Question

In a group of 3 girls and 4 boys, there are two boys \( B_1 \) and \( B_2 \). The number of ways in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \( B_1 \) and \( B_2 \) are not adjacent to each other, is:

• A. 144
• B. 120
• C. 72
• D. 96

Hint:

For problems involving arrangements with restrictions: - Start by calculating the total number of arrangements without any restrictions. - Then, subtract the cases where the restricted condition is violated (e.g., when \( B_1 \) and \( B_2 \) are adjacent). - Use the principle of inclusion-exclusion if necessary.

Answer 1

The Correct Option is C

We start by treating the girls as a single block since all the girls must stand together. Thus, we have 5 objects to arrange: the girls block and the 4 boys. The total number of ways to arrange these 5 objects is \( 5! \), but we must consider that \( B_1 \) and \( B_2 \) should not be adjacent. First, calculate the total arrangements where all 5 objects are arranged: \[ 5! = 120. \] Next, calculate the number of ways in which \( B_1 \) and \( B_2 \) are adjacent. If they are adjacent, treat them as a single block, so now we have 4 objects to arrange. The total number of ways to arrange these 4 objects is \( 4! \), and within the \( B_1 B_2 \) block, there are \( 2! \) ways to arrange \( B_1 \) and \( B_2 \). Thus, the number of ways in which \( B_1 \) and \( B_2 \) are adjacent is: \[ 4! \times 2! = 24 \times 2 = 48. \] The number of ways in which \( B_1 \) and \( B_2 \) are not adjacent is: \[ 120 - 48 = 72. \] Thus, the answer is \( 72 \).

Answer 2

Step 1: Treat groups as single units.
There are two main groups: one of 3 girls (as a single block) and one of 4 boys (as another block). These two groups can be arranged in: \[ 2! = 2 \text{ ways (Girls first or Boys first).} \]

Step 2: Arrange the girls within their group.
The 3 girls can be arranged among themselves in: \[ 3! = 6 \text{ ways.} \]

Step 3: Arrange the boys with the condition.
The 4 boys can be arranged among themselves in \( 4! = 24 \) ways in total.
However, \( B_1 \) and \( B_2 \) should not be adjacent.

Step 4: Number of arrangements where \( B_1 \) and \( B_2 \) are together.
Treat \( B_1 \) and \( B_2 \) as a single unit ⇒ now we have 3 units \( (B_1B_2), B_3, B_4 \). They can be arranged in \( 3! = 6 \) ways. Within the block \( (B_1B_2) \), \( B_1 \) and \( B_2 \) can swap positions in \( 2! = 2 \) ways. Hence, number of arrangements where \( B_1 \) and \( B_2 \) are together: \[ 3! \times 2 = 12. \]

Step 5: Number of arrangements where \( B_1 \) and \( B_2 \) are not together.
\[ 4! - 12 = 24 - 12 = 12. \]

Step 6: Combine all arrangements.
Total number of possible arrangements: \[ (\text{arrangements of groups}) \times (\text{arrangements of girls}) \times (\text{arrangements of boys}) \] \[ = 2 \times 6 \times 12 = 144. \]

Step 7: Check adjacency restriction correctly.
Each arrangement of boys already ensures \( B_1 \) and \( B_2 \) are not adjacent, so the total valid arrangements are \( 144 \).

Step 8: But total arrangement includes both group orders; let's recheck:
Yes, the girls’ group and boys’ group can stand in either order (2 ways), and all internal arrangements are counted correctly.

Final Answer:

\[ \boxed{144} \]