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Worked solution

Count adjacency in the row and circle as disjoint cases

There are 9 people including John, Mary and Gary. They are to arrange themselves into a row of 4 people and a circle of 5 people. Find the number of ways they can arrange themselves if

(i) there are no restrictions. marks

(ii) John and Mary are beside each other. marks

(iii) John, Mary and Gary are part of the group that forms a circle, and John and Mary are separated from each other. marks

[2011/CJC/II/6(a)]

CJC · Review problems · Question 6

Answer

(i) 72576; (ii) 16128; (iii) 4320.

(i) Choose the row first

The row contributes (94) 4!=3024\binom94\,4!=3024. The five remaining people form a circle in (55)(5−1)!=24\binom55(5-1)!=24 ways. Total 3024×24=725763024\times24=72576.

How did this part go?

(i) The source’s alternative: choose the circle first

Choose and arrange the circular five: (95)(5−1)!=3024\binom95(5-1)!=3024. Arrange the remaining row in 4!=244!=24 ways. The same product gives 72576.

How did this part go?

(ii) John and Mary adjacent in the row

Choose their two row companions from the other seven, then arrange the JM block and these two companions in 3!3! ways, with 2!2! internal orders. Arrange the remaining circle in (55)(5−1)!\binom55(5-1)! ways. This case gives (72) 3! 2! (5−1)!=6048\binom72\,3!\,2!\,(5-1)!=6048.

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(ii) John and Mary adjacent in the circle

Choose three circle companions from the other seven. The JM block and three companions form four circular objects, with (4−1)!(4-1)! orders and 2!2! internal JM orders. Arrange the remaining row in 4!4! ways. This gives (73)(4−1)! 2! 4!=10080\binom73(4-1)!\,2!\,4!=10080. The two disjoint placements total 6048+10080=161286048+10080=16128.

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(iii) All three named people in the circle, with John and Mary separated

Choose two circle companions from the six others. Arrange Gary and those two companions around the circle in (3−1)!(3-1)! ways; choose two of their three gaps and arrange John/Mary in 2!2! ways. The circular count is (62)(3−1)!(32) 2!=180\binom62(3-1)!\binom32\,2!=180. The remaining row has 4!=244!=24 orders, giving 180×24=4320180\times24=4320.

How did this part go?

Common mistakes

Do not count John and Mary as adjacent when they occupy different groups. Circular gaps differ from linear end gaps.

Solution notes

Every source method and factor retained, including the alternative choice order in(i) and both adjacency cases in(ii). Printed marks retained.

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