Question

A person invites a party of 10 friends at dinner and placed of them are
(i) 5 at one round table, 5 at the other round table.
(ii) 4 at one round table, 6 at the other round table.
Then the ratio of number of circular permutations of case (i) and case (ii) is
(A) \[\dfrac{24}{13}\]
(B) \[\dfrac{25}{24}\]
(C) \[\dfrac{24}{25}\]
(D) \[\dfrac{13}{24}\]

Answer 1

Hint: In the \[{{1}^{st}}\] case, it is given that 5 persons are at one round table and the remaining 5 persons are at the other round table. Calculate the number of ways to select 5 persons for the first table by using the formula, \[^{n}{{C}_{m}}\] . Now, calculate the number of possible circular rearrangements of 5 persons sitting at the first table using the formula \[\left( n-1 \right)!\] . Similarly, calculate the number of possible circular rearrangements of the remaining 5 persons sitting at the second table. Now, get the
total number of circular permutations in the \[{{1}^{st}}\] case. In the \[{{2}^{nd}}\] case, it is given that 6 persons are at one round table and the remaining 4 persons are at the other round table. Calculate the number of ways to select 6 persons for the first table by using the formula, \[^{n}{{C}_{m}}\] . Now, calculate the number of possible circular rearrangements of 6 persons sitting at the first table using the formula \[\left( n-1 \right)!\] . Similarly, calculate the number of possible circular rearrangements of the remaining 4 persons sitting at the second table. Now, get the total number of circular permutations in the \[{{2}^{nd}}\] case. Finally, calculate the ratio of the circular permutations in the \[{{1}^{st}}\] case and \[{{2}^{nd}}\] case.

Complete step-by-step answer:
According to the question, it is given that a person invites a party of 10 friends at dinner. Here, we have two cases.
In the \[{{1}^{st}}\] case, it is given that
5 persons are at one round table and the remaining 5 persons are at the other round table.

We know the formula to select m items out of n items, \[^{n}{{C}_{m}}\] .
Here, we have to select 5 persons for the first table out of 10 persons.
The number of ways to select 5 persons for the first table = \[^{10}{{C}_{5}}\] ……………………………..(1)
We know that the possible number of circular rearrangements for n items is \[\left( n-1 \right)!\] .
The number of possible rearrangements of 5 persons sitting on the round table = \[\left( 5-1 \right)!=4!\] ………………………………..(2)
Now, we have the remaining 5 persons for the second round table.
The number of possible rearrangements of the remaining 5 persons sitting on the round table = 4! ………………………………..(3)
The total number of circular permutations in the \[{{1}^{st}}\] case = \[^{10}{{C}_{5}}\times 4!\times 4!\] …………………………(4)
In the \[{{2}^{nd}}\] case, it is given that
6 persons are at one round table and the remaining 4 persons are at the other round table.

We know the formula to select m items out of n items, \[^{n}{{C}_{m}}\] .
Here, we have to select 6 persons for the first table out of 10 persons.
The number of ways to select 6 persons for the first table = \[^{10}{{C}_{6}}\] ……………………………..(5)
We know that the possible number of circular rearrangements for n items is \[\left( n-1 \right)!\] .
The number of possible rearrangements of 6 persons sitting on the round table = \[\left( 6-1 \right)!=5!\] ………………………………..(6)
Now, we have the remaining 4 people for the second round table.
The number of possible rearrangements of the remaining 4 persons sitting on the round table = \[\left( 4-1 \right)!=3!\] ………………………………..(7)
The total number of circular permutations in the \[{{2}^{nd}}\] case = \[^{10}{{C}_{6}}\times 5!\times 3!\] …………………………(8)
From equation(4) and equation (8), we have the total number of permutations for \[{{1}^{st}}\] case and \[{{2}^{nd}}\] case.
The ratio of the circular permutation in \[{{1}^{st}}\] case and \[{{2}^{nd}}\] case = \[\dfrac{^{10}{{C}_{5}}4!\times 4!}{^{10}{{C}_{6}}5!\times 3!}\] = \[\dfrac{\dfrac{10\times 9\times 8\times 7\times 6}{1\times 2\times 3\times 4\times 5}\times 4!\times 3!\times 4}{\dfrac{10\times 9\times 8\times 7\times 6\times 5}{1\times 2\times 3\times 4\times 5\times 6}\times 5\times 4!\times 3!}=\dfrac{6\times 4}{5\times 5}=\dfrac{24}{25}\] .
Hence, the correct option is (C).

Note: In this question, one might calculate the number of possible rearrangements using the formula \[n!\] . We cannot apply this formula here because here the persons are sitting on the round table. So, the number of possible rearrangements can be calculated using the formula for circular rearrangements, \[\left( n-1 \right)!\] where n is the total number of elements for the rearrangements.