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Nov 2023 p51 q6
2826
In a restaurant, the tables are rectangular. Each table seats four people: two along each of the longer sides of the table (see diagram). Eight friends have booked two tables, X and Y. Rajid, Sue, and Tan are three of these friends.
The eight friends will be divided into two groups of 4, one group for table X and one group for table Y.
Find the number of ways in which this can be done if Rajid and Sue must sit at the same table as each other and Tan must sit at the other table.
Solution
First, we need to choose a table for Rajid and Sue. There are 2 choices: either they sit at table X or table Y.
Once Rajid and Sue are seated at a table, Tan must sit at the other table. This leaves 5 friends to be divided into two groups of 3 and 2 to fill the remaining seats at each table.
We choose 2 additional friends to sit with Rajid and Sue. The number of ways to choose 2 friends from the remaining 5 is given by the combination:
\(\binom{5}{2} = 10\)
Since Rajid and Sue can sit at either table, we multiply by 2:
\(10 \times 2 = 20\)
Thus, there are 20 ways to arrange the seating under the given conditions.