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A hostess must seat 10 people around 2 circular tables.

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A hostess must seat 10 people around 2 circular tables. [#permalink]

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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!
[Reveal] Spoiler: OA
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Re: A hostess must seat 10 people around 2 circular tables. [#permalink]

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debbiem 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!


Hi debbiem,

solve by taking one step at a time..
1)since there are two tables, the hostess has to choose 6 out of 10 and the remaining 4 will go to other table..
so WAYS= 10C6=10!/6!4!=210..
2)now 6 on the Table 1 can be seated in (6-1)! ways
3)and 4 on the Table 1 can be seated in (4-1)! ways

TOTAL ways=210*5!*3!
D

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Re: A hostess must seat 10 people around 2 circular tables. [#permalink]

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Re: A hostess must seat 10 people around 2 circular tables. [#permalink]

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New post 12 May 2017, 07:40
chetan2u wrote:
debbiem 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!


Hi debbiem,

solve by taking one step at a time..
1)since there are two tables, the hostess has to choose 6 out of 10 and the remaining 4 will go to other table..
so WAYS= 10C6=10!/6!4!=210..
2)now 6 on the Table 1 can be seated in (6-1)! ways
3)and 4 on the Table 1 can be seated in (4-1)! ways

TOTAL ways=210*5!*3!
D


Can someone explain steps 2 and 3 please?
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Re: A hostess must seat 10 people around 2 circular tables. [#permalink]

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New post 12 May 2017, 09:23
vmelgargalan wrote:
chetan2u wrote:
debbiem 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!


Hi debbiem,

solve by taking one step at a time..
1)since there are two tables, the hostess has to choose 6 out of 10 and the remaining 4 will go to other table..
so WAYS= 10C6=10!/6!4!=210..
2)now 6 on the Table 1 can be seated in (6-1)! ways
3)and 4 on the Table 1 can be seated in (4-1)! ways

TOTAL ways=210*5!*3!
D


Can someone explain steps 2 and 3 please?



Because in a circular arrangement (using table 1 as an example) the seating arraignments for 123456 and 234561 are considered to be the same compared to a linear arraignment. Subtracting 1 removes the "double dipping".
Re: A hostess must seat 10 people around 2 circular tables.   [#permalink] 12 May 2017, 09:23
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A hostess must seat 10 people around 2 circular tables.

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