Question

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

Answer 1

We have been given seven digits, namely 1, 2, 3, 4, 3, 2, 1.

So, we have to from 7-digit numbers, so that odd digits occupy odd places.

Every 7-digit number has 4 odd places.

Given four odd digits are 1, 3, 3, 1 out of which 1 occurs 2 times and 3 occurs 2 times.

So, the number of ways to fill up 4 odd place = $\frac{4!}{(2!)\times (2!)}=6.$

Given three even digits are 2, 4, 2 in which 2 occurs 2 times and 4 occurs 1 times.

So, the number of ways to fill up 3 even places = $\frac{3!}{2!}=3.$

Hence, the required number of numbers = $(6\times 3)=18.$