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

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

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

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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