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12 Nov 2019, 02:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:04) correct
32%
(02:01)
wrong
based on 329
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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. \(5!*5!*3!\)
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
16 Jan 2020, 06:07
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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!
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!
General Discussion
12 Nov 2019, 03:03
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first people are seated on table 1 = 5! ways
and then for table 2 we have 10c4 *3! ways
IMO D
\(5!*3!*210\)
and then for table 2 we have 10c4 *3! ways
IMO D
\(5!*3!*210\)
Bunuel
