A hostess must seat 10 people around 2 circular tables. Table 1 can ho

Question

A hostess must seat 10 people around 2 circular tables. Table 1 can hold 6 people, table 2 can hold 4 people. If the hostess always seats people at Table 1 first and 2 seating arrangements are said to be different only when the positions of people are different relative to each other at the table, how many seating arrangements are possible?

A. \(5*210\)

B. \(15!*210\)

C. \(120*3!\)

D. \(5!*3!*210\)

E. \(5!*5!*3!\)

Are You Up For the Challenge: 700 Level Questions

A.  5*210

B.  15!*210

C.  120*3!

D.  5!*3!*210

E.  5!*5!*3!

Are You Up For the Challenge: 700 Level Questions

CrackverbalGMAT · Accepted Answer

The concepts being tested here are those of Circular Permutations and those of distribution into groups.

The number of circular permutations of n objects is given by (n-1)! The number of ways in which 6 people can sit on the first table will be 5! Similarly, the number of ways in which 4 people can sit on the second table will be 3!
The answer should have 5! * 3!. Based on this, options A, B and C can be eliminated. The possible answers at this stage are D or E.

The number of ways in which (m+n) objects can be distributed into two groups in such a way that one of the groups contains m objects and the other contains n objects is given by  (m+n)_C_m  ways.

In our question, m = 6 and n = 4. Therefore, the number of ways of distributing 10 people into two groups in such a way that 6 people go to one table and 4 go to the other =  10_C_6  ways = 210 ways. 
For each of these 210 selections, the number of permutations on the first and the second tables will be 5!*3!. 
Therefore, the total number of ways in which the 10 persons can be seated = 210 * 5! * 3!.

The correct answer option is D.

Hope that helps!

Archit3110 · Answer

first people are seated on table 1 = 5! ways
and then for table 2 we have 10c4 *3! ways
IMO D 
  5!*3!*210

allkagupta · Answer

How did you get 210 ?

AjitPatil · Answer

lets break this problem 10 people , 2 circular chairs and , capacity of chairs C1--> 6 , C2--> 4
Condition C1 should be filled first.

first person has 5 ways on chair 1 
second person has 5 ways on chair 1
3rd has 4 ways 
4th has 3 ways
5th has 2 ways 
6th has 1 way
therefore 5 *5 ! ways now fill chair 1 and chair 2 , implying chair 2 has
4 seats , 4 people remaining, therefore 
1 person has 3 ways 
2 nd has 3 ways , 3rd has 2 ways to seat, 4th has 1 ways--> 3*3! ways 
chair 1 and chair 2
(5*5!) * (3*3!)= 5!*3!*15 ways 
correct me if am missing something??because we do not have that answer choice

Archit3110 · Answer

allkagupta

circular arrangements can be arranged in ( n-1)! ways
so in this case total people = 10
and for 1st table which can hold 6 people we can arrange in 5! ways
now for 2nd table we are left with 4 people who can be arranged in 10c4 ways or say 210 
and total ways to make sit 4 people in circular table = 3! 
answer ; 
5!*3!*210

hope this helps

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A hostess must seat 10 people around 2 circular tables. Table 1 can ho

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A hostess must seat 10 people around 2 circular tables. Table 1 can ho

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards