Question

Given \[\xi = \{ a,{\text{ }}b,\;{\text{ }}c,\;{\text{ }}d,\;{\text{ }}e,\;{\text{ }}f,\;{\text{ }}g,{\text{ }}h,{\text{ }}i,{\text{ }}j,{\text{ }}k\} \], \[C = \{ b,{\text{ }}e,{\text{ }}a,{\text{ }}d\} \], \[D = \{ j,\;a,\;d,\;e\} \] and \[E = \{ b,\;a,\;d,\;g,\;e\} \]. Determine,
1. \[C \cap D \cap E\]
2. \[C \cup D \cup E\]
3. \[n[(C \cup D) \cap E']\]
4. \[n[(C \cup E)' \cap D]\]

Answer 1

Hint: Here we have to find the required set operations by using the given sets. This problem comes under sets and functions. Set is a collection of well-defined objects. The basic operation for this problem is union and intersection. The symbol \[ \cup \] represents union which means combining all elements of sets together. The symbol \[ \cap \] represents intersection which means combining common elements of sets. This problem requires both the operation. For that, first we need to solve it by a complete step-by-step explanation.

Complete step-by-step answer:
1. \[C \cap D \cap E\]
First we consider \[C \cap D\]
\[ \Rightarrow C = \{ b,e,a,d\} \cap D = \{ j,a,d,e\} \]
Combining same elements of two sets we get
\[ \Rightarrow C \cap D = \{ a,e,d\} \]
Now consider \[C \cap D \cap E\]
We know the elements of \[C \cap D\] then now combine common elements with \[E\]
\[ \Rightarrow (C \cap D) \cap E = \{ a,e,d\} \cap \{ b,a,d,g,e\} \]
\[ \Rightarrow C \cap D \cap E = \{ a,e,d\} \]
$\therefore $ Thus we find (i) \[C \cap D \cap E = \{ a,e,d\} \]
2. \[C \cup D \cup E\]
For this we should solve do union operation, now we need to combine all elements
Now same like before we need to solve separately
First we consider \[C \cup D\]
\[ \Rightarrow C = \{ b,e,a,d\} \cup D = \{ j,a,d,e\} \]
\[ \Rightarrow C \cup D = \{ b,e,a,j,d\} \]
We know the elements of \[C \cup D\] now combine all elements with \[E\]
\[ \Rightarrow (C \cup D) \cup E = \{ b,e,a,j,d\} \cup \{ b,a,d,g,e\} \]
\[ \Rightarrow C \cup D \cup E = \{ b,a,e,d,g,j\} \]
$\therefore $ Thus we find \[C \cup D \cup E = \{ b,a,e,d,g,j\} \]
3. \[n[(C \cup D) \cap E']\]
For this, we also need to know about complement, this means that the complement of a set A contains everything that is not in the set A and also we want to find a number of elements.
First we consider \[C \cup D\]
\[ \Rightarrow C = \{ b,e,a,d\} \cup D = \{ j,a,d,e\} \]
\[ \Rightarrow C \cup D = \{ b,e,a,j,d\} \]
Now find \[E'\]
\[ \Rightarrow E' = \xi - E\]
\[ \Rightarrow E' = \{ a,b,c,d,e,f,g,h,i,j,k\} - \{ b,a,d,g,e\} \]
\[ \Rightarrow E' = \{ c,f,h,i,j,k\} \]
Now we know the elements of \[(C \cup D)\]now combine common elements with \[E'\]
\[ \Rightarrow (C \cup D) \cap E' = \{ b,e,a,j,d\} \cap \{ c,f,h,i,j,k\} \]
\[ \Rightarrow (C \cup D) \cap E' = \{ j\} \]
Now we need to find number of elements in\[(C \cup D) \cap E'\]
\[ \Rightarrow n[(C \cup D) \cap E'] = 1\]
Thus we find 3.
Now we consider 4. \[n[(C \cup E)' \cap D]\]
This is also like previous subdivision, first we need to find \[(C \cup E)\]
\[ \Rightarrow C \cup E = \{ b,e,a,d\} \cup \{ b,a,d,g,e\} \]
\[ \Rightarrow C \cup E = \{ b,a,d,g,e\} \]
Now find\[(C \cup E)'\]
\[ \Rightarrow (C \cup E)' = \xi - (C \cup E)\]
\[ \Rightarrow (C \cup E)' = \{ a,b,c,d,e,f,g,h,i,j,k\} - \{ b,a,d,g,e\} \]
\[ \Rightarrow (C \cup E)' = \{ c,f,h,i,j,k\} \]
Now we know the elements of \[(C \cup E)'\]now combine common elements with \[D\]
\[ \Rightarrow (C \cup E)' \cap D = \{ c,f,h,i,j,k\} \cap \{ j,a,d,e\} \]
\[ \Rightarrow (C \cup E)' \cap D = \{ j\} \]
Now, we find the number of elements in the set
\[n[(C \cup E)' \cap D] = 1\]
$\therefore $ Thus, we find 4. \[n[(C \cup E)' \cap D] = 1\]

Note: This solution basically comes under three operation union, intersection, and complement more than that basic set function concept and knows to combine common elements as well as all elements of a set together. The number element is to count the number of elements after the set operations are done.