1. **State the problem:** We have a universal set $U$ with $n(U) = 100$ elements and two subsets $A$ and $B$ with $n(A) = 20$, $n(B) = 60$, and their intersection $n(A \cap B) = 10$. We want to find the number of elements in each of the four disjoint subsets formed by $A$ and $B$: $A \cap B'$, $A \cap B$, $A' \cap B$, and $A' \cap B'$.\n\n2. **Recall the formulas and rules:**\n- $n(A \cap B')$ is the number of elements in $A$ but not in $B$.\n- $n(A' \cap B)$ is the number of elements in $B$ but not in $A$.\n- $n(A' \cap B')$ is the number of elements in neither $A$ nor $B$.\n- The sets $A \cap B'$, $A \cap B$, $A' \cap B$, and $A' \cap B'$ are disjoint and their union is $U$.\n\n3. **Calculate $n(A \cap B')$:**\nSince $n(A) = n(A \cap B) + n(A \cap B')$, we have\n$$n(A \cap B') = n(A) - n(A \cap B) = 20 - 10 = 10.$$\n\n4. **Calculate $n(A' \cap B)$:**\nSimilarly, $n(B) = n(A \cap B) + n(A' \cap B)$, so\n$$n(A' \cap B) = n(B) - n(A \cap B) = 60 - 10 = 50.$$\n\n5. **Calculate $n(A' \cap B')$:**\nSince the four subsets partition $U$,\n$$n(U) = n(A \cap B') + n(A \cap B) + n(A' \cap B) + n(A' \cap B').$$\nPlugging in known values:\n$$100 = 10 + 10 + 50 + n(A' \cap B').$$\nSolving for $n(A' \cap B')$:\n$$n(A' \cap B') = 100 - (10 + 10 + 50) = 30.$$\n\n**Final answers:**\n- $n(A \cap B') = 10$\n- $n(A \cap B) = 10$\n- $n(A' \cap B) = 50$\n- $n(A' \cap B') = 30$