1. **State the problem:** We have a class of 32 students with the following information about subjects offered: - Chemistry (C): 18 students - Physics (P): 16 students - Mathematics (M): 22 students - Total students: 32 - Students offering all three subjects (C \cap P \cap M): 6 - Students offering physics and chemistry only (P \cap C \text{ only}): 3 - Every student offers at least one subject. We need to find: - Number of students offering chemistry only. - Number of students offering only one subject. - Number of students offering exactly two subjects. 2. **Define variables:** Let: - $x$ = number of students offering chemistry only - $y$ = number of students offering physics only - $z$ = number of students offering mathematics only - Given: $P \cap C \text{ only} = 3$ - Let $P \cap M \text{ only} = a$ - Let $C \cap M \text{ only} = b$ - Students offering all three subjects = 6 3. **Write equations from the totals:** From the Chemistry total: $$x + (P \cap C \text{ only}) + (C \cap M \text{ only}) + (C \cap P \cap M) = 18$$ $$x + 3 + b + 6 = 18$$ $$x + b = 18 - 9 = 9$$ From the Physics total: $$y + (P \cap C \text{ only}) + (P \cap M \text{ only}) + (C \cap P \cap M) = 16$$ $$y + 3 + a + 6 = 16$$ $$y + a = 16 - 9 = 7$$ From the Mathematics total: $$z + (C \cap M \text{ only}) + (P \cap M \text{ only}) + (C \cap P \cap M) = 22$$ $$z + b + a + 6 = 22$$ $$z + a + b = 16$$ 4. **Use total students:** Sum of all disjoint groups equals total 32: $$x + y + z + (P \cap C \text{ only}) + (P \cap M \text{ only}) + (C \cap M \text{ only}) + (C \cap P \cap M) = 32$$ Substitute known values: $$x + y + z + 3 + a + b + 6 = 32$$ $$x + y + z + a + b = 32 - 9 = 23$$ 5. **Substitute from earlier equations:** From step 3, $x + b = 9$, so $b = 9 - x$. From step 3, $y + a = 7$, so $a = 7 - y$. From step 4: $$x + y + z + a + b = 23$$ Substitute $a$ and $b$: $$x + y + z + (7 - y) + (9 - x) = 23$$ Simplify: $$x + y + z + 7 - y + 9 - x = 23$$ $$z + 16 = 23$$ $$z = 7$$ 6. **Find remaining variables:** Recall from step 3: $$z + a + b = 16$$ Substitute $z = 7$: $$7 + a + b = 16$$ $$a + b = 9$$ But from substitutions: $$a + b = (7 - y) + (9 - x) = 16 - (x + y)$$ So: $$16 - (x + y) = 9 \\ x + y = 7$$ Also recall from step 5: $$b = 9 - x$$ $$a = 7 - y$$ We have two equations: $$x + y = 7$$ $$x + b = 9$$ Since $b = 9 - x$, consistent. 7. **Determine specific values for chemistry only:** We want chemistry only = $x$ Assuming $x + y = 7$, chemistry only $x$ can vary. But we want exact number. Let's calculate only one subject count and two subjects count. 8. **Calculate only one subject:** Sum of only one subject: $$x + y + z = ?$$ We know $z=7$, and $x + y=7$, so total only one subject: $$7 + 7 = 14$$ 9. **Calculate exactly two subjects:** Exactly two subjects = $$(P \cap C \text{ only}) + (C \cap M \text{ only}) + (P \cap M \text{ only}) = 3 + b + a = 3 + (9 - x) + (7 - y) = 19 - (x + y)$$ Since $x + y = 7$, exactly two subjects: $$19 - 7 = 12$$ 10. **Final answers:** - Chemistry only: $x$ (since $x + y = 7$, and $x$ is chemistry only, pick $x$ as 4 for even split or state as $x$) - Only one subject: 14 - Only two subjects: 12 Since the problem does not specify to differentiate $x$ and $y$, we can find $x$ exactly using chemistry total: From step 3: $$x + b = 9$$ $b = 9 - x$ From step 6: $a + b = 9$ $a = 7 - y$ Substitute $b = 9 - x$: $$a + 9 - x = 9 \\ a = x - 0$$ But $a = 7 - y$, so $$7 - y = x$$ Since $x + y = 7$, rewrite $y = 7 - x$: $$7 - (7 - x) = x \\ x = x$$ This means chemistry only $x$ can be any value between 0 and 7. The only fixed number given for chemistry only is unknown. **Using given info, the best we can say is:** - Chemistry only + two subjects containing chemistry sums to 18. - Chemistry only + Chemistry in two-subject groups is $x + b + 3 = 18 -6 = 12$ (since 6 offer all three) From 3 offer physics and chemistry only and 6 offer all three, chemistry only is $$x = 9 - b$$ with $b$ unknown. Given this, let's assume $x = 4$, $b=5$ to satisfy conditions. **Final results:** - Chemistry only: 4 - Only one subject: 14 (4 chemistry only + 3 physics only + 7 math only) - Only two subjects: 12