Tessellate S.T.E.M.S. (2019) - Computer Science - School - Set 3 - Objective Problem 2

Probability Level 4

There are 4 couples attended a ceremony. Each couple had 2 children, a son and a daughter. How many ways can these 16 people be seated in a circle such that the following hold:

  • All the 8 children have to sit in adjacent seats.
  • No couple can sit in adjacent seats.
  • No two people of the same gender can sit in adjacent seats.

Note: Clockwise and anti-clockwise sitting are treated the same. Rotations as well.

This problem is a part of Tessellate S.T.E.M.S (2019)

( 4 ! ) 3 (4!)^3 3 ( ( 8 ! ) 2 ) 3((8!)^2) ( 3 ( 4 ! ) 3 ) (3(4!)^3) ( 8 ! ) 2 (8!)^2

Tessellate Stems Computer Science by Tessellate STEMS Compute… , 27

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1 solution

Mark Hennings
Nov 23, 2018

The adults must sit in a block of 8 8 , and then the children sit in a block of eight. There are 16 16 choices of where the first (counting clockwise around the table) adult can sit.

If the first adult is a man, the seating arrangements, starting with that man, are MWMWMWMWBGBGBGBG. There are 4 ! 4! ways of arranging the four men, 4 ! 4! ways of arranging the four boys and 4 ! 4! ways of arranging the four girls. For any arrangement of the four men, there are 3 3 possible arrangements of the wives: M 1 W 3 M 2 W 4 M 3 W 1 M 4 W 2 M 1 W 3 M 2 W 4 M 3 W 2 M 4 W 1 M 1 W 4 M 2 W 1 M 3 W 2 M 4 W 3 M_1W_3M_2W_4M_3W_1M_4W_2 \hspace{0.5cm} M_1W_3M_2W_4M_3W_2M_4W_1 \hspace{0.5cm} M_1W_4M_2W_1M_3W_2M_4W_3 If the first adult is a woman, we have exactly the same arguments with M and W swapped, and B and G swapped.

Thus the total number of seating arrangements is 16 × 2 × 4 ! × 4 ! × 4 ! × 3 = 32 × 3 × ( 4 ! ) 3 16 \times 2 \times 4! \times 4! \times 4! \times 3 \; = \; 32 \times 3 \times (4!)^3

If you ignore reflections and rotations, we divide the above answer by 16 × 2 = 32 16 \times2 = 32 to get the answer 3 ( 4 ! ) 3 \boxed{3(4!)^3} .

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