Question

The number of natural numbers which lie between ${10}^8$ and ${10}^9$ which have products of their digits as 6 is___.

Answer 1

${10}^8$<n<${10}^9$

n is a 9 digit number

Product of 6 is possible in the following cases

Exactly one digit is 6 and the remaining digits are 1

Number of possible numbers= 9

One digit is 2 and the other is 3, remaining are 1 = ${}^9{P}_2$ = 72

Total possible numbers = 72+9 = 81