10 people are to be seated around two identical 5-person tables. In ho

Question

10 people are to be seated around two identical 5-person tables. In how many ways this can be done?

A. 10!/5^2
B. 10!/5!
C. 10!/(5!5!)
D. 10!/(5!4!)
E. 10!/(4!4!)

Mona2019 · Answer

5 people can sit around either table.
Ways to select people for the two tables = 10C5*5C5
These people can be arranged around the each of the table in (n-1)! =4!*4! Ways

So total possible arrangements = 10C5*4!*4!

IMO A is the answer

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CrackverbalGMAT · Answer

Number of ways of arranging n people in a circle = (n - 1)!

Since both tables are identical, we select 5 people out of 10 first in 10C5 ways.

These 5 can be arranged in (5 - 1)! = 4! ways.

The remaining 5 people go to the other table and can be arranged in 4! ways.

Total number of ways = 10C5 * 4! * 4! =  \frac{

Option A

Arun Kumar

Stanindaw · Answer

Please Help!
The formula i remember for forming two groups of five members out of 10 people will be = 10C2x5C5/2!

Stormik1 · Answer

Hello  Bunuel ,
The given answer 10!/25 is correct only if the table are non identical.
Since the tables are identical, the correct answer should be 10!/50 (correct answer not given in options)

egmat · Answer

Hi Stormik1,

Your divide-by- 2  logic is actually spot on  for the case you're imagining . If the two tables were truly indistinguishable - same object, no way to tell one from the other - then "group X here, group Y there" and "group Y here, group X there" would be the  same  physical outcome, and you'd divide by  2  to kill that double count. So your reasoning about the mechanics is correct.

Where the intended reading differs:  the word "identical" here is describing the  type  of table - both are  5 -seaters, each holding exactly five people - not saying the two tables are the same object.

Two real tables still sit in two different spots in the room. So even though they look alike, they are  distinguishable by position : putting a group at the left table is a genuinely different seating from putting that same group at the right table. Because they can be told apart by location, there's  nothing to divide out , and the count stays:

-  10 C 5  x  4 ! x  4 ! =  10!/52  - choice  A

That's the assumption the answer choices are built on - notice the options don't even offer a version with an extra / 2 , which is the setter's signal that the tables are being treated as two separate places.

A quick way to feel the difference:  picture seating  4  people at two  2 -person tables.

- If the tables are two spots in a room (distinguishable) - no divide by  2 .
- If you're instead splitting them into two  unlabeled  groups on paper (truly interchangeable) - then you divide by  2 .

Same people, same split - the only thing that changes the / 2  is whether the two "tables" can be told apart. Here they can, so  A  holds.

Answer: A

10 people are to be seated around two identical 5-person tables. In ho

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10 people are to be seated around two identical 5-person tables. In ho

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