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

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

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12 Nov 2019, 02:30
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75% (hard)

Question Stats:

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

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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. $$5!*5!*3!$$

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

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

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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
Joined: 04 Jun 2019
Posts: 1
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

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

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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!
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
Joined: 12 Apr 2019
Posts: 365
Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho  [#permalink]

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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!
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Re: A hostess must seat 10 people around 2 circular tables. Table 1 can ho   [#permalink] 16 Jan 2020, 06:07
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