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At a circular table for eight will sit six adults

Expert replies
by MartyMurray » Mon Apr 25, 2016 8:19 pm
At a circular table for eight will sit six adults and two small children. The two children will sit next to each other, and the two parents of the children will sit next to the children so that there is one parent on each side of the two children. If rotating their positions around the table is not considered changing the arrangement, in how many different ways can the eight people be arranged around the table.

(A) 24

(B) 96

(C) 1440

(D) 5040

(E) 40320
Last edited by MartyMurray on Tue Apr 26, 2016 6:14 am, edited 1 time in total.
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Source: — Problem Solving |

by chetan.sharma » Tue Apr 26, 2016 5:30 am
Marty Murray wrote:At a circular table for eight will sit six adults and two small children. The two children will sit next to each other, and the two parents of the children will sit next to the children so that there is one parent on each side of the two children. If rotating their positions around the table is not considered changing the arrangement, in how many different ways can the eight people be arranged around the table.

(A) 24

(B) 96

(C) 1440

(D) 5040

(E) 40320

I'll post the OA by Wednesday.
HI,
lets take two kids and their parents as one identity, so we have 5 people sitting at the table..
ways they can be arranged on a circular table = (5-1)! = 4! = 24..
the two kids can be arranged in 2! and parents in another 2! ways.
ans = 24*2!*2! = 96
B
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by GMATGuruNY » Tue Apr 26, 2016 5:50 am
At a circular table for eight will sit six adults and two small children. The two children will sit next to each other, and the two parents of the children will sit next to the children so that there is one parent on each side of the two children. If rotating their positions around the table is not considered changing the arrangement, in how many different ways can the eight people be arranged around the table?

(A) 24

(B) 96

(C) 1440

(D) 5040

(E) 40320
For circular arrangements, count the number of ways to arrange the REMAINING people after one person has been seated.

After one of the 2 children has been seated:
Number of options for the second child = 2. (To the left or right of the first child.)
Number of options for the first parent = 2. (To the left or right of the two children.)
Number of options for the second parent = 1. (The remaining seat next to the two children.)
Number of ways to arrange the remaining 4 people = 4!.
To combine these options, we multiply:
2*2*1*4! = 96.

The correct answer is B.
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by Matt@VeritasPrep » Wed Apr 27, 2016 2:22 pm
I'd treat the kids as one person, fix their seat, and assign everyone else around them.

In their one seat, the kids can have two arrangements (older kid left, younger kid right, or vice versa).

On either side of them, the parents can have two arrangements (Mom left, Dad right, or vice versa).

The other four adults can be arranged in 4! ways.

The product of all the arrangements is thus 2 * 2 * 4!, or 96.
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