2018/19 AHSMC Part 1 Problem 13

Problem 13 of 16

There are seven children belonging to three families. The first family has three children, and the other two families have two children each. In how many ways can these seven children be arranged in a circle so that no two children from the same family are next to each other? Notice that seating arrangements that differ by a rotation are considered the same.

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Solution

Let $A_1$ , $A_2$ , $A_3$ be the children of the first family, $B_1$ , $B_2$ of the second family and $C_1$ , $C_2$ of the third family. Arrange $A_1$ , $A_2$ , $A_3$ in a circle with two unoccupied spots between $A_1$ and $A_2$ , one unoccupied spot between $A_2$ and $A_3$ and one unoccupied spot between $A_3$ and $A_1$ .

The two free spots between $A_1$ and $A_2$ , in the direction from $A_1$ toward $A_2$ , can be occupied by the components of each of the following ordered pairs: $(B_1,C_1),\ (B_1,C_2),\ (B_2,C_1),\ (B_2,C_2),\ (C_1,B_1),\ (C_1,B_2),\ (C_2,B_1),\ (C_2,B_2).$ For each pair that occupies the first two spots there are two possibilities to fill in the last two unoccupied spots. Therefore there are $16$ possibilities to fill in the unoccupied spots. Since there are $3! = 6$ ways to place the three-children family, it comes out that the children can be arranged in $16 \times 6 = 96$ ways. The answer is (D).

University of Alberta

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