Match List-I with List-II

List - I	List - II
(A) If X and Y are two sets such that $n(X)=17$, $n(Y)=23$, $n(X \cup Y)=38$, then $n(X \cap Y)$ is	I. 20
(B) If $n(X)=28$, $n(Y)=32$, $n(X \cap Y)=10$, then $n(X \cup Y)$ is	II. 10
(C) If $n(X)=10$, then $n(7X)$ is	III. 50
(D) If $n(Y)=20$, then $n\!\left(\tfrac{Y}{2}\right)$ is	IV. 2

Choose the correct answer from the options given below:

1. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

2. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

3. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

4. (A) - (IV), (B) - (II), (C) - (I), (D) - (III)

(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.

(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.

(C) If n(X) = 10, then n(?(X)) (power set) = 2¹⁰ = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 2¹⁰ = 1024. ? But given options map (C) with II = 10, so they mean **n(?(X)) = 2^n(X)** was NOT intended. They likely meant n(X) itself. So (C) → II.

(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.

Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: Option 1