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Question: A gentleman invites a party of \[m+n\left( m\ne n \right)\] friends to a dinner and places m at one ...

A gentleman invites a party of m+n(m≠n)m+n\left( m\ne n \right) friends to a dinner and places m at one table T1{{T}_{1}} and n at another table T2{{T}_{2}}, the table being round. If not all people shall have the same neighbour in any two arrangement, then the number of ways in which he can arrange the guests, is
(a) (m+n)!4mn\dfrac{\left( m+n \right)!}{4mn}
(b) 12(m+n)!mn\dfrac{1}{2}\dfrac{\left( m+n \right)!}{mn}
(c) 2(m+n)!mn2\dfrac{\left( m+n \right)!}{mn}
(d) none

Explanation

Solution

In this question, we need to first need select m people out of m+nm+n and the rest will be going to the next table which can be done using the formula given by nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}. Then we need to arrange the m people in a round table and then n people which can be done using the formula (k−1)!\left( k-1 \right)!. Here, as mentioned that they don't have the same neighbour in any arrangement we need to consider the clockwise and anti clockwise as same which is given by (k−1)!2\dfrac{\left( k-1 \right)!}{2} and simplify further to get the result.

Complete step by step answer:
COMBINATIONS:
Each of the different groups or selections which can be made by some or all of a number of given things without reference to the order of the things in each group is called a combination.
The number of combinations of n different things taken r at a time is
nCr=n!(n−r)!r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}, 0≤r≤n0\le r\le n
Number of circular permutations of k different things taken all at a time, when clockwise or anti clockwise order is not different is given by the formula
(k−1)!2\dfrac{\left( k-1 \right)!}{2}
Now, given that there are m+nm+n people in which m people should be placed in the first table and the remaining people should be placed in the second table
Here, we first need to select m people out of m+nm+n people so that remaining n people go to the next one which can be done by using the formula
⇒nCr\Rightarrow {}^{n}{{C}_{r}}
Now, on substituting the respective values in the formula we get,
⇒m+nCm\Rightarrow {}^{m+n}{{C}_{m}}
Now, the remaining n people can be selected in only a single way as they go to the next table.
Now, we need to arrange the selected m people in the first table and the n people in the second table.
Here, as mentioned that all people shall not have the same neighbour in any two arrangement so we considered clockwise and anti clockwise as same
Now, the m people can be arranged in a round table in
⇒(m−1)!2\Rightarrow \dfrac{\left( m-1 \right)!}{2}
Now, the n people can be arranged in a round table in
⇒(n−1)!2\Rightarrow \dfrac{\left( n-1 \right)!}{2}
Now, the total number of ways in which guests can be arranged is given by
⇒m+nCm×(m−1)!2×(n−1)!2\Rightarrow {}^{m+n}{{C}_{m}}\times \dfrac{\left( m-1 \right)!}{2}\times \dfrac{\left( n-1 \right)!}{2}
Now, this can be further written as
⇒(m+n)!m!n!×(m−1)!2×(n−1)!2\Rightarrow \dfrac{\left( m+n \right)!}{m!n!}\times \dfrac{\left( m-1 \right)!}{2}\times \dfrac{\left( n-1 \right)!}{2}
Now, on cancelling out the common terms in the numerator and denominator we get,
⇒(m+n)!4mn\Rightarrow \dfrac{\left( m+n \right)!}{4mn}
Hence, the correct option is (a).

Note:
Instead of first selecting m people out of m+nm+npeople we can also solve it by first selecting n people after which m people can be automatically sent to the next table and then arrange them in a round table accordingly. Both the methods give the same result.
It is important to note that we need to consider that the clockwise and anti clockwise arrangement are the same as already mentioned that all people shall not have the same neighbour in any two arrangements. If we consider them as different then we get incorrect results which may be given in the options.