Q. 74.5( 2 Votes )

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?

Answer :

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.


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