Solution. (i). \(\mathbb {W}\) is the set of whole numbers \(\{0,1,2,\ldots \}\), so \(E\) collects the whole numbers up to and including \(13\): \[E=\{0,1,2,3,4,5,6,7,8,9,10,11,12,13\}.\]

For \(D\), run through \(k=1,2,3,\ldots \) and keep the values of \(2k+1\) that fall strictly between \(1\) and \(10\): \[k=1\to 3,\quad k=2\to 5,\quad k=3\to 7,\quad k=4\to 9,\quad k=5\to 11\ (\text {too big}).\] \[D=\{3,5,7,9\}.\]

For \(G\), keep the values of \(2k\) with \(1\leq 2k<13\): \[k=1\to 2,\quad k=2\to 4,\quad \ldots ,\quad k=6\to 12,\quad k=7\to 14\ (\text {too big}).\] \[G=\{2,4,6,8,10,12\}.\]

(ii). Work outwards from the innermost bracket. First the complement, taken inside \(E\): \[G^c=E-G=\{0,1,3,5,7,9,11,13\}.\] Then the union: \[C\cup G^c=\{5,6,7,8,9\}\cup \{0,1,3,5,7,9,11,13\}=\{0,1,3,5,6,7,8,9,11,13\}.\] Finally intersect with \(A=\{1,2,3,4,5\}\), keeping only what appears in both: \[A\cap \left (C\cup G^c\right )=\{1,3,5\}.\]

(iii). \(A-B\) means the elements of \(A\) that are not in \(B\). From \(A=\{1,2,3,4,5\}\) remove \(4\) and \(5\): \[A-B=\{1,2,3\}.\]

(iv). First \(A\cap B=\{4,5\}\). Now remove anything also in \(C=\{5,6,7,8,9\}\), which removes the \(5\): \[(A\cap B)-C=\{4\}.\]

(v). First \(A\cap D=\{1,2,3,4,5\}\cap \{3,5,7,9\}=\{3,5\}\), so \[(A\cap D)^c=E-\{3,5\}=\{0,1,2,4,6,7,8,9,10,11,12,13\}.\] Now take from \(F=\{1,5,9\}\) the elements not in that set. Both \(1\) and \(9\) are in it, and \(5\) is not: \[F-(A\cap D)^c=\{5\}.\]

Solution. Listing \(A\). With \(k=1,2,3,\ldots \) the values \(2k-2\) are \(0,2,4,6,\ldots \), stopping before \(20\): \[A=\{0,2,4,6,8,10,12,14,16,18\},\] the even numbers from \(0\) to \(18\). The universal set is \(E=\{0,1,2,\ldots ,21\}\).

The Venn diagram. Placing each of the \(22\) elements in exactly one region:

The eight counts are \(7+3+3+0+2+1+1+5=22\), which is the size of \(E\) — the check that nothing has been placed twice or left out.

(i). \(A^c\) is everything in \(E\) outside \(A\), and \(B^c\) everything outside \(B\). Their intersection is what lies outside both: \[A\cup B=\{0,1,2,3,4,6,8,10,12,14,16,18,19,20\},\] \[A^c\cap B^c=E-(A\cup B)=\{5,7,9,11,13,15,17,21\}.\]

(ii). \((A\cup B)\cap C\) keeps the elements of \(C\) that appear in \(A\) or in \(B\). Testing each element of \(C=\{1,2,6,8,9,11,15\}\) in turn: \(1\in B\), \(2\in A\), \(6\in A\), \(8\in A\), while \(9\), \(11\) and \(15\) are in neither. \[(A\cup B)\cap C=\{1,2,6,8\}.\]

(iii). \(B\cup C=\{1,2,3,6,8,9,11,15,19,20\}\). Removing these from \(A\): \[A-(B\cup C)=\{0,4,10,12,14,16,18\}.\]

(iv). \((A\cup B)^c=E-(A\cup B)=\{5,7,9,11,13,15,17,21\}.\)

Solution. The power set \(P(S)\) is the set of all subsets of \(S\), and it always contains both \(\emptyset \) and \(S\) itself.

(i). The empty set has exactly one subset — itself. So \[P(A)=\{\emptyset \},\] a set with one element. Note that \(P(A)\) is not empty: it is a box containing an empty box.

(ii). List the subsets by size — none, one element, then two: \[P(B)=\Big \{\ \emptyset ,\ \{1\},\ \{2\},\ \{1,2\}\ \Big \},\] which has \(4\) elements.

(iii). \(V=\{a,e,i,o,u\}\), with \(n(V)=5\), so \[n\left (P(V)\right )=2^5=32.\] Grouped by size, the \(32\) subsets are:

Solution. (a) \(A\subset B\) means every element of \(A\) is also in \(B\): the circle for \(A\) sits entirely inside the circle for \(B\).

(i). Since all of \(A\) lies in \(B\), the overlap is the whole of \(A\): \[A\cap B=A\quad \implies \quad (A\cap B)^c=A^c.\]

(ii). \(A\cup B^c\) is everything that is in \(A\) or outside \(B\). What is missing is exactly what is in \(B\) but not in \(A\), so \[A\cup B^c=(B-A)^c.\] Checking with De Morgan: \((B-A)^c=\left (B\cap A^c\right )^c=B^c\cup A\), which is the same set.

(iii). \(B\cap A^c\) is the part of \(B\) outside \(A\), namely \(B-A\). Adding \(A\) back restores the whole of \(B\): \[\left (B\cap A^c\right )\cup A=(B-A)\cup A=B.\]

(b) Disjoint means the two circles do not touch: \(A\cap B=\emptyset \).

(i). That is the definition itself: \[A\cap B=\emptyset .\]

(ii). \(A\cap B^c\) is the part of \(A\) lying outside \(B\). Since none of \(A\) is inside \(B\), that is all of it: \[A\cap B^c=A.\]

(iii). By De Morgan and part (i): \[A^c\cup B^c=(A\cap B)^c=\emptyset ^c=E.\]

Solution. (i). The distributive law lets \(A\) be taken outside: \[(A\cup B)\cap \left (A\cup B^c\right )=A\cup \left (B\cap B^c\right )=A\cup \emptyset =A.\]

(ii). Apply De Morgan to the outer complement first: \[\left [(X\cap Y)^c\cup (X-Y)\right ]^c=(X\cap Y)\cap (X-Y)^c.\] Now \(X-Y=X\cap Y^c\), so by De Morgan again \((X-Y)^c=X^c\cup Y\). Hence \[=(X\cap Y)\cap \left (X^c\cup Y\right ) =\left [(X\cap Y)\cap X^c\right ]\cup \left [(X\cap Y)\cap Y\right ].\] The first bracket is empty, because nothing can be in \(X\) and in \(X^c\) at once. The second is \(X\cap Y\), because \(X\cap Y\) is already inside \(Y\). So \[=\emptyset \cup (X\cap Y)=X\cap Y.\]

(iii). Distribute \(B^c\) across the union: \[B^c\cap (A\cup B)=\left (B^c\cap A\right )\cup \left (B^c\cap B\right ) =\left (A\cap B^c\right )\cup \emptyset =A-B.\]

(iv). De Morgan on the outer complement: \[\left [A^c\cup \left (A\cap B^c\right )\right ]^c=A\cap \left (A\cap B^c\right )^c =A\cap \left (A^c\cup B\right ).\] Distributing: \[=\left (A\cap A^c\right )\cup (A\cap B)=\emptyset \cup (A\cap B)=A\cap B.\]

Solution. (i). Reading the expression: take \(B\), and remove from it everything that lies in \(A\) but not in \(C\). The set \(A-C\) is the part of \(A\) outside \(C\); intersecting \(B\) with its complement keeps the rest of \(B\).

Of the four regions making up \(B\), only one — the part in \(A\) but not in \(C\) — is in \(A-C\). So the shading is all of \(B\) except that piece: \(B\) alone, \(B\cap C\) alone, and the centre \(A\cap B\cap C\).

(ii). Here \(C\) lies wholly inside \(B\), and \(A\) overlaps both. The expression takes \(B\) and removes everything belonging to \(A\) or to \(C\), so what is left is the part of \(B\) lying outside both circles.

Solution. Step 1 — write down what is known. Every one of the \(70\) uses at least one method, so \[n(\text {SM}\cup \text {SN}\cup \text {CH})=70,\] \[n(\text {SM})=56,\quad n(\text {SN})=56,\quad n(\text {CH})=50,\] \[n(\text {SM}\cap \text {SN})=48,\quad n(\text {SM}\cap \text {CH})=42,\quad n(\text {SN}\cap \text {CH})=44.\]

Step 2 — use the inclusion–exclusion rule for three sets. Writing \(x\) for the number using all three, \[n(\text {SM}\cup \text {SN}\cup \text {CH})=n(\text {SM})+n(\text {SN})+n(\text {CH}) -n(\text {SM}\cap \text {SN})-n(\text {SM}\cap \text {CH})-n(\text {SN}\cap \text {CH})+x.\]

Step 3 — substitute and solve. \[70=56+56+50-48-42-44+x\] \[70=162-134+x=28+x\] \[\implies \quad x=42.\]

\[\therefore \quad 42 \text { people used all three methods}.\]

Step 4 — check by filling in every region. Working from the centre outwards:

Solution. Write \(M\), \(E\) and \(D\) for the three sets.

(a). Always start at the centre and work outwards, subtracting what has already been placed.

(b). “Exactly one” means the three outer regions only: \[25+10+5=40.\]

\[\therefore \quad 40 \text { students like exactly one course}.\]

(c). “Neither Economics nor Demography” is \((E\cup D)^c\). First \[n(E\cup D)=n(E)+n(D)-n(E\cap D)=40+45-25=60,\] so \[n\left ((E\cup D)^c\right )=100-60=40.\]

\[\therefore \quad 40 \text { students like neither Economics nor Demography}.\]

Solution. (a). With \(x\) the number taking all three, inclusion–exclusion gives \[80=62+55+48-40-35-42+x\] \[80=165-117+x=48+x\] \[\implies \quad x=32.\]

\[\therefore \quad 32 \text { customers take all three}.\]

(b). “Exactly two” excludes those taking all three, so each pairwise overlap must have the \(32\) removed from it: \[\text {chips and soda only}=40-32=8,\] \[\text {chips and candy only}=35-32=3,\] \[\text {soda and candy only}=42-32=10,\] \[\text {total}=8+3+10=21.\]

\[\therefore \quad 21 \text { customers take exactly two snacks}.\]

Check. The single-snack regions are \[62-40-35+32=19,\qquad 55-40-42+32=5,\qquad 48-35-42+32=3,\] and the eight regions total \[19+5+3+8+3+10+32=80.\ \checkmark \]