Answer :

Given, that there are 4 different books each having 3 instances, having a total of 12 books.

Lets assume that there are 4 different books B1, B2, B3, and B4. So we have 12 books as -

B1, B1, B1, B2, B2, B2, B3, B3, B3, B4, B4, B4.

To find: Number of arrangements of these 12 books in such a way that all arrangements must be distinct.

The problem can now be rephrased as to find number of permutations of 12 objects in which 3 objects are of one type, 3 objects are of another type, 3 objects are of a third type, and remaining 3 objects are of different type.

Since we know, Permutation of n objects taking r at a time is ^{n}P_{r},and permuation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the number of repeated objects of same type are in denominator multiplication with factorial.

The number of permutation of 12 objects with repeating books in factor of 3

= 369600

Hence, total number of permutation of given 12 books will be equals to 369600.

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