The number of 5 -digit natural numbers, such that the product of...

Let the five digit number be $abcde$

$∵ 36 = {2}^{2} \times {3}^{2}$

Case I: When exactly one digit is $1$

So, number of permutations of \left(2, 2, 3, 3, 1\right)$will be</p><p>$= \dfrac{5 !}{2 ! 2 !} = 30$</p><p><strong>Case II:</strong> When exactly two digits are$1$</p><p>So, number of permutations of \left(4, 3, 3, 1, 1\right), \left(6, 2, 3, 1, 1\right) or \left(9, 2, 2, 1, 1\right) will be</p><p>=\dfrac{5 !}{2 ! 2 !} + \dfrac{5 !}{2 !} + \dfrac{5 !}{2 ! 2 !} =$120$</p><p><strong>Case III:</strong> When exactly three digits are$1

So, number of $permutations of \left(4, 9, 1, 1, 1\right) or \left(6, 6, 1, 1, 1\right) will be</p><p>=\dfrac{5 !}{3 !} + \dfrac{5 !}{3 ! 2 !} =30$

Hence, the required numbers$= 30 + 120 + 30 = 180$