In how many ways can $7^{13}$ be represented as product of $3$ natural numbers? How i solved it:
all possible non-distinct groups $(a,b,c)$ are,
$a = 0 \Rightarrow (b,c) = (0,13)(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$
$a = 1 \Rightarrow (b,c) = (1,11)(2,10)(3,9)(4,8)(5,7)(6,6)$
$a = 2 \Rightarrow (b,c) = (2,9)(3,8)(4,7)(5,6)$
$a = 3 \Rightarrow (b,c) = (3,7)(4,6)(5,5)$
$a = 4 \Rightarrow (b,c) = (4,5)$
Thus, $7^{13}$ can be written as product of $3$ natural numbers in $7+6+4+3+1 = 21$ ways.
Though this gives the required solution, it takes time and is a lengthy way. Is there an alternate method to tackle such problems? (May be by using the "bars and stars" method with some adjustments? I tried but failed to get the correct answer that way.) Please help me out and share your method!
Thanks a lot!

What you are doing, essentially, is partitioning the exponent $13$, and this amounts to solving the equation $a + b + c = 13$, where $(a, b, c)$ is a 3-tuple of non-negative integers. 
Hence, the classic Stars-and-Bars Problem comes into play:
Using the fact that there are $$\binom{n + k - 1}{k}$$ distinct n-tuples of non-negative integers whose sum is $k$, we calculate the number of distinct 3-tuples of non-negative integers whose sum is $13$. In the case at hand, the number of distinct 3-tuples given by $(a, b, c)$ such that $7^a\times 7^b \times 7^c = 7^{a + b + c} = 7^{13}$ is given by $$\binom{3 + 13 - 1}{13} = \binom{15}{13} = \binom{15}{2} = \frac{15!}{13!2!}= 105$$ 
