Question

How many numbers can be formed using all the digits of the number 1234321 such that odd digits occupy odd places only?

Answer 1

In the number 1234321 the odd digits are 1, 3, 3, 1 and even digits are 2, 4, 2.

4 odd digits can occupy odd places in \(\frac{41}{2!2!}\) ways and 3 even digits can occupy the remaining even places in \(\frac{3!}{2!}\) ways.

∴ Total permutations = \(\frac{4!}{2!2!}×\frac{3!}{2!}\)

= \(\frac{24}{2×2}×\frac{6}{2}\)

= 6 × 3

= 18