The number of 5 -digit natural numbers, such that the product of their digits is 36 , is

Step-by-step solution

Let the five digit number be a b c d e ∵       36 = 2 2 × 3 2 Case I: When exactly one digit is 1 So, number of permutations of 2 , 2 , 3 , 3 , 1 will be = 5 ! 2 ! 2 ! = 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 = 5 ! 2 ! 2 ! + 5 ! 2 ! + 5 ! 2 ! 2 ! = 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 = 5 ! 3 ! + 5 ! 3 ! 2