Question

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is _______.

Answer 1

17280

Answer 2

Case I (All boys together):

• Treat the 5 boys as a single block. Along with 4 girls, there are $5$ units.
• Arranging units: $5! = 120$ ways.
• Arranging boys within the block: $5! = 120$ ways.
• Total ways: $5! \times 5! = 120 \times 120 = 14400$.

Case II (No two boys together):

• First arrange 4 girls: $4! = 24$ ways.
• This creates $5$ gaps (_ G _ G _ G _ G _) where 5 boys can be placed, one per gap.
• Arranging boys: $5! = 120$ ways.
• Total ways: $4! \times 5! = 24 \times 120 = 2880$.

Combined total:

$$14400 + 2880 = 17280$$