Question

A person invites 15 guests for dinner and wishes to arrange them at two round tables that can accommodate 8 persons and 7 persons respectively. In how many ways can he arrange the guests? 

(a) 7 ! × 6 ! 

(b) 2 × 8 ! × 7 ! 

(c) \(\frac{15!}{56}\)

(d) \(15C_2\times\frac{13!}{56}\)

Answer 1

(c) \(\frac{15!}{56}\)

First the 15 guests have to be divided into two groups of 8 persons and 7 persons. 

∴ Number of ways of dividing the guests = \(\frac{15!}{8!7!}\) 

Now 8 persons can be seated in one round table in (8 – 1)! = 7! ways 

Also 7 persons can be seated in another round table in (7 – 1)! = 6! ways

∴ Number of ways of arranging the guests =

= \(\frac{15!}{8\times7}\) = \(\frac{15!}{56}\)