MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)
Question:
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, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]
Solution:
(i) (10-1)!=9!=362880
(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:
Total =
(iv)
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 =
Singapore Maths Tuition 
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)!
(QED)
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