An expert question on combinations for P5–P6, with a step-by-step solution, diagrams and the answer. Counting arrangements and choices appears in the Hua Xia Cup from the P1 and P2 first rounds to the P5 finals.
The question
At break, 3 children (Ming, May and Keung) pass a ball, starting with Ming.
Each time, whoever has the ball passes it to any one of the others (never to themself).
After pass 3 the ball is with May, and after 5 passes it is back with Ming. How many ways of passing are there?
Show the hints (3)
💡Hint 1: Split the passing in two: Ming to May (3 passes), then May back to Ming (2). How do the two legs' counts combine?
💡Hint 2: Each leg is too long to list. Keep two counts, "with Ming" and "with one particular other", pass by pass: new "with Ming" = 2 × old "with one other"; new "with one other" = old "with Ming" + 1 × old "with one other".
💡Hint 3: From May back to Ming has as many ways as from Ming to May (swap their names).
Worked solution
Have a go first, then read the solution step by step.
1.Let A = the ways the ball is with Ming, B = with one particular other child. One more pass: new A = 2 × B (from one of the 2 others); new B = A + 1 × B (from Ming or one of the other 1).
2.From A = 1, B = 0: after 1: A = 0, B = 1; after 2: A = 2, B = 1; after 3: A = 2, B = 3.
3.First leg: Ming to May in 3 passes, 3 ways. Second leg: May back to Ming in 2; swapping names, it is like Ming to one particular child in 2: 1 way.
4.Every first leg goes with every second leg: 3 × 1 = 3.
⚠️Common mistake: adding the two legs (3 + 1); or using "with Ming after 2", 2, for the second leg (that starts from Ming).
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