Solution

The correct answer is 5

Explanation:

To determine how many items will be in the set for GOTO(I₀, $) given the grammar:

Grammar:
1. S → A
2. A → $B$
3. A → id
4. B → B, A
5. B → A

Step 1: Compute the CLOSURE of I₀ = CLOSURE({[S → .A]})

The initial item set I₀ consists of the item [S → .A] .

Applying the Closure:
1. From S → .A , we look for productions for A :
    A → $B$
    A → id

   So, we add:
    [A → $B$]
    [A → id]

2. We also have B productions, since B can be derived from A :
    B → B, A
    B → A

Thus, the closure will contain the following items:

• [S → .A]
• [A → $B$]
• [A → id]
• [B → B, A]
• [B → A]

Closure Result:

• So, I₀ = {[S → .A], [A → $B$], [A → id], [B → B, A], [B → A]} contains 5 items.

Step 2: Compute GOTO(I₀, $)

Now we compute GOTO(I₀, $) .

For each item in I_0 :
  1. Item: [S → .A]
      This item has the dot before A , and since we are doing GOTO(I₀, $) , we look for a production that has A followed by a $.
  
  2. Item: [A → $B$]
      This item is already at $, and it does not produce anything new.
  
  3. Item: [A → id]
      This item does not contain any $.
  
  4. Item: [B → B, A]
      This item does not contain any $.
  
  5. Item: [B → A]
      This item does not contain any $.

The only transition will happen from the first item:

• From [S → .A] to [S → A.] if we were to look at it followed by $ but as we are considering directly into GOTO to $ it will transition to the completion of handling B .

Conclusion for GOTO(I₀, $) :

• After considering all items:
• The item resulting from [S → .A] will yield [S → A.] which finalizes based on A and only B transitions over.

Resulting Items in GOTO Set:

• Thus we gather:
  • [S → A cdot]
  • [A → $B$]
  • [A → id]
  • [B → B, A]
  • [B → A]

This retains the closure set yet leading us to assess from the transitions that all items still remain as is thus maintaining a total of 5 items.

Final Answer:
The correct answer is 3) 5.