Explanation:

Let's break down the clues:

- Six friends: E, R, K, S, A, G

- Birthdays: 11th or 18th of March, May, November

- R has birthday on 18th

- E doesn't have her birthday last

- Number of people before K = Number after R

- Only two have birthday before G

- E’s birthday is in a month with 30 days (only May, since March and November have 31 and 30 days, respectively)

- No more than two birthdays after S

- R & A have birthdays in different months

Let’s arrange them stepwise:

- Possible dates, in order: March 11, March 18, May 11, May 18, Nov 11, Nov 18.

- If only two before G, G must be third: ( _, _, G, _, _, _ )

- E's birthday is in May. E can’t be last, so E is May 11 or 18.

- R is on an 18th; Possible positions: 2, 4, or 6.

- G must be March or May (since 3rd position is May 11, possibly G).

- If K has the same people before and after R, they are symmetrical: K in spot 3 (if R is 4) or K in 4 (if R is 3).

Trying to fit all, the only way number of people before K equals people after R is K 3rd and R 4th. Sequence: _, _, K, R, _, _.

Thus, K has 3 people after him.

Correct answer:

- Option 3: Three

Summary:

- Three people have birthdays after K (positions 4, 5, 6).

- Details: G is third, E is not last and in May, R is fourth, only two after S, etc., lead to this sequence.

Key takeaways:

- K is in the third position; so after K: positions 4, 5, 6 (that’s three people).

- Your answer (option 1: five) is not correct.