Question

21 friends were invited for a party. Two round tables can accommodate 12 and 9 friends each, The number of ways of the seating arrangements of friends is .....

• A. \(11! \times 8!\)
• B. \(12! \times 9!\)
• C. \(\frac{35}{9} \times 19!\)
• D. \(\frac{20!}{12!8!} \times 11! \times 9!\)

Hint:

For two round tables of different sizes:
• first choose who sits at each table,
• then use circular arrangement formula \((n-1)!\) for each table. e}

Answer 1

The Correct Option is C

Concept:
For seating around a round table, the number of arrangements of \(n\) people is: \[ (n-1)! \] Here, the two tables are different because their capacities are different: one seats \(12\) and the other seats \(9\). ip

Step 1: Choose the 12 friends for the larger table.
Number of ways: \[ \binom{21}{12} \] ip

Step 2: Arrange friends around the two round tables.
For the table with \(12\) seats: \[ (12-1)! = 11! \] For the table with \(9\) seats: \[ (9-1)! = 8! \] So total number of seatings: \[ \binom{21}{12}\cdot 11!\cdot 8! \] ip

Step 3: Simplify the expression.
\[ \binom{21}{12}\cdot 11!\cdot 8! = \frac{21!}{12!9!}\cdot 11!\cdot 8! \] \[ = \frac{21!}{12\cdot 9} \] \[ = \frac{21\cdot 20 \cdot 19!}{108} \] \[ = \frac{35}{9}\cdot 19! \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{35}{9}\times 19!} \]