Question

There are three identical books on English, 4 identical books on Hindi, 2 identical books on mathematics. In how many distinct ways can they be arranged on a shelf?

Answer 1

Hint:
If we have to arrange $n$ objects in $n$ places, then there can be $n!$ of arranging it. But, if there are ${n_1}$ identical objects, then we will divide by ${n_1}!$ and so on. Hence, on substituting the values, we will get the required answer.

Complete step by step solution:
We are given that there are 3 three identical books on English, 4 identical books on Hindi, 2 identical books on mathematics.
This means we have a total $3 + 4 + 2 = 9$ books on the shelf.
We know that if we have to arrange $n$ objects in $n$ places, then there can be $n!$ of arranging it.
But, we will also divide the above number by the ${n_1}!$, where ${n_1}$ identical objects are there.
Here, there are 3 types of books which are identical, then the number of ways are,
$\dfrac{{n!}}{{{n_1}!{n_2}!{n_3}!}}$, where $n$ is the total books, ${n_1}$ is the number of English books, ${n_2}$ is the number of Hindi and ${n_3}$ is the number of Mathematics books.
Therefore, the total number of ways are,
 $\dfrac{{9!}}{{3!4!2!}} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4! \times 2 \times 1}} = 1260$

Hence, the number of ways of arranging books on shelf is 1260.

Note:
Since, the order of the books matters, we cannot use combinations. We have used the concept of permutation in this question when certain objects are repeated. Also, one must know that $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1$ to simplify the question.