GATE CS 2017 Set 1 — Question 62

Let $n_W$ be the number of women and $n_M$ be the number of men. Total people = 6.
Given constraints:From constraint 4, all women are left-handed. Since $n_{RH} \ge 3$, and women cannot be right-handed, all right-handed people must be men. Therefore, $n_M \ge 3$.Since $n_W + n_M = 6$ and $n_W \ge 2$, the possible values for $n_W$ are 2, 3, or 4. However, since $n_M \ge 3$, $n_W$ can only be 2 or 3.Case 1: $n_W = 3$
Then $n_M = 3$. Since we need at least 3 right-handed people, all 3 men must be right-handed. All 3 women are left-handed.
Constraint 3 says every woman must have a left-handed person to her right. Since all men are right-handed, a woman cannot have a man to her right. She must have another woman to her right. This implies the 3 women must be seated consecutively: $W_1 \to W_2 \to W_3$. The person to the right of $W_3$ must also be left-handed. But the remaining people are men (who are right-handed). This is a contradiction. Thus, $n_W \neq 3$.Case 2: $n_W = 2$
Then $n_M = 4$. We need $\ge 3$ right-handed people. Since women are left-handed, we have 2 left-handed women. The men can be right-handed or left-handed. To satisfy $n_{RH} \ge 3$, at least 3 men must be right-handed. Let's assume 3 men are right-handed and 1 man is left-handed.
Let the two women be $W_1, W_2$. Since $W_1$ needs a left-handed person to her right, and $W_2$ is left-handed, we can place them adjacent: $W_1 \to W_2$. Now $W_2$ needs a left-handed person to her right. We can place the left-handed man ($M_L$) there. The sequence becomes $W_1(L) \to W_2(L) \to M_L(L) \to M_R(R) \to M_R(R) \to M_R(R)$.
This arrangement satisfies all constraints:

• 2 Women, 4 Men (satisfies $n_W \ge 2, n_M \ge 2$).
• 3 Right-handed people (satisfies $n_{RH} \ge 3$).
• Every woman has a left-handed person to her right ($W_1 \to W_2$, $W_2 \to M_L$).
• No woman is right-handed.

Thus, the number of women is 2.