Q. 154.8( 6 Votes )

# There are three c

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 nPr,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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Four letters E, KRD Sharma - Mathematics

. There are 6 iteRD Sharma - Mathematics

How many permutatRD Sharma - Mathematics

Find the total nuRD Sharma - Mathematics

In how many ways RD Sharma - Mathematics

From among the 36RD Sharma - Mathematics

How many differenRD Sharma - Mathematics

How many three-diRD Sharma - Mathematics

In how many ways RD Sharma - Mathematics

How many three-diRD Sharma - Mathematics