First slide

Combinations

Question

A gentleman invites 13 guests to a dinner and places 8 of them at one table and remaining 5 at the other, the tables being round. The number of ways he can arrange the guests is

Moderate

A

$\frac{11!}{40}$
B

9!
C

$\frac{12!}{40}$
D

$\frac{13!}{40}$

Solution

The number of ways in which 13 guests may be divided into groups of 8 and 5 = $^{13} C_{5} = \frac{13!}{5! 8!}$
Now, corresponding to one such group, the 8 guests may be seated at one round table in (8 – 1)! i.e., 7! ways and the five guests at the other table in (5 – 1)! i.e., 4! ways.
But each way of arranging the first group of 8 persons can be associated with each way of arranging the second group of 5, therefore, the two processes can be performed together in 7! × 4! ways.
Hence, required number of arrangements
$= \frac{13!}{5! 8!} \times 7! \times 4! = \frac{13!}{5 \cdot 4! 8 \cdot 7!} \times 7! \times 4! = \frac{13!}{40}$

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