MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, 
(ii) the men and women are seated alternately, 
(iii) members of the same family are seated together and the two other women must be seated separately, 
(iv) members of the same family are seated together and the seats are numbered. 
(ii) First fix the men’s sitting arrangement: (5-1)!
Then the remaining five women’s total number of arrangements are: 5!
Total=4! x 5!=2880
(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!
(3! to permute each family)
By drawing a diagram, the two women have 4 slots to choose from, where order matters:
We first find the required number of ways by treating the seats as unnumbered:
Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways =
One thought on “MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)”
Rather than remember the formula (10-1)! in question (i), I prefer going back to first principle:
Permutation of 10 persons = 10!
Since they sit around a circle, there are 10 repetitions which gives actual possible arrangement = 10!/10 = 9! = (10-1)!