Question

If n is even number and there are $n$ neighbours, then the number of ways of arranging $m$ persons if each person has exactly one neighbour is \[\]
A. $\left( ^{n}{{P}_{\dfrac{n}{2}}} \right)\left( ^{m+n-2}{{P}_{\dfrac{n}{2}}} \right)$\[\]
B. $\left( ^{n}{{P}_{n}} \right)\left( ^{m+n-2}{{P}_{\dfrac{n}{2}}} \right)$\[\]
C. $\left( ^{n}{{P}_{\dfrac{n}{2}}} \right)\left( ^{m+n-2}{{P}_{n}} \right)$\[\]
D. None of these \[\]

Answer 1

Hint: We assume $n=2k,k\in Z$ and the total number of non-neighbours is $n-2k$. We assume the number of non-neighbours placed left of first pair of neighbour ${{A}_{1}}=\left( {{N}_{1}},{{N}_{2}} \right)$ as ${{x}_{o}}$ and placed right of last pair of neighbour ${{A}_{k}}$ as ${{x}_{k}}$. We assume the number of non-neighbours placed in between any two pairs ${{A}_{i}},{{A}_{i+1}}$ as ${{x}_{i}},i=1,2,3,..k-1$. The answer is the product of number of positive integral positions of the equation ${{x}_{o}}+{{x}_{1}}+...+{{x}_{k}}=m-2k$ and number self-arrangements among the neighbour themselves.

Complete step-by-step solution:
We know from the combination that the number of positive integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+...+{{x}_{r}}=a$ is $^{a-1}{{C}_{r-1}}$.\[\]
We are given that there is a $n$ number of neighbors and there is a total $m$ number of persons and we have to place exactly 1 neighbor in the seat arrangement. We are also given that the number $n$ is even. So let us assume $n=2k$ where $k$ any positive integer is.
We can place two neighbors ${{N}_{1}},{{N}_{2}}$ together close to each other. We denote the pair of neighbors ${{N}_{1}},{{N}_{2}}$ as ${{A}_{1}}$. So there will be a total $\dfrac{n}{2}=k$ number of pairs like ${{A}_{1}}$ and we denote them as ${{A}_{1}},{{A}_{2}},{{A}_{2}},...{{A}_{k}}$ and also we observe that in between $k$ pairs of neighbor there will be $\left( k-1 \right)$ gaps with any number of seats between them.\[\]
Let us assume number of seats between the pairs of neighbour ${{A}_{i}},{{A}_{i+1}}$ be ${{x}_{i}}$ where $i=1,2,3,..k$ . We also assume that fill the seats number of seats on left the pair ${{A}_{1}}$ as ${{x}_{0}}$ and the number of seats on the right of the pair ${{A}_{k}}$ as ${{x}_{k}}$. We can fill all these gaps ${{x}_{o}},{{x}_{1}},{{x}_{2}},...,{{x}_{k-1}},{{x}_{k}}$ with a person who is not a neighbour for which we have total $m-n=m-2k$ number of persons to choose from. So we have
\[{{x}_{o}}+{{x}_{1}}+{{x}_{2}}+...+{{x}_{k-1}}+{{x}_{k}}=m-2k\]
We see that the solutions for ${{x}_{o}},{{x}_{1}},{{x}_{2}},...,{{x}_{k-1}},{{x}_{k}}$ can only be a positive integers as the solutions are the number of persons. We use the combination formula with $a=m-2k,r=k+1$ and find the number of solutions of the above linear equation as
\[^{m-2k-1}{{C}_{k+1-1}}{{=}^{m-2k-1}}{{C}_{k}}\]
We can also arrange the number of neighbours within themselves in $n!=\left( 2k \right)!$ ways. So we use rule of product and find the total number of arrangements for $n$ persons as,
\[\begin{align}
& ^{m-2k-1}{{C}_{k}}\left( 2k! \right)=\dfrac{\left( m-2k-1 \right)!}{k!\left( m-2k-1-k \right)!}\left( 2k! \right)=\dfrac{\left( 2k! \right)}{k!}\times \dfrac{\left( m-2k-1 \right)!}{\left( m-2k-1-k \right)!} \\
\end{align}\]
We use the permutation formula $^{a}{{P}_{r}}=\dfrac{a!}{\left( a-r \right)!}$ and replace $2k=n$ to have ,
\[\dfrac{\left( 2k! \right)}{k!}\times \dfrac{\left( m-2k-1 \right)!}{\left( m-2k-1-k \right)!}=\left( ^{2k}{{P}_{k}} \right)\left( ^{m-2k-1}{{P}_{k}} \right)=\left( ^{n}{{P}_{\dfrac{n}{2}}} \right)\left( ^{m-n-1}{{P}_{\dfrac{n}{2}}} \right)\]
So the correct option is A.

Note: The formula is we used here that the number of integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+...+{{x}_{r}}=a$ is $^{a-1}{{C}_{r-1}}$. If none of them are equal values then the number of positive integral solutions is $^{a-1}{{P}_{r-1}}$. If all of the variables would have been non-negative the number of non-negative integral solutions is $^{a+r-1}{{C}_{r-1}}$.