Maths : Basic Numeracy

Q 264 / 370

UPSC CSE Prelims 2017

If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?

EXPLANATION

Correct Option (3)

To determine the number of arrangements for 2 boys (B) and 2 girls (G) in a row such that the girls are not next to each other, two primary combinatorial methods can be applied.

Method 1: Complementary Counting Principle

  1. Total unrestricted arrangements:
    • There are 4 distinct individuals (2 boys and 2 girls).
    • The total number of ways to arrange these 4 individuals is 4! (4 factorial).
    • 4! = 4 × 3 × 2 × 1 = 24 arrangements.
  2. Arrangements where the two girls ARE next to each other:
    • Treat the two girls (G1, G2) as a single unit (GG).
    • Now, we are arranging 3 units: (GG), B1, B2. These 3 units can be arranged in 3! ways.
    • 3! = 3 × 2 × 1 = 6 ways.
    • Within the (GG) unit, the two girls can be arranged in 2! ways (G1G2 or G2G1).
    • 2! = 2 × 1 = 2 ways.
    • Therefore, the total number of arrangements where the girls are together is 3! × 2! = 6 × 2 = 12 arrangements.
  3. Arrangements where the girls are NOT next to each other:
    • Subtract the arrangements where girls are together from the total unrestricted arrangements.
    • Number of arrangements = (Total unrestricted arrangements) - (Arrangements where girls are together)
    • = 24 - 12 = 12 arrangements.

Method 2: Gap Method

  1. Arrange the boys first:
    • Arrange the 2 boys (B1, B2) in 2! = 2 ways (e.g., B1 B2 or B2 B1).
  2. Create positions for girls:
    • Arranging the boys creates 3 potential positions (gaps) where the girls can be placed so they are not adjacent: _ B _ B _
  3. Place the girls:
    • To ensure girls are not next to each other, they must occupy two of these three available positions.
    • The number of ways to choose 2 positions out of 3 is C(3, 2) = 3! / (2! × 1!) = 3 ways.
    • Once the 2 positions are chosen, the 2 girls (G1, G2) can be arranged in those 2 positions in 2! = 2 ways.
  4. Total arrangements:
    • Multiply the possibilities from each step:
    • Total arrangements = (Arrangements of boys) × (Ways to choose positions for girls) × (Arrangements of girls in chosen positions)
    • = 2! × C(3, 2) × 2! = 2 × 3 × 2 = 12 arrangements.

Both methods yield 12 possible arrangements.

Incorrect Options:

  • Option 1 (3): This value does not represent a complete or valid calculation for the given problem. It might correspond to an intermediate step like C(3,2) if other permutations were ignored.
  • Option 2 (6): This value represents 3! (the number of ways to arrange 3 units, such as the (GG) block and the two boys) but does not account for the internal arrangement of the girls within their block (2!) or the total permutations.
  • Option 4 (24): This value represents the total number of arrangements of 4 distinct individuals without any restrictions, thus failing to satisfy the condition that the girls are not next to each other.