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 "between any two boys there is a girl and between any two girls there is a boy" implies that the arrangement must be strictly alternating.

Step 2: Detailed Explanation:
1. Determine the pattern:
Since there are 4 girls ($G$) and 3 boys ($B$), the only way to alternate without placing two girls together is:
\[ G \ B \ G \ B \ G \ B \ G \]
If a boy were at the start, there would eventually be two girls sitting together at the end.
2. Arrange the individuals:
- The 4 girls can be arranged in the 4 fixed positions in $4!$ ways.
- The 3 boys can be arranged in the 3 fixed positions in $3!$ ways.
3. Calculate total ways:
\[ \text{Total Ways} = 4! \times 3! \]
\[ \text{Total Ways} = 24 \times 6 = 144 \]

Step 3: Final Answer:
The total number of ways is 144.