1. **State the problem:** We are given the cardinalities of various intersections and unions of three sets $X$, $Y$, and $Z$ inside a universal set $\xi$ with $n(\xi) = 62$. We need to find the number of elements in each region of the Venn diagram labeled $a, b, c, d, e, f, g, h$. 2. **Recall the regions:** - $a$: only in $X$ - $b$: in $X \cap Y$ but not $Z$ - $c$: only in $Y$ - $d$: in $X \cap Z$ but not $Y$ - $e$: in $X \cap Y \cap Z$ - $f$: in $Y \cap Z$ but not $X$ - $g$: only in $Z$ - $h$: outside $X, Y, Z$ but inside $\xi$ 3. **Given values:** - $n(X \cap Y \cap Z) = e = 4$ - $n(X' \cap Y \cap Z) = b + f = 2$ (since $X'$ means not in $X$, so intersection with $Y$ and $Z$ excludes $X$) - $n(X \cap Z) = d + e = 15$ - $n(Z) = d + e + f + g = 20$ - $n([X \cup Y \cup Z']') = h = 9$ (complement of union of $X$, $Y$, and complement of $Z$ is outside all three sets) - $n(X' \cap Y \cap Z') = c = 12$ (in $Y$ only, not in $X$ or $Z$) - $n(X \cap Y) = b + e = 12$ 4. **Find $b$:** From $n(X' \cap Y \cap Z) = b + f = 2$ and $n(X \cap Y) = b + e = 12$ with $e=4$, so $b = 12 - 4 = 8$. 5. **Find $f$:** From $b + f = 2$, $f = 2 - b = 2 - 8 = -6$ which is impossible. Re-examine $n(X' \cap Y \cap Z) = 2$ means elements in $Y$ and $Z$ but not in $X$, so $f = 2$ (since $b$ is in $X$ and $Y$ but not $Z$ is not correct, $b$ is $X \cap Y$ but not $Z$; $f$ is $Y \cap Z$ but not $X$). So $b$ is $X \cap Y \cap Z'$ and $f$ is $X' \cap Y \cap Z$. Given $n(X' \cap Y \cap Z) = f = 2$. 6. **Correct $b$:** From $n(X \cap Y) = b + e = 12$ and $e=4$, $b=8$. 7. **Find $d$:** From $n(X \cap Z) = d + e = 15$, $d = 15 - 4 = 11$. 8. **Find $g$:** From $n(Z) = d + e + f + g = 20$, substitute $d=11$, $e=4$, $f=2$, so $g = 20 - (11 + 4 + 2) = 3$. 9. **Find $c$:** Given $n(X' \cap Y \cap Z') = c = 12$. 10. **Find $a$:** From $n(X) = a + b + d + e$. We don't have $n(X)$ directly, but we can find it from total. 11. **Find $h$:** Given $h = 9$. 12. **Find total elements in $X \cup Y \cup Z$:** $$n(X \cup Y \cup Z) = n(\xi) - h = 62 - 9 = 53$$ 13. **Sum all parts inside $X \cup Y \cup Z$:** $$a + b + c + d + e + f + g = 53$$ Substitute known values: $$a + 8 + 12 + 11 + 4 + 2 + 3 = 53$$ $$a + 40 = 53$$ $$a = 13$$ 14. **Summary of values:** - $a = 13$ - $b = 8$ - $c = 12$ - $d = 11$ - $e = 4$ - $f = 2$ - $g = 3$ - $h = 9$ **Final answer:** $$\boxed{a=13, b=8, c=12, d=11, e=4, f=2, g=3, h=9}$$