Question

There are 4 mathematics books, Junior inter volume 1 & 2, Senior inter volume 1 & 2 among 20 books. They are arranged in a row at random. The probability that all the mathematics books are arranged in that order not necessarily side by side is:
A. $\dfrac{1}{16}$
B. $\dfrac{1}{12}$
C. $\dfrac{1}{24}$
D. $\dfrac{1}{8}$

Answer 1

Hint:Here we need to find the probability that the 4 given books remain in the same order when the 20 books are arranged such that those 4 may or may not be side by side. For this, we will use the formula of the probability of the event given by $\text{Probability=}\dfrac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$ and then we will find our total number of outcomes using the formula for arranging ‘n’ distinct objects in a row given by ‘n!’. Then, we will find the number of favorable outcomes by assuming the 4 given books to be identical and then using the formula of arranging ‘n’ objects out of which ‘r’ are identical given by $\dfrac{n!}{r!}$. Then we will keep both these values in the formula for the probability of an event and hence we will get our answer.

 Complete step by step answer:
Now, we need to find the probability that these 4 books are arranged in the same order, not necessarily side by side.
We know that the probability of any event is given by the formula:
$\text{Probability=}\dfrac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$
We will now first find the total number of outcomes.
Now, we know that ‘n’ distinct objects can be arranged in ‘n!’ ways.
Thus, the total number of ways in which these 20 distinct books can be arranged is given by:
$20!$
Thus, there are a total of $20!$ ways in which the 20 books can be arranged.
Thus, the total number of outcomes=20!
Now, we will find the number of favorable outcomes, i.e. the number of ways in which these 20 books can be arranged such that the 4 books- junior inter volume 1 and 2 and senior inter volume 1 and 2 can be arranged in the same order such that these books need not to be side by side.
Now, we can do this by considering those 4 books as 4 identical items as they need to be arranged in the same order but it is not necessary for them to be side by side. This is explained as follows:
Let us consider the example of 5 numbers out of which 2 are the same and the rest are distinct. Now, these numbers can be arranged in $\dfrac{5!}{2!}$, and in all of those arrangements, there will be two similar numbers which may or may not occur side by side.
Here, we have to the same with our books as their order has to remain intact. If it was necessary for them to be side by side, then we’d have to make a unit of those 4 books and hence those 4 books will be considered as 1 single book.
Now, we know that ‘n’ different objects out of which ‘r’ are identical, can be arranged in the following ways:
$\dfrac{n!}{r!}$
Here, we have 20 different books out of which we have considered 4 books to be identical.
Hence, here we can say that:
$\begin{align}
  & n=20 \\
 & r=4 \\
\end{align}$
Thus, the number of ways in which these 20 books can be arranged such that these 4 books remain in the same order is given by:
$\begin{align}
  & \dfrac{n!}{r!} \\
 & \Rightarrow \dfrac{20!}{4!} \\
\end{align}$
Thus, the number of favorable outcomes is equal to $\dfrac{20!}{4!}$
Now, we have both, number of favorable outcomes and the number of total outcomes.
We know that probability of any event is given by:
$\text{Probability=}\dfrac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$
Hence, the probability of the required event is given as:
$\text{Probability=}\dfrac{\dfrac{20!}{4!}}{20!}$
Solving this we get:
$\begin{align}
  & \text{Probability=}\dfrac{\dfrac{20!}{4!}}{20!} \\
 & \Rightarrow \text{Probability=}\dfrac{20!}{4!20!} \\
 & \Rightarrow \text{Probability=}\dfrac{1}{4!} \\
\end{align}$
Now, we know that 4! Is given by:
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 4!=24 \\
\end{align}$
Thus, we get the required probability as:
$\begin{align}
  & \text{Probability=}\dfrac{1}{4!} \\
 & \Rightarrow \text{Probability}=\dfrac{1}{24} \\
\end{align}$
Hence, the required probability is $\dfrac{1}{24}$
Thus, option (C) is the correct option.

Note:
Don’t solve the factorials separately as they are factorials of large numbers and hence will take up a very good amount of time to be solved. Even then, there is no guarantee that the value of the factorial will be correct. When factorials of big numbers come into play, we should keep them in the same form and solve only when all the factorials are put into the required formula. Many values will be canceled out when that will happen and hence make the solution easy and less prone to mistakes.