In a class of 65 students, 2 were absent. 28 had sculpture books, 30 graphics, and 32 leather work books. 11 had sculpture and graphics, 12 sculpture and leather, 9 graphics and leather. Find (a) stud

The formula is:

$$n(S \cup G \cup L) = n(S) + n(G) + n(L) - n(S \cap G) - n(S \cap L) - n(G \cap L) + n(S \cap G \cap L)$$

Substitute the given values:

$$63 = 28 + 30 + 32 - 11 - 12 - 9 + n(S \cap G \cap L)$$ $$63 = 90 - 32 + n(S \cap G \cap L)$$ $$63 = 58 + n(S \cap G \cap L)$$

Solve for $n(S \cap G \cap L)$:

$$n(S \cap G \cap L) = 63 - 58$$ $$n(S \cap G \cap L) = 5$$

The number of students who had all three books is $\boxed{\text{5}}$.

Number of students with Sculpture and Graphics books ONLY:

$$n(S \cap G \text{ only}) = n(S \cap G) - n(S \cap G \cap L) = 11 - 5 = 6$$

Number of students with Sculpture and Leather books ONLY:

$$n(S \cap L \text{ only}) = n(S \cap L) - n(S \cap G \cap L) = 12 - 5 = 7$$

Number of students with Graphics and Leather books ONLY:

$$n(G \cap L \text{ only}) = n(G \cap L) - n(S \cap G \cap L) = 9 - 5 = 4$$

The number of students who had exactly two books is the sum of these values:

$$6 + 7 + 4 = 17$$

The number of students who had exactly two books is $\boxed{\text{17}}$.

Number of students with Sculpture books ONLY:

$$n(S \text{ only}) = n(S) - n(S \cap G \text{ only}) - n(S \cap L \text{ only}) - n(S \cap G \cap L)$$ $$n(S \text{ only}) = 28 - 6 - 7 - 5 = 28 - 18 = 10$$

Number of students with Graphics books ONLY:

$$n(G \text{ only}) = n(G) - n(S \cap G \text{ only}) - n(G \cap L \text{ only}) - n(S \cap G \cap L)$$ $$n(G \text{ only}) = 30 - 6 - 4 - 5 = 30 - 15 = 15$$

Number of students with Leather books ONLY:

$$n(L \text{ only}) = n(L) - n(S \cap L \text{ only}) - n(G \cap L \text{ only}) - n(S \cap G \cap L)$$ $$n(L \text{ only}) = 32 - 7 - 4 - 5 = 32 - 16 = 16$$

The number of students who had only one book is the sum of these values:

$$10 + 15 + 16 = 41$$

The number of students who had only one book is $\boxed{\text{41}}$.