Let the five digit number be abcde
∵36=22×32
Case I: When exactly one digit is 1
So, number of permutations of (2,2,3,3,1) will be
=2!2!5!=30
Case II: When exactly two digits are 1
So, number of permutations of (4,3,3,1,1),(6,2,3,1,1) or (9,2,2,1,1) will be
=2!2!5!+2!5!+2!2!5!=120
Case III: When exactly three digits are 1
So, number of permutations of (4,9,1,1,1) or (6,6,1,1,1) will be
=3!5!+3!2!5!=30
Hence, the required numbers=30+120+30=180