Question

There are 12 seats in a row of a cinema theatre, which is nearer to the entrance. 4 persons enter the theater and occupy the seats in that row (before their entry the seats are vacant). If ${n_1}$ is the number of ways in which they occupy the seats, such that no two persons are together and ${n_2}$ is the number of ways in which each person has exactly one neighbor ${n_1}:{n_2}$ is
A.$7:2$
B.$3:2$
C.$5:4$
D.None of these

Answer 1

Here, we will first use the formula of permutations is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $n$ is the total number of object and $r$ is the number required and then assume the four persons and pair them using the given condition, ${P_1}{P_2}:{P_3}{P_4}$. So we will assume ${x_1}$ be the number of empty seats before ${P_1}{P_2}$, ${x_2}$ be the number of empty seats between ${P_1}{P_2}$ and ${P_3}{P_4}$ and ${x_3}$ be the number of empty seats after ${P_3}{P_4}$ .Then we will use the values ${y_1} = {x_1}$, ${y_2} + 1 = {x_2}$ and ${y_3} = {x_3}$ in the equation, ${x_1} + {x_2} + {x_3} = 8$. The use the formula of combinations, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to find required ratio.

Complete step-by-step answer:
Given that there are 12 numbers of seats.
If ${n_1}$ is the number of ways in which the 4 persons occupy the seats, such that no two persons will sit together.
So we will have 8 empty seats and 9 gaps in between these persons.
We know that the formula of permutations, that is, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $n$ is the total number of object and $r$ is the number required.
Using this formula of permutations without repetition to find the value of ${n_1}$, we get

$${n_1} = {}^9{P_4} \\\ = \dfrac{{9!}}{{\left( {9 - 4} \right)!}} \\\ = \dfrac{{9!}}{{5!}} \\\$$

Calculating the factorials of the above fraction, we get

$$\dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} = 9 \times 8 \times 7 \times 6 \\\ = 3024 \\\$$

We have given that ${n_2}$ is the number of ways in which each person has exactly one neighbour.
Let us assume that the four persons are ${P_1}$, ${P_2}$, ${P_3}$ and ${P_4}$.
The four persons can be paired like ${P_1}{P_2}:{P_3}{P_4}$ (say).
Let us take ${x_1}$ be the number of empty seats before ${P_1}{P_2}$, ${x_2}$ be the number of empty seats between ${P_1}{P_2}$ and ${P_3}{P_4}$ and ${x_3}$ be the number of empty seats after ${P_3}{P_4}$.
Then we have ${x_1} + {x_2} + {x_3} = 8$, where ${x_1} \geqslant 0$, ${x_2} \geqslant 0$, ${x_3} \geqslant 0$.
Putting ${y_1} = {x_1}$, ${y_2} + 1 = {x_2}$ and ${y_3} = {x_3}$ in the above equation, we get

$$\Rightarrow {y_1} + {y_2} + 1 + {y_3} = 8 \\\ \Rightarrow {y_1} + {y_2} + {y_3} = 8 - 1 \\\ \Rightarrow {y_1} + {y_2} + {y_3} = 7 \\\$$

Now we will find the number of solutions using the combinations of ${y_1} + {y_2} + {y_3} = 7$, where ${y_1} \geqslant 0$, ${y_2} \geqslant 0$, ${y_3} \geqslant 0$.
${}^9{C_2}$
For each pair 2 can be selected in ${}^4{C_2}$ ways, each pair is arranged in $2!$ ways.
We will now find the value of ${n_2}$ from the above values.
${n_2} = {}^9{C_2} \cdot {}^4{C_2} \cdot 2! \cdot 2!$
Using the formula of combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ in the above equation, we get

$$\Rightarrow \dfrac{{9!}}{{2! \cdot \left( {9 - 2} \right)!}} \cdot \dfrac{{4!}}{{2! \cdot \left( {4 - 2} \right)!}} \cdot 2! \cdot 2! \\\ \Rightarrow \dfrac{{9!}}{{2! \cdot 7!}} \cdot \dfrac{{4!}}{{2! \cdot 2!}} \cdot 2! \cdot 2! \\\$$

Calculating the factorials of the above fraction, we get

$$\Rightarrow \dfrac{{9 \times 8 \times 7!}}{{2! \cdot 7!}} \cdot \dfrac{{4 \times 3 \times 2!}}{{2! \cdot 2!}} \cdot 2! \cdot 2! \\\ \Rightarrow \dfrac{{9 \times 8}}{{2 \times 1}} \cdot 12 \cdot 2 \\\ \Rightarrow \dfrac{{1728}}{2} \\\ \Rightarrow 864 \\\$$

Substituting these values of ${n_1}$ and ${n_2}$ in ${n_1}:{n_2}$ to find the ratio, we get
$\dfrac{{3024}}{{864}} = \dfrac{7}{2}$
Thus, the value of ${n_1}:{n_2}$ is $7:2$.
Hence, the option A is correct.

Note: In solving these types of questions, you should be familiar with the formula to find the permutations, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , where $n$ is the total number of object and $r$ is the number required and combinations, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the total number of object and $r$ is the number required. We will assume the ratio of the four persons carefully and use them to substitute in the equation ${x_1} + {x_2} + {x_3} = 8$ to find the required value.