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

Let the five digit number be $abcde$

$\because 36=2^2\times 3^2$

Case I: When exactly one digit is $1$

So, number of permutations of $\left(2,2,3,3,1\right)$ will be

 $=\frac{5!}{2!2!}=30$

Case II: When exactly two digits are $1$

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

$=\frac{5!}{2!2!}+\frac{5!}{2!}+\frac{5!}{2!2!}=120$

Case III: 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

$=\frac{5!}{3!}+\frac{5!}{3!2!}=30$

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