Question

Let ${{P}_{n}}$ denotes the number of ways in which three people can be selected out of ‘n’ people sitting in a row, if no two of them are consecutive. If ${{P}_{n+1}}-{{P}_{n}}=15$ then the value of ‘n’ is?

Answer 1

Hint: We will first start by depicting the situation by using a rough sketch. Then we will divide the total people in four parts in which three people act as points of division. Then we will let the people in between two people as x variables and form an equation as the given data. Finally we will use the method of counting non – negative integral solutions to solve the problem.

Complete step-by-step answer:
Now, we have been given that ${{P}_{n}}$ is the way of selecting 3 people out of n peoples in which the three people are non – consecutive. So, we can depict one of such way as,
$\left. \underset{---}{\mathop{{{x}_{1}}}}\, \right|\left. \underset{---}{\mathop{{{x}_{2}}}}\, \right|\left. \underset{---}{\mathop{{{x}_{3}}}}\, \right|\underset{---}{\mathop{{{x}_{4}}}}\,$
Where $\left| {} \right.$ denote the people selected. So, in total we have four partition and we let the people in each partition as ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$. Now, since the three people are non-consecutive. So, we have,
${{x}_{2}}\ge 1\ and\ {{x}_{2}}\ge 3$. So, we let,
$\begin{align}
  & {{x}_{2}}=x_{2}^{1}+1......\left( 1 \right) \\
 & {{x}_{3}}=x_{3}^{1}+1......\left( 2 \right) \\
\end{align}$
Where $x_{2}^{1},x_{3}^{1}$ are variable.
Now, we have from the question that,
${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=n-3$
Now, using (1) and (2) we have,
$\begin{align}
  & {{x}_{1}}+x_{2}^{1}+x_{3}^{1}+1+{{x}_{4}}=n-3 \\
 & {{x}_{1}}+x_{2}^{1}+x_{3}^{1}+{{x}_{4}}=n-5 \\
\end{align}$
Now, we know that each of the variable ${{x}_{1}},x_{2}^{1},x_{3}^{1},{{x}_{4}}$ is non – negative. So, we use the formula of finding the number of non – negative integral solutions of the equation.
${{x}_{1}}+{{x}_{2}}+........+{{x}_{n}}=k$ is ${}^{k+n-1}{{C}_{n-1}}$
Now, we have the equation as,
${{x}_{1}}+x_{2}^{1}+x_{3}^{1}+{{x}_{4}}=n-5$
Therefore,
 $\begin{align}
  & {{P}_{n}}={}^{n-5+4-1}{{C}_{n-1}} \\
 & {{P}_{n}}={}^{n-2}{{C}_{3}} \\
\end{align}$
Now, we have been given that,
${{P}_{n+1}}-{{P}_{n}}=15$
So, using ${{P}_{n}}={}^{n-2}{{C}_{3}}$ we have,
${}^{n-1}{{C}_{3}}-{}^{n-2}{{C}_{3}}=15$
Now, we know that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.
$\dfrac{\left( n-1 \right)!}{\left( n-4 \right)!\times 3!}-\dfrac{\left( n-2 \right)!}{\left( n-5 \right)!\times 3!}=15$
Now, we know that,
$\begin{align}
  & \left( n-1 \right)!=\left( n-4 \right)!\times \left( n-3 \right)\left( n-2 \right)\left( n-1 \right) \\
 & \left( n-2 \right)!=\left( n-5 \right)!\times \left( n-4 \right)\left( n-3 \right)\left( n-2 \right) \\
 & \Rightarrow \dfrac{\left( n-4 \right)!\left( n-3 \right)\left( n-2 \right)\left( n-1 \right)}{\left( n-4 \right)!\times 3!}-\dfrac{\left( n-5 \right)!\left( n-4 \right)\left( n-3 \right)\left( n-2 \right)}{\left( n-5 \right)!\times 3!}=15 \\
 & \dfrac{\left( n-3 \right)\left( n-2 \right)\left( n-1 \right)}{6}-\dfrac{\left( n-4 \right)\left( n-3 \right)\left( n-2 \right)}{6}=15 \\
\end{align}$
Now, we cross – multiply and take $\left( n-3 \right)\left( n-2 \right)$ as common,
$\begin{align}
  & \left( n-3 \right)\left( n-2 \right)\left\{ n-1-n+4 \right\}=15\times 6 \\
 & \left( n-3 \right)\left( n-2 \right)\times 3=15\times 6 \\
\end{align}$
Now, on expanding LHS we have,
$\begin{align}
  & {{n}^{2}}-3n-2n+6=30 \\
 & {{n}^{2}}-5n-24=0 \\
 & {{n}^{2}}-8n+3n-24=0 \\
 & n\left( n-8 \right)+3\left( n-8 \right)=0 \\
 & \left( n+3 \right)\left( n-8 \right)=0 \\
 & either\ x+3=0\ or\ n-8=0 \\
\end{align}$
Since n can’t be negative. Therefore, the value of n is 8.

Note: It is important to note that we have used a fact that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$. Also, it is important to remember that the number of non – negative integral solution of equation,
${{x}_{1}}+{{x}_{2}}+........+{{x}_{n}}=k$ is ${}^{k+n-1}{{C}_{n-1}}$.