+0  
 
0
13
2
avatar+82 

Three adults and three children are to be seated at a circular table. In how many different ways can they be seated if each child must be next to two adults? (Two seatings are considered the same if one can be rotated to form the other.)

 Oct 4, 2023
 #1
avatar+617 
-1

To solve this problem, you can use the concept of permutations and the principle of symmetry in a circular arrangement.

Let's first consider the adults as A1, A2, and A3, and the children as C1, C2, and C3. You want each child to be seated between two adults, which means the arrangement should be in the order: A-C-A-C-A-C.

Now, we can treat this as a linear arrangement. There are 6 seats in a circle, so if we arrange A1, A2, A3, C1, C2, and C3 in a line, we have 6! (6 factorial) ways to arrange them. However, we need to account for the circular arrangement, so we must divide by 6 to eliminate the rotations.

So, the total number of different circular seatings where each child is next to two adults is:

6! / 6 = 720 / 6 = 120

So, there are 120 different ways to seat three adults and three children at a circular table, where each child is seated between two adults.

 Oct 4, 2023
 #2
avatar+118604 
0

Put one adult anywhere.

then there are 2 posibilities for how to sit the other 2 adults 

Tthen 3! = 6 ways to seat the chiildren.

 

So that is  2*6=12 ways

 Oct 6, 2023

1 Online Users