As given $n_0 = n = p_1^{e_1}p_2^{e2}...p_k^{e_k}$.
To get the next number $n_1$ in the relevant sequence, we've to choose $1$ prime number out of $e_1 + e_2 + ... + e_k$ prime numbers and the product of remaining prime numbers will be our $n_1$ and the value $n_0/n_1$ will be that chosen prime number.
We've to repeat this (choose and remove $1$ prime number out from $n_i$ to get $n_{i+1}$) till we exhaust all the prime numbers and we'll get only $n_t = 1$ where $t = e_1 + e_2 + ... + e_k$.
Now, for each relevant sequence there exists an unique sequence of removed prime numbers. So, there is a bijection between relevant sequences and sequences of removed prime numbers.
Consider an example, $n_0 = 60 = 2^2 3^1 5^1$.
Relevant sequence $= 60, 30, 10, 5, 1$.
Corresponding sequence of removed prime numbers $= 2, 3, 2, 5$.
Relevant sequence $= 60, 12, 4, 2, 1$.
Corresponding sequence of removed prime numbers $= 5, 3, 2, 2$.
Total number of prime numbers $= e_1 + e_2 + ... + e_k$.
Of which $e_1$ prime numbers are identical ($p_1$), $e_2$ prime numbers are identical ($p_2$), and so on.
$\therefore$ number of sequences of removed prime numbers $= \frac{(e_1+e_2+...+e_k)!}{(e_1)! + (e_2)! + ... + (e_k)!}$.
Answer :- D.