Question

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 1

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in $\frac{4!}{2!2!}$ways.
The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in $\frac{3!}{2!}$ways.
By fundamental principle of counting:
Required number of arrangements = $\frac{4!}{2!2!}$$\times$$\frac{3!}{2!}$ = 18