5 boys and 4 girls stand in a row. The number of ways such that all 5 boys stand together OR no two boys stand adjacent to each other is
Step-by-Step Solution
Key Concept: The two events are mutually exclusive (all boys together vs. no two adjacent), so use $|A|+|B|$ directly.
**Event $A$ — all 5 boys together:** Treat the 5-boy block as one unit; arrange 5 units (block + 4 girls) in $5!$ ways, then arrange boys within the block in $5!$ ways:
$|A|=5!\times5!=120\times120=14400$.
**Event $B$ — no two boys adjacent:** First arrange 4 girls in $4!$ ways. This creates 5 gaps (including ends). Place 5 boys one per gap (all gaps used): $5!$ ways to assign boys to gaps:
$|B|=4!\times5!=24\times120=2880$.
**$A\cap B=\emptyset$** (5 boys together $\Rightarrow$ at least 4 pairs of adjacent boys).
Total $=14400+2880=17280$.
<div class="key-concept"><strong>Key Concept:</strong> The two events are mutually exclusive (all boys together vs. no two adjacent), so use $|A|+|B|$ directly.</div>
<div class="trap-box"><strong>Trap:</strong> Assuming the two events can overlap (they cannot: if 5 boys are all together, any two consecutive ones are adjacent, violating no-two-adjacent).</div>
Correct Answer: 17280