`{:(A , B , C , D , E , F , G , H , I , J , K , L , M , N , O , P , Q , R ,S , T ,U , V , W ,X , Y , Z), (1 ,2 , 3 , 4 ,5, 6 , 7,8,9,10 , 11 , 12, 13, 14, 15 , 16 , 17 , 18 , 19, 20 , 21 ,22 , 23 , 24 , 25 , 26):}`
ASTN: ZTSO :: MSUB :?

Options

A. LRRC
B. LTTA
C. NTVC
D. LTTC

Explanation

To solve the analogy problem given, we need to analyze the pattern used in the first pair (ASTN: ZTSO) and apply the same pattern to the second pair (MSUB: ?).

### Step-by-Step Solution:

1. **Identify the Pattern in the First Pair (ASTN: ZTSO)**:
- A → Z: Move back 1 position in the alphabet (A is the 1st letter, Z is the 26th).
- S → T: Move forward 1 position in the alphabet (S is the 19th letter, T is the 20th).
- T → S: Move back 1 position in the alphabet (T is the 20th letter, S is the 19th).
- N → O: Move forward 1 position in the alphabet (N is the 14th letter, O is the 15th).

The pattern alternates between subtracting 1 and adding 1.

2. **Apply the Same Pattern to the Second Pair (MSUB)**:
- M → L: Move back 1 position (M is the 13th letter, L is the 12th).
- S → T: Move forward 1 position (S is the 19th letter, T is the 20th).
- U → T: Move back 1 position (U is the 21st letter, T is the 20th).
- B → C: Move forward 1 position (B is the 2nd letter, C is the 3rd).

3. **Combine the Results**:
- From M we get L.
- From S we get T.
- From U we get T.
- From B we get C.

Therefore, the resulting letters are L, T, T, C.

4. **Final Answer**:
- The answer to the analogy MSUB: ? is LTTCC.