At a circular table for eight will sit two children, their two parents

Question

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

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

Source: Self Made

Resad95 · Accepted Answer

Define Children as C , Parents of Children as M , and Parents of Parents as P . We know that children must sit next to each other so it will be CC Next we know that each parent must sit next to child, one on either side of the two children, it becomes M CC M Next we add parents of each parent, that must sit next to each other and arrangement becomes PP M CC M PP . From this order we know that we can change order of Eldest parents among themselves PP (parent of parent) which must be next to its child M (parent of child), can be arranged in 2 ways and since there are 2 pairs of eldest parents ( PP MccM PP ) we multiply again by 2. (2x2 till here) We can change the order of eldest with their child(parents of child) PPM cc MPP 2 times ( Parent of child can sit next to each child) (here it becomes 2x2x2) Finally we can change arrangement of children between themselves cc 2 times So it becomes 2x2x2x2=16 Appreciate Kudos if you liked it, put a great effort

Nevernevergiveup · Answer

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.

MartyTargetTestPrep · Answer

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  B .

Nevernevergiveup · Answer

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

law258 · Answer

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.

bumpbot · Answer

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At a circular table for eight will sit two children, their two parents

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