Questions: Select your answer (13 out of 14) At a rehearsal dinner the night before a wedding, the bride and groom need to assign 10 people to two tables of five people. How many different seating arrangements are there?

Solution Steps

To solve this problem, we need to determine the number of ways to divide 10 people into two groups of 5. This is a combinatorial problem where we first choose 5 people out of 10 to sit at the first table, and the remaining 5 will automatically sit at the second table. Since the order of selection does not matter, we use combinations. However, since the two tables are indistinguishable, we divide the result by 2 to avoid counting the same arrangement twice.

To find the number of ways to assign 10 people to two tables of 5, we first calculate the number of ways to choose 5 people from 10. This is given by the combination formula:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Substituting \( n = 10 \) and \( k = 5 \):

\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

Since the two tables are indistinguishable, we must divide the total combinations by 2 to avoid double counting:

\[ \text{Distinct Arrangements} = \frac{\binom{10}{5}}{2} = \frac{252}{2} = 126 \]

Final Answer