# How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Given, digits 1, 2, 3, 4, 3, 2, 1.

To find: Number of numbers formed from given digits such that odd digits always occupies odd positions/places.

Since we know, Permutation of n objects taking r at a time is nPr,and permutation 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.

A total number of ways of arranging 3 even digits = 3! / 2! since there is repetition of a digit (2).

A total number of ways of arranging 4 odd digits = 4! / (2! x 2!) since there are twice repetition of digits (1 and 3).

Total number of ways of arranging the digits such that odd digits always occupies odd places =

= 18

Hence, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Circular permutations61 mins
Challenging Quiz on P&C | Test Yourself55 mins
Permutations & Combinations | Analyze your learningFREE Class
Understand Permutations like never before60 mins
Check Your progress Part 2| Interactive Quiz: Permutation & CombinationFREE Class
Lecture on Combinations49 mins
Interactive Quiz on Division and distribution of objects17 mins
Interactive Quiz on Combinations-0253 mins
Interactive Quiz on Combinations50 mins
Division and distribution of objects58 mins
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