Seven persons – A, B, C, D, E, F and G go to watch a puppet show one after another but not necessarily in the same order.

D goes to the puppet show three persons before C. One person goes between C and G. As many persons go after G, as before F. F doesn’t go first to watch the puppet show. B goes immediately before A.

Who is the last person to go to watch the puppet show?

The correct answer is: The one who goes two persons after G

Step-by-step Explanation:

Let us determine the order of the seven persons (positions 1 to 7) going to watch the puppet show based on the given clues:

1. D goes to the puppet show three persons before C:
This means there are 2 persons between D and C (i.e., D _ _ C).

2. One person goes between C and G:
So, G can be 2 positions before C or 2 positions after C.

3. As many persons go after G as before F:
This means the position of F from the start (1st) is symmetric to the position of G from the end (7th). That is, Position(F) + Position(G) = 8.

4. F doesn't go first:
So, F cannot be in Position 1, which also means G cannot be in Position 7.

5. B goes immediately before A:
This means B and A must form a contiguous block: (B, A).

Now, let us analyze the possible positions for D and C:

- Case 1: D is in Position 1, C is in Position 4.
If C is in 4, G can be in Position 2 or Position 6.
- Subcase 1a: If G is in Position 2, then 5 persons go after G, so 5 persons must go before F, putting F in Position 6. Now the positions are: 1:D, 2:G, 3:_, 4:C, 5:_, 6:F, 7:_. The remaining empty slots are 3, 5, 7. We need adjacent slots for (B, A), but no two empty slots are adjacent. Thus, this subcase fails.
- Subcase 1b: If G is in Position 6, then 1 person goes after G, so 1 person must go before F, putting F in Position 2. Now the positions are: 1:D, 2:F, 3:_, 4:C, 5:_, 6:G, 7:_. The available empty slots are 3, 5, 7. Again, no adjacent slots are available for (B, A). Thus, this subcase fails.

- Case 2: D is in Position 2, C is in Position 5.
If C is in 5, G can be in Position 3 or Position 7.
- G cannot be in Position 7 (since F cannot be 1st). So G must be in Position 3.
- If G is in Position 3, 4 persons go after G, so 4 persons must go before F, putting F in Position 5. But C is already in Position 5! So this case fails.

- Case 3: D is in Position 3, C is in Position 6.
If C is in 6, G can be in Position 4 (since 2 positions after C would be 8, which is invalid).
- If G is in Position 4, 3 persons go after G, so 3 persons must go before F, placing F in Position 4. But G is already in Position 4! So this case fails.

- Case 4: D is in Position 4, C is in Position 7.
If C is in 7, G must be in Position 5 (one person between C and G).
- Since G is in Position 5, 2 persons go after G (positions 6 and 7). Therefore, 2 persons must go before F, placing F in Position 3.
- Now the placement so far is: 1:_, 2:_, 3:F, 4:D, 5:G, 6:_, 7:C.
- The remaining empty positions are 1, 2, and 6.
- Since B goes immediately before A, they must occupy the adjacent empty slots 1 and 2 (Position 1 = B, Position 2 = A).
- The only remaining person is E, who takes the empty slot at Position 6.

Final Arrangement:

1. B
2. A
3. F
4. D
5. G
6. E
7. C

Conclusion:
The last person to go to watch the puppet show is C (Position 7).
G is at Position 5. Two persons after G is Position $5+2=7$, which is C.
Therefore, the last person (C) is The one who goes two persons after G.