GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 28 Jan 2020, 15:57

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 60727
A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 12 Nov 2019, 02:30
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

50% (02:06) correct 50% (01:33) wrong based on 34 sessions

HideShow timer Statistics

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

_________________
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5748
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 12 Nov 2019, 03:03
first people are seated on table 1 = 5! ways
and then for table 2 we have 10c4 *3! ways
IMO D
\(5!*3!*210\)

Bunuel wrote:
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
Intern
Intern
avatar
B
Joined: 16 Dec 2019
Posts: 16
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 11 Jan 2020, 11:40
Archit3110 wrote:
first people are seated on table 1 = 5! ways
and then for table 2 we have 10c4 *3! ways
IMO D
\(5!*3!*210\)

Bunuel wrote:
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



How did you get 210 ?
Intern
Intern
avatar
B
Joined: 04 Jun 2019
Posts: 1
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 14 Jan 2020, 00:48
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
GMAT Club Legend
GMAT Club Legend
User avatar
V
Joined: 18 Aug 2017
Posts: 5748
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 14 Jan 2020, 00:59
1
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


Quote:
Archit3110 wrote:
first people are seated on table 1 = 5! ways
and then for table 2 we have 10c4 *3! ways
IMO D
\(5!*3!*210\)

Bunuel wrote:
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



How did you get 210 ?
CrackVerbal Quant Expert
User avatar
G
Joined: 12 Apr 2019
Posts: 365
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

Show Tags

New post 16 Jan 2020, 06:07
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!
_________________
GMAT Club Bot
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho   [#permalink] 16 Jan 2020, 06:07
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne