How many numbers can be formed with the digits 2, 3, 4, 5, 4, 3, 2 so that the odd digits occupy the odd places?

Answer

18

Explanation

There are three odd digits, namely 3, 5, 3 out of which 3 occurs twice and 5 occurs once.
Number of ways to fill the odd places `=(3!)/(2!)=3.`
There are four even digits namely 2, 4, 4, 2 out of which 2 occurs twice and 4 occurs twice.
Number of ways to fill the even places`=(4!)/((2!)xx(2!))=6.`
Required number of numbers `=(3xx6)=18.`