## Thinking Mathematically (6th Edition) First, we find the four intersections of the provided sets: \begin{align} & A\cap B=\left\{ {{x}_{3}},\ {{x}_{9}} \right\}\cap \left\{ {{x}_{1}},\ {{x}_{2}},\ {{x}_{3}},\ {{x}_{5}},\ {{x}_{6}} \right\} \\ & =\left\{ {{x}_{3}} \right\} \end{align} \begin{align} & B\cap C=\left\{ {{x}_{1}},\ {{x}_{2}},\ {{x}_{3}},\ {{x}_{5}},\ {{x}_{6}} \right\}\cap \left\{ {{x}_{3}},\ {{x}_{4}},\ {{x}_{5}},\ {{x}_{6}},\ {{x}_{9}} \right\} \\ & =\left\{ {{x}_{3}},\ {{x}_{5}},\ {{x}_{6}} \right\} \end{align} \begin{align} & A\cap C=\left\{ {{x}_{3}},\ {{x}_{9}} \right\}\cap \left\{ {{x}_{3}},\ {{x}_{4}},\ {{x}_{5}},\ {{x}_{6}},\ {{x}_{9}} \right\} \\ & =\left\{ {{x}_{3}},\ {{x}_{9}} \right\} \end{align} \begin{align} & A\cap B\cap C=\left\{ {{x}_{3}},\ {{x}_{9}} \right\}\cap \left\{ {{x}_{1}},\ {{x}_{2}},\ {{x}_{3}},\ {{x}_{5}},\ {{x}_{6}} \right\}\cap \left\{ {{x}_{3}},\ {{x}_{4}},\ {{x}_{5}},\ {{x}_{6}},\ {{x}_{9}} \right\} \\ & =\left\{ {{x}_{3}} \right\} \end{align} Now, place the elements in the regions formed by the above intersections: First, place the elements of $A\cap B\cap C=\left\{ {{x}_{3}} \right\}$ in the innermost region V. Then, place those elements of $A\cap B$in the region II, which do not belong to the region V, which are 5 and 6. Place the elements of $A\cap C$in the region IV, which donot belong to the region V, this region is empty since 4 already came in region V. Then, place the elements of $B\cap C$in the region VI, which do not belong to the region V. So, it contains 7 only. Then, put the remaining elements of sets A, B, and C in the regionsI, III, and VII, respectively. So region I contains 8, region III contains 1 and 2, and region VII contains 3 only. Finally, region VIII contains the remaining elements of the set U. So, all the elements of the set U are used up. So, it contains only 9.