SSC Steno Grade C and D 6 Aug 2025 Shift 1 Paper

Solution:This is a problem of permutations with constraints. Since no two women can sit together, we must use the "Gap Method" (or "Spacing Method"). We first arrange the people who have no restrictions (the men) and then place the restricted people (the women) into the gaps created by the unrestricted group.Step 1: Arrange the Men (M)We have 3 distinct men ($M_1, M_2, M_3$). The number of ways to arrange 3 distinct items is given by $3!$.$\text{Arrangements of Men} = 3! = 3 \times 2 \times 1 = 6$When the 3 men are seated, they create 4 possible gaps (including the spaces at the two ends) where the women can sit:$\_ M \_ M \_ M \_$(Gap 1, Gap 2, Gap 3, Gap 4)Step 2: Place the Women (W)We have 2 distinct women ($W_1, W_2$) and 4 available gaps. To ensure no two women sit together, we must choose 2 different gaps out of the 4 available gaps and place one woman in each.This is a permutation problem because the women are distinct, and the order in which we place them matters. We are selecting 2 positions from 4 and arranging the 2 women in those positions, which is $P(4, 2)$.$\text{Ways to place women} = P(4, 2) = \frac{4!}{(4-2)!} = 4 \times 3 = 12$Step 3: Calculate the Total ArrangementsThe total number of arrangements is the product of the ways to arrange the men and the ways to place the women.$\text{Total Arrangements} = (\text{Arrangements of Men}) \times (\text{Ways to place women})$$\text{Total Arrangements} = 3! \times P(4, 2)$$\text{Total Arrangements} = 6 \times 12 = 72$Therefore, there are 72 possible seating arrangements.Answer: D. 72***Shortcut:1. Arrange the 3 men: $3! = 6$ ways.2. The 3 men create 4 gaps.3. Place the 2 women in 2 of the 4 gaps: $P(4, 2) = 12$ ways.4. Total arrangements = $3! \times P(4, 2) = 6 \times 12 = 72$.