JEE Mathematics 2024 Question with Solution

Given:

• $|M| = 20$
• $|P| = 25$
• $|C| = 16$
• $|M \cup P \cup C| = 40$
• $|M \cap P| \le 11$
• $|P \cap C| \le 15$
• $|M \cap C| \le 10$

Find: The maximum value of $|M \cap P \cap C|$.

Use the principle of inclusion-exclusion for three sets $M$, $P$, and $C$:

$$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |M \cap C| + |M \cap P \cap C|$$

Substitute the given values using the maximum allowed pairwise intersections:

$$40 = 20 + 25 + 16 - 11 - 15 - 10 + x$$ $$x = 10$$

Thus, the maximum number of students who passed in all three subjects is $10$.

Given:

• Total students in at least one subject $= 40$
• $n(M) = 20, \; n(P) = 25, \; n(C) = 16$
• $n(M \cap P) \le 11$
• $n(P \cap C) \le 15$
• $n(M \cap C) \le 10$

Find: The maximum possible value of $x = n(M \cap P \cap C)$.

By inclusion-exclusion,

$$40 = 20 + 25 + 16 - n(M \cap P) - n(P \cap C) - n(M \cap C) + x$$ $$40 = 61 - \left[n(M \cap P) + n(P \cap C) + n(M \cap C)\right] + x$$

So,

$$n(M \cap P) + n(P \cap C) + n(M \cap C) = 21 + x$$

Let

• $d$ = students in Mathematics and Physics only,
• $e$ = students in Physics and Chemistry only,
• $f$ = students in Mathematics and Chemistry only.

Then

$$n(M \cap P) = d + x, \quad n(P \cap C) = e + x, \quad n(M \cap C) = f + x$$

Therefore,

$$(d+x) + (e+x) + (f+x) = 21 + x$$ $$d + e + f + 3x = 21 + x$$ $$d + e + f = 21 - 2x$$

Since the number of students in each region cannot be negative,

$$d + e + f \ge 0$$

Hence,

$$21 - 2x \ge 0$$ $$x \le \frac{21}{2} = 10.5$$

Since $x$ must be an integer, the maximum possible value is $10$.

Verification for $x = 10$:

• $d+e+f = 21 - 20 = 1$, which is possible.
• For example, take $d=1, e=0, f=0$. Then,

$$n(M \cap P) = 1 + 10 = 11$$ $$n(P \cap C) = 0 + 10 = 10$$ $$n(M \cap C) = 0 + 10 = 10$$

All constraints are satisfied.

Therefore, the maximum number of students who passed in all three subjects is $10$.