<s>[INST] 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!
 [/INST] If you consider different orders equivalent, so that $7^87^37^2$ is considered the same as $7^27^37^8$, you are asking about the number of partitions of $13$ into $3$ parts, while the stars-and-bars answers give you the number of compositions of $13$ into $3$ parts. In the partition page, it states that the number of partitions of $n$ into $1,2,3$ parts is the nearest integer to $\frac {(n+3)^2}{12}$, here $21$.  As you area are allowing $0$, which means you include $1$ or $2$ parts, that is your answer.
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