The image contains formulas related to set theory, including formulas for the union, intersection, and complement of sets involving two and three sets. It appears to be a reference sheet for mathematical formulas, not a specific question to be solved.

a(i): Prove $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Step 1: Elements in $A \cup B$ are in $A$ only, $B$ only, or both.

Step 2: Substitute $n(A \setminus B) = n(A) - n(A \cap B)$.

Step 3: Substitute $n(B \setminus A) = n(B) - n(A \cap B)$.

Step 4: Simplify.

n(A \cup B) = n(A) + n(B) - n(A \cap B)

a(ii): Same as a(i), standard inclusion-exclusion principle for two sets.

n(A \cup B) = n(A) + n(B) - n(A \cap B)

a(iii): Prove using rearrangement from a(i).

From $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, rearrange.

But listed as for union, perhaps alternative form.

n(A \cup B) = n(A) + n(B) - n(A \cap B)

b(i): Prove $n(A \cap B) = n(A) + n(B) - n(A \cup B)$.

Step 1: $n(A) = n(A \setminus B) + n(A \cap B)$.

Step 2: $n(B) = n(B \setminus A) + n(A \cap B)$.

Step 3: Add the formulas.

Step 4: Note $n(A \cup B) = n(A \setminus B) + n(B \setminus A) + n(A \cap B)$.

Step 5: Substitute into sum.

Step 6: Solve for intersection.

n(A \cap B) = n(A) + n(B) - n(A \cup B)

c(i): Likely $n(A \cup B) = n(A \cap B) + n(A \setminus B) + n(B \setminus A)$, which is equivalent to a(i).

Step 1: Partition $A \cup B$.

n(A \cup B) = n(A \cap B) + n(A \setminus B) + n(B \setminus A)

d(i): Prove $n(A - B) = n(A) - n(A \cap B)$.

Step 1: $A - B = A \setminus B = A \cap B^c$, elements in $A$ not in $B$.

Step 2: $n(A) = n(A \setminus B) + n(A \cap B)$.

Step 3: Solve for difference.

n(A - B) = n(A) - n(A \cap B)

e: Both sides equal: $n(A - B) = n(A) - n(A \cap B)$.

Verified as in d(i).

n(A - B) = n(A) - n(A \cap B)

f: Both: Alternative proof $n(A - B) = n(A \cap B^c)$.

Assuming finite sets, count elements in $A$ excluding $A \cap B$.

Same as d(i).

n(A - B) = n(A) - n(A \cap B)

g: Of the two: $n(A - B) = n(A \cap B')$.

Yes, by definition $A - B = A \cap B^c$, where $B'$ is complement relative to universe, but cardinality same.

n(A - B) = n(A \cap B')

h: Prove $n(A \cup B \cup C) = n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$.

Step 1: Use inclusion-exclusion principle for three sets.

Divide into 7 disjoint regions:

• Only A: $n(A \setminus (B \cup C))$

• Only B: $n(B \setminus (A \cup C))$

• Only C: $n(C \setminus (A \cup B))$

• A and B only: $n((A\cap B) \setminus C)$

• A and C only: $n((A\cap C) \setminus B)$

• B and C only: $n((B\cap C) \setminus A)$

• All three: $n(A\cap B \cap C)$

Step 2: $n(A \cup B \cup C)$ = sum of all 7.

Step 3: Express each:

$n(A \setminus (B \cup C)) = n(A) - n(A\cap B) - n(A\cap C) + n(A\cap B \cap C)$

(Subtract pairwise, add back triple.)

Similarly for others.

Step 4: Add all:

Full expansion:

$n(A) + n(B) + n(C) - 2n(A\cap B) - 2n(A\cap C) - 2n(B\cap C) + 3n(A\cap B\cap C)$

No, standard derivation:

From two-set: $n((A\cup B)\cup C) = n(A\cup B) + n(C) - n((A\cup B)\cap C)$

Step 5: $n(A\cup B) = n(A)+n(B)-n(A\cap B)$

Step 6: $n((A\cup B)\cap C) = n( (A\cap C) \cup (B\cap C) ) = n(A\cap C) + n(B\cap C) - n(A\cap B \cap C)$

Step 7: Substitute.

Step 8: Simplify.

n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)