Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is

Let the total number of people be $6$, where $M$ is the number of men and $W$ is the number of women.

1. Given constraints:

   • $W \ge 2$ and $M \ge 2$

   • Right-handed persons $\ge 3$

   • None of the women are right-handed $\implies$ All women are left-handed.

   • Therefore, all right-handed persons must be men, which implies $M \ge 3$.

   • Since $M + W = 6$ and $M \ge 3$, the possible values for $W$ are $2$ or $3$.

2. Case 1: If $W = 3$

   • Then $M = 3$. Since there are at least 3 right-handed people, all 3 men must be right-handed.

   • This means the only left-handed people at the table are the 3 women.

   • The condition states that *every woman has a left-handed person to her immediate right*. Hence, the person to the immediate right of every woman must be another woman.

   • In a circular table of 6 people with 3 men and 3 women, if every woman must have a woman to her right, all women would have to be in an unbroken cycle of women, which would require all 6 people to be women (impossible since $M = 3$). A block of 3 consecutive women $W_1 - W_2 - W_3$ leaves the last woman ($W_3$) with a man (who is right-handed) to her immediate right, violating the condition.

   • Thus, $W = 3$ is not possible.

3. Case 2: If $W = 2$

   • Then $M = 4$. Exactly 3 men can be right-handed and 1 man can be left-handed ($M_L$).

   • Seating arrangement (clockwise): $W_1 \to W_2 \to M_L \to M_R \to M_R \to M_R$.

   • Right of $W_1$ is $W_2$ (left-handed).

   • Right of $W_2$ is $M_L$ (left-handed).

   • This satisfies all conditions.

Thus, the number of women at the table is 2.