Worked solution
Keep a grouped IP pair and separately place the JC2 members
An Art Club has 10 members:2 IP1,2 IP2,3 JC1 and 3 JC2. They take a group photo. Find the number of arrangements with all members
(i) in a row with the 2 IP1 members together and no JC2 members together, marks
(ii) in two equal rows, one in front of the other, with the 4 IP members at the four corners. marks
Afterwards all members sit at a round table. Find the probability that the 2 IP1 members are seated together. marks
[2011/VJC/II/9]
CJC · Review problems · Question 10
Answer
(i) 302400; (ii) 17280; circular probability .
(i) Group IP1 and form the base arrangement
Group the two IP1 members as one object, leaving nine objects in total. Temporarily remove the three JC2 members. The remaining six objects are the IP1 block, two IP2 people and three JC1 people. Arrange them in ways and the IP1 pair internally in ways.
(i) Insert JC2 members into different gaps
The six base objects have seven gaps, including both ends. Choose three different gaps and order the distinct JC2 members in ways. Total . This interprets “no JC2 members together” as pairwise separation, as in the printed gap construction.
(ii) Fix all four corners
Two rows of five have four fixed corner positions. Arrange the four distinct IP members there in ways, then the six remaining members in the other positions in ways. Total .
Circular seating probability
With the IP1 pair grouped, nine circular objects have arrangements. Without restrictions, ten distinct people have circular arrangements. The probability is . The photo restrictions do not carry over to this separate seating question.
Common mistakes
The JC2 gap construction needs different gaps; simply excluding one triple block would not prevent adjacent pairs.
Solution notes
All three printed allocations and complete source factor interpretations retained, including the unnumbered final circular probability ask.