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!
 A: 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 $$\binom{3 + 13 - 1}{13} = \binom{15}{13} = \binom{15}{2} = \frac{15!}{13!2!}= 105$$ 
A: 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.
A: Writing $7^{13}$ as a product of $3$ natural numbers becomes $7^{13}=7^{a}\times7^{b}\times7^{c}$
where $a,b,c$ are nonnegative integers with $a+b+c=13$. In fact
number $13$ must be split up in three parts. Do that by writing $15$
zero's on a row followed by picking out $2$ of them that are changed
into $1$. Then $13$ zeros are split up. There are $\left({15\atop 2}\right)=105$
choices when it comes to picking out $2$ from $15$.
Example: 
$000000000000000$ becomes $000100000100000$ that stands for possibility
$13=3+5+5$
$3$ on the left, $5$ in the middle and $5$ on the right.
A: The most natural interpretation is as stated by Ross Millikan above - that the order of parts is not important (although it would be better if the problem specified it). In this case, it is not a stars and bars problem.
Also, you must consider partitions of 13 in at most 3 parts (which would allow $7^0 \times 7^0 \times 7^{13}$ as a possibility, for example). If you were asked: how many ways can you write 13 as a sum of 3 positive natural numbers, you would consider partitions of 13 in exactly 3 parts. If you were asked: how many ways can you write 13 as a sum of 3 natural numbers, you would consider partitions of 13 in at most 3 parts (since we consider 0 to be a natural number).
A: In  this case, you are distributing 13 into a + b +c.
This can be done by 15C2.
But 15C2 i.e 105 will be considering 1 case more than one time because it will give you ordered pairs or solutions. So, you have to subtract redundant pairs.
You cant do that by dividing it by 3! as every single pair is not present 6 times. For ex - (a,a,b) is present only 3 times. So remove all these cases and we will be left with only (a,b,c) pairs. 
(13C2 - 3*7)/3! 
This will give you unordered pair of (a,b,c). Now add 7 pairs which we have removed from them.
(13C2 - 3*7)/3! + 7 = 21
