GATE CS 2017 Set 1 QGA-7 — General Aptitude

Let $M$ and $W$ represent the number of men and women, respectively. We are given $M + W = 6$.

1. Analyze the constraints:

• There are at least two men and two women: $2 \le W \le 4$ and $2 \le M \le 4$.
• There are at least three right-handed ($R$) persons.
• None of the women are right-handed, which means all women are left-handed ($L$).
• Since only men can be right-handed, there must be at least 3 right-handed men ($M_R \ge 3$). This implies $M \ge 3$.
• Since $W + M = 6$ and $M \ge 3$, we must have $W \le 3$. Thus, $W$ can only be $2$ or $3$.

2. Test $W = 3$:

• If $W = 3$, then $M = 3$. Since $M_R \ge 3$, all 3 men must be right-handed.
• This means the only left-handed people at the table are the 3 women.
• The rule states: "Every woman has a left-handed person to her immediate right." Since only women are left-handed, every woman must have another woman to her immediate right. This circular dependency requires all 6 seats to be occupied by women, which is a contradiction. Thus, $W = 3$ is impossible.

3. Test $W = 2$:

• If $W = 2$, then $M = 4$. We can have 3 right-handed men and 1 left-handed man ($M_L$).
• The left-handed group consists of 2 women ($W_1, W_2$) and 1 man ($M_L$).
• We can arrange them such that $W_2$ is to the immediate right of $W_1$, and $M_L$ is to the immediate right of $W_2$. Both women now have a left-handed person to their immediate right, satisfying all conditions.

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