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

Step 1: Understanding the Concept:
This problem requires two stages: first, choosing which friends go to which table, and second, arranging them at those tables.
Since the tables are round, we must use the formula for circular permutations.
Step 2: Key Formula or Approach:
1. Combination formula for choosing \(r\) from \(n\): \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\).
2. Circular permutation formula: \((n-1)!\) ways for \(n\) people.
Step 3: Detailed Explanation:
First, we select 12 friends out of 21 for the first table. The remaining 9 will automatically go to the second table:
\[ \text{Ways to select groups} = \binom{21}{12} = \frac{21!}{12! \cdot 9!} \] Next, we arrange the 12 friends at the first round table: \((12-1)! = 11!\).
Then, we arrange the 9 friends at the second round table: \((9-1)! = 8!\).
Total number of ways = \(\frac{21!}{12! \cdot 9!} \cdot 11! \cdot 8!\).
Let's simplify this expression to match the options:
\[ = \frac{21! \cdot 11! \cdot 8!}{(12 \cdot 11!) \cdot (9 \cdot 8!)} = \frac{21!}{12 \cdot 9} \] \[ = \frac{21 \cdot 20 \cdot 19!}{108} = \frac{420 \cdot 19!}{108} \] Dividing numerator and denominator by 12:
\[ = \frac{35}{9} \times 19! \] Step 4: Final Answer:
The total number of seating arrangements is \(\frac{35}{9} \times 19!\).