Question

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

Answer 1

The odd digits having two 1‘s alike and two 3’s alike can be arranged in four odd places in \(\frac{4!}{2!2!}=6\) ways. 

The three even digits having two 2’s alike can be arranged in the three even places in \(\frac{3!}{2!}\) = 3 ways. 

∴ The number of numbers = 6 × 3 = 18.