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At a circular table for eight will sit two children, their two parents [#permalink]
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Notice, the table is circular, but since every person to be seated has to be seated in relation to another person, and since the arrangement works from the center, where the two children are, to the outside, where the two sets of grandparents sit, the answer is the same as it would be were we arranging a row of eight people.

You can start this one in multiple ways. Maybe the best way is to start with the children.

The children can be arranged in 2 ways, C₁C₂ or C₂C₁.

The parents of the children have to sit next to the children and can also be arranged in 2 ways, P₁CCP₂ or P₂CCP₁.

The somewhat tricky part is understanding the implications of the fact that the pairs of grandparents are always at the sides of their respective children.

For instance, the grandparents that go with P₁ will always be next to P₁, whether P₁ is to the left or to the right of the children.

We could make G₁ and G₂ the grandparents that go with P₁ and G₃ and G₄ the grandparents that go with P₂.

One possible arrangement is the following.

G₁G₂P₁C₁C₂P₂G₃G₄

If P₁ and P₂ were to swap sides, their respective sets of grandparents would go with them.

G₃G₄P₂C₁C₂P₁G₁G₂

So the positions of the pairs of grandparents are dictated by the positions of the parents that the grandparents sit beside.

Therefore the only way that the grandparents themselves can be arranged in different ways is by reversing them within their own pairs, for instance, the grandparents with P₁ could be arranged G₁G₂P₁ or G₂G₁P₁.

So the parents and children can be arranged in 2 x 2 = 4 ways, and then for each of those four ways each pair of grandparents can be arranged in two different ways, giving us 2 x 2 = 4 ways to arrange the grandparents.

So the total number of ways in which they all can be arranged is 2 x 2 x 2 x 2 = 16 ways.

The correct answer is .

Originally posted by MartyTargetTestPrep on 25 Nov 2016, 18:44.
Last edited by MartyTargetTestPrep on 27 Nov 2016, 22:47, edited 2 times in total.
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Re: At a circular table for eight will sit two children, their two parents [#permalink]
Nevernevergiveup wrote:
MartyMurray wrote:
At a circular table for eight will sit two children, their two parents, and the two parents of each of the children's parents, eight people in total. The children will sit together, the children's two parents will sit next to the children, one on either side of the two children, and the parents of each of the children's parents will sit next to each other beside the person who is their child. Given the restrictions, in how many ways can the people be arranged around the table.

(A) 8
(B) 16
(C) 64
(D) 96
(E) 1440


Source: Self Made


I believe we should place children first and we can do it in \(7C1*2!\) ways

and rest looks same for remaining arrangement of parents i.e., \(2*2*2\) ways.

MartyMurray please explain. :?


I got my mistake. We can select first child in any location and second child is either left or right(2 places).

and we can do it in \(1*2\)=2 ways
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Re: At a circular table for eight will sit two children, their two parents [#permalink]
PP1 PP2 Pa C1 C2 Pb PP1 PP2
PP2 PP1 ---------- PP2 PP1
PP2 PP1 ---------- PP1 PP2
PP1 PP2 ---------- PP2 PP1

-4 combinations moving grandparents around

PP1 PP2 Pb C1 C2 Pa PP1 PP2
PP2 PP1 ---------- PP2 PP1
PP2 PP1 ---------- PP1 PP2
PP1 PP2 ---------- PP2 PP1

-4 combinations from switching parents-

PP1 PP2 Pa C2 C1 Pb PP1 PP2
PP2 PP1 ---------- PP2 PP1
PP2 PP1 ---------- PP1 PP2
PP1 PP2 ---------- PP2 PP1

-4 combinations switching children-

PP1 PP2 Pb C2 C1 Pa PP1 PP2
PP2 PP1 ---------- PP2 PP1
PP2 PP1 ---------- PP1 PP2
PP1 PP2 ---------- PP2 PP1

-4 combinations switching children and parents-

16 total - B.
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Re: At a circular table for eight will sit two children, their two parents [#permalink]
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Re: At a circular table for eight will sit two children, their two parents [#permalink]
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