Question

Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits

Answer 1

Digits are 3, 4, 5, 6, 7, 8, 9

Odd places are 1^st, 3^rd, 5^th and 7^th and odd digits are 3, 5, 7, 9. Even places are 2^nd, 4^th and 6^th and even digits are 2, 4, 6.

The four odd digits occupy 4 odd places in ⁴P₄ ways and three even digits can occupy the remaining 3 even places in ³P₃ ways.

∴ the total number of distinct numbers that can be formed

= ⁴P₄ × ³P₃ = 4!3!

= 4 × 3 × 2 × 1 × 3 × 2 × 1

= 144.