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 =