Question

There are $5$ volumes of Mathematics among $25$ books. They are arranged on a shelf in random order. The probability that the volumes mathematics stand in increasing order from left to right (the volumes are not necessarily kept side by side) is
$1)$ $\dfrac{1}{5!}$
$2)$ $\dfrac{50!}{55!}$
$3)$ $\dfrac{1}{{{50}^{5}}}$
$4)$ None of these

Answer 1

Hint: For getting the probability of arrangement of books. Firstly, we will find the total number of possible ways of arrangement and then we will find the favourable number of possible arrangement of books with the use of formula ${}^{n}{{P}_{r}}$ , where $n$ denotes total value and $r$ denotes required value. Then, we will calculate the ratio of favourable number of possible arrangements and total number of arrangements that is required probability also.

Complete step-by-step solution:
Since, for the total number of possible ways of arrangement where we can arrange all books and volumes in any ways, will be:
$\Rightarrow {}^{n}{{P}_{r}}$
Here, we have a total $25$ and we can arrange them in any way. So, the required value will be $1$ for every will be arranged in its specific way. Thus, the total number of possible arrangements is:
$\Rightarrow {}^{25}{{P}_{1}}$
The expansion of above formula will be as:
$\Rightarrow \dfrac{25!}{1!}$
Now, we will arrange the $5$ volumes of mathematics book. So, we have required value as $5$ . Thus, the favourable number of possible arrangements is:
\[\Rightarrow {}^{25}{{P}_{5}}\]
And its expansion will be as:
$\Rightarrow \dfrac{25!}{5!}$
Since, we got the both values the total number of possible arrangements and the favourable number of possible arrangements. So, we will calculate its ratio to find the required probabilities as:
$\Rightarrow \dfrac{Favourable\text{ }Number\text{ }of\text{ }Possible\text{ }Arrangements}{Total\text{ }Number\text{ }of\text{ }possible\text{ }Arrangements}$
After applying the values, we will have the above step as:
$\Rightarrow \dfrac{\dfrac{25!}{5!}}{\dfrac{25!}{1!}}$
We can write the above step below as:
\[\Rightarrow \dfrac{25!}{5!}\times \dfrac{1}{25!}\]
Since, $25!$ will be canceling out. So, we will get:
\[\Rightarrow \dfrac{1}{5!}\]
Hence, the required probability is \[\dfrac{1}{5!}\] .

Note: Here, we will use the method of permutation to find the required probability of the given question because it denotes the possible ways of arrangement. So, we will use the formula of permutation that is ${}^{n}{{P}_{r}}$ and the expansion it will be $\dfrac{n!}{r!}$ , where, $n!$ and $r!$ denotes the multiplication of natural numbers up to $n$ and $r$ respectively.