Question

Prove that only 18 numbers can be formed by using all the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places.

Answer 1

There are 4 odd places and there are 4 odd numbers (2 are alike i.e. 1, 1; and 2 are alike i.e. 3, 3). These can be arranged in four places in
$\frac{4!}{2!2!}=\frac{4\times 3}{2}=6$ ways.
There will be 3 even places namely 2nd, 4th and 6th in which 3 even numbers (2 are alike i.e. 2, 2). These can be arranged in
$\frac{3!}{2!}=\frac{3\times 2}{2}=3$ ways.
Hence the total number of the numbers thus formed is $6\times 3=18$.