# Find the number of ways to express 3600 as a product of three factors(triplets) [closed]

Ok so I came upon this question, and I have absolutely no idea how do we proceed.

I had done a similar kind of question which was :-

Find the number of ways to express 12 as a product of 2 factors

In this case it was much simpler, in which I was only required to find out the total number of factors and then divide it by 2 in order to obtain the answer.

In case it had been a perfect square , then I would be required to take the smallest integer function of $$\frac{Number\ of\ factors}{2} +1$$.

However what do we do for expressing numbers as triplets? I tried thinking about considering various perfect squares and perfect cubes in the factors and trying to deduce something from that , however I could not do that.

• There is no way to express $3600$ as a product of three prime factors.
– user
Commented Mar 12 at 18:29
• The answer to the posed question -"Find the number of ways to express 3600 as a product of three prime factors(triplets)" - is "zero ways." Key word: prime. Commented Mar 12 at 18:30
• Actually, $3600=p^nq^mr^k$ for three different primes $p,q,r$ (which are indeed $2,3,5$). Perhaps the question has a misunderstanding. Commented Mar 12 at 18:35
• Sorry.... accidentally wrote the word "prime" Commented Mar 12 at 18:45
• Does the order of factors matter?
– user
Commented Mar 12 at 18:47

Since $$3600=2^4\cdot3^2\cdot5^2$$ then assuming that the order of factors matters, the number in question is: $$\binom{4+2}2\binom{2+2}2\binom{2+2}2=15\cdot6\cdot6=540,$$ where binomial coefficients $$\binom{m+2}2$$ stay for the number of ways to distribute $$m$$ indistinguishable objects in $$3$$ bins (see stars and bars).
If the order of factors does not matter one can compute the number using $$540=3N_2+6N_3,$$ where $$N_k$$ is the number of combinations with $$k$$ distinct factors ($$N_1=0$$ since $$3600$$ is not a perfect cube). $$N_2$$ can be computed as number of divisors of $$2^2\cdot3\cdot5$$ in a similar way as above: $$N_2=\binom{2+1}1\binom{1+1}1\binom{1+1}1=12,$$ so that the number of combinations is $$N_2+N_3=96$$.
• I do not know the name of the method. But generally the full number of solutions where the order matters is the sum over unordered solutions multiplied by multiplicity of their permutations. The multiplicity is the multinomial coefficient $\frac{n!}{n_1!n_2!\cdots n_k!}$ where $n_i$ is the number of objects of $i$-th type ($\sum n_i=n$).