Question

There are 3 boys and 4 girls in a group. The number of ways they can sit in a row so that between any two boys there is a girl and between any two girls there is a boy, is

• A. 88
• B. 96
• C. 124
• D. 144
• E. 288

Hint:

When counts differ by exactly 1 (like 4 and 3), the arrangement is unique (starting with the larger group). Just multiply the factorials of the counts.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The condition strictly dictates an alternating seating arrangement.
Since there are 4 girls and 3 boys, the arrangement must start and end with a girl.
Step 2: Key Formula or Approach:
The only possible alternating pattern is G B G B G B G.
Count the permutations of girls in the 'G' slots and boys in the 'B' slots.
Step 3: Detailed Explanation:
The girls must occupy the 1st, 3rd, 5th, and 7th positions.
Number of ways to arrange the 4 girls = \(4!\).
The boys must occupy the 2nd, 4th, and 6th positions.
Number of ways to arrange the 3 boys = \(3!\).
Total number of ways = \(4! \times 3!\).
\[ 4! = 24 \] \[ 3! = 6 \] \[ \text{Total} = 24 \times 6 = 144 \] Step 4: Final Answer:
The number of ways is 144.