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.