Question

There are three copies each of four different books. The number of ways in which they can be arranged in a shelf is:
(A). $\dfrac{12!}{{{\left( 3! \right)}^{4}}}$
(B). $\dfrac{12!}{{{\left( 4! \right)}^{3}}}$
(C). $\dfrac{21!}{{{\left( 3! \right)}^{4}}4!}$
(D). $\dfrac{12!}{{{\left( 4! \right)}^{3}}3!}$

Answer 1

- Hint: First assume that different books have 4 variables. Assume the copies of books as the same variable. Now we know the arrangement of n objects is \[n!\] . Use this to find a number of arrangements. Now try to find the number of ways which are similar arrangements. As there are copies of objects you might as well count repetitions. So, try to remove those repetitions by any means

Complete step-by-step solution -

Factorial: - In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinatorics.

Look at the factorization definition, you see the product which works as factorial. Keep that aside.
Now, if we need to arrange n objects in n places we follow the following method:
The number of ways /places for ${{1}^{st}}$ object = n
The number of places remaining for next =n-1
                                  :
                                  :
The number of places remaining for the last object =1. So, Total number of ways of arranging n objects is given by $n\left( n-1 \right)...\left( 3 \right)\left( 2 \right)\left( 1 \right)$.
By factorization definition, we can say value of number of ways is $\left( n \right)!$
Here we have 12 objects. So, number of way will be: $\left( 12 \right)!$
But out of those 12 there are 4 types, assume them as A, B, C, D.
For example: if $-A-A-A-$
A’s are placed like this now we can arrange these as in $3!$ ways. Which are all the same.
Similarly we get for B, C, D, too.
Now, in $12!$ ways we have these all repetitions.
So, we divide the value $12!$ by these all number effect due to A is compensated by doing the step:
$\dfrac{12!}{3!}$ .
Effect due to B is compensated by doing the step:
$\dfrac{12!}{3!3!}$ .
Effect due to C is compensated by doing the step:
$\dfrac{12!}{3!3!3!}$ .
Effect due to D is compensated by doing the step:
$\dfrac{12!}{3!3!3!3!}$ .
This can be written as $\dfrac{12!}{{{\left( 3! \right)}^{2}}{{\left( 3! \right)}^{2}}}$
By simplifying, we can write it as $\dfrac{12!}{{{\left( 3! \right)}^{4}}}$
Therefore option (a) is correct.

Note: Generally students between combinations, permutation. If you have to arrange then use permutation. It is a very nice trick to use. Don’t forget to consider all possibilities or else you might reach the wrong answer. Do it carefully, For example: here if you miss taking effect of repetitions you get value as $12!$ but it is wrong.