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?

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 $=\frac{4 !}{2 ! 2 !} \times \frac{3 !}{2 !}=18$