3} \right\}\] and $C = \left\{ {1,2} \right\}$
$A \cap C = \left\{ {1,2} \right\}$
$\left( {A - B} \right) \times \left( {A \cap C} \right) = \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {3,1} \right),\left( {3,2} \right)} \right\}$
As we can see, there are 6 elements.
Therefore, the number of elements in the set $\left( {A - B} \right) \times \left( {A \cap C} \right)$ is 6.

Note: If $M = \left\{ {p,q} \right\}$, $N = \left\{ {r,s,t} \right\}$, $X = \left\{ {a,b,c,d,e} \right\}$, $Y = \left\{ {c,d,e,f,g} \right\}$ and $Z = \left\{ {d,e,f,g} \right\}$then,
 $X - Y = \left\{ {a,b} \right\}$, i.e., we remove all the common elements of $Y$ from $X$.
$X \cap Z = \left\{ {d,e} \right\}$, i.e., we only include the elements which are common in both $X$ and $Z$.
$M \times N = \left\{ {\left( {p,r} \right),\left( {q,r} \right),\left( {p,s} \right),\left( {q,s} \right),\left( {p,t} \right),\left( {q,t} \right)} \right\}$, i.e., we multiply each element of $M$ with each element of $N$.