Hi Tanishq Jain,
To find the number of factors of a number, prime factorize the number i.e., express the number as a product of the powers of its divisors that are prime numbers. For example, $$120 = 2^3*3*5$$.

Now, the factors of 120 can be constructed by selecting one number from $${ 2^0, 2^1, 2^2, 2^3 }$$, one number from $${ 3^0, 3^1 }$$ and one number from $${ 5^0, 5^1}$$. For example, if $$2^1$$ is selected from the first set, $$3^1$$ is selected from the second set, and $$5^0$$ is selected from the third set, then the factor will be $$2^1*3^1*5^0$$ = 6. So, the number of factors of 120 is $$^4C_1*^2C_1*^2C_1$$ = 4*2*2 = 16.

In general, if the number N can be factorized as $$p^a*q^b*r^c*…$$, where p , q, r, … are prime factors of N, then the number of factors of N is (a+1)*(b+1)*(c+1)…
I hope this helps!