Question

Fifteen persons, among whom are A and B, sit down at random at a round table. The probability that there are 4 persons between A and B is
A.$\dfrac{1}{7}$
B.$\dfrac{2}{{17}}$
C.$\dfrac{3}{{17}}$
D.$\dfrac{4}{{17}}$

Answer 1

Hint: We should use this concept that $r$ distinct objects from \[n\] distinct objects can be done in ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ ways. The number of ways to arrange n persons at a round table is given by $\left( {n - 1} \right)!$.
Also, the probability of the event can be given as $\dfrac{{{\rm{Favorable cases}}}}{{{\rm{Total number of cases}}}}$.

Complete step-by-step answer:
We know that the number of ways to arrange n persons at a round table is given by $\left( {n - 1} \right)!$.
In the question, it is given that there are 15 persons that are to be seated at a round table.
The total number of ways to arrange 15 persons at the round table is
$\Rightarrow \left( {15 - 1} \right)! = 14!$.
As per the question, there are two persons A and B included in those 15 persons. Question wants 4 persons to be seated between A and B.
We know that we can arrange $n$ seats in $n!$ ways.

As shown in the figure, 4 persons, ${{\rm{P}}_1}{\rm{, }}{{\rm{P}}_2}{\rm{, }}{{\rm{P}}_3}{\rm{, }}{{\rm{P}}_4}$ are seated between A and B.
To find the number of ways in which 4 persons between A and B are arranged when 15 persons are arranged, first we have to find the number of ways to choose and arrange those 4 persons.
We know that to choose $r$ distinct objects from \[n\] distinct objects can be done in ${}^n{C_r}$ ways.
The 4 persons can be selected and arranged as
$\Rightarrow {}^{13}{C_4} \cdot 4!$, A and B are removed from 15 persons and the 4 persons are selected from 13 persons.
A and B can be arranged in $2!$ ways. Also, the remaining people $15 - 2 - 4 = 9$, can be arranged in $9!$ ways.
All the factors are to be multiplied when they are to be considered simultaneously.
In this question, 15 persons are arranged simultaneously.
The number of cases such that 4 persons are seated and arranged between A and B are $\Rightarrow {}^{13}{C_4} \cdot 4! \cdot 2! \cdot 9!$
The probability of the event can be given as $\dfrac{{{\text{Favorable cases}}}}{{{\text{Total number of cases}}}}$.
So, the probability that there are 4 people between A and B is $\dfrac{{{}^{13}{C_4} \cdot 4! \cdot 2! \cdot 9!}}{{14!}}$
The probability is given by
$
\Rightarrow {\rm{Probability}} = \dfrac{{{}^{13}{C_4} \cdot 4! \cdot 2! \cdot 9!}}{{14!}}\\
 = \dfrac{{\dfrac{{13!}}{{4!\left( {13 - 4} \right)!}} \cdot 4! \cdot 2! \cdot 9!}}{{14!}}\\
 = \dfrac{{\dfrac{{13!}}{{4!9!}} \cdot 4! \cdot 2! \cdot 9!}}{{14!}}\\
 = \dfrac{{13! \cdot 2!}}{{14!}}
$
$
\Rightarrow {\rm{Probability}} = \dfrac{2}{{14}}\\
 = \dfrac{1}{7}
$
Therefore, the probability that there are 4 persons between A and B is $\dfrac{1}{7}$.

So, the correct answer is “Option A”.

Note: Students should take care of the language used in this question. There is a possibility that students might misunderstand the language while calculating the value of probability.
When $r$ objects out of $n$ are selected, the total number of ways to do so is given by ${}^n{C_r}$. Students should take care while using this formula, they often mistake the value to be ${}^r{C_n}$ instead of ${}^n{C_r}$.