Selecting 2 cups without handle and 3 cups with handle There are several tea cups in the kitchen, some with handle and the others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly $1200$. What is the maximum possible number of cups in the kitchen?
Here's what I did:
I assumed cups with handle are $x$ and without handle are $y$. Now ways of selecting three cups with a handle are $^xC_3$ and ways of selecting $2$ cups without a handle are $^yC_2$. So, $x(x-1)y(y-1)(y-2)=14400$ .Now I am stuck here. How do I proceed from here? Is the only way hit and trial? Please help me out.
 A: I think it is practically necessary to do at least a little trial and error for this problem. But you can make the number of trials small.
To start with, you can easily find the prime factorization of $14400.$
From that you can generate a list of all factors.
In the equation $x(x−1)y(y−1)(y−2)=14400$ we see that there are three consecutive numbers, $y-2,$ $y-1,$ $y,$ each of which is a factor of $14400.$
So already the only possible candidates for $y$ occur when you find three consecutive integers in the list of factors of $14400.$ This only happens a few times.

You might also notice that $y(y−1)(y−2)$ grows like $y^3$ while $x(x−1)$ grows only like $x^2,$ so you might guess that every time you increase $y$ by $1$ you force $x$ to decrease by more than $1,$ and therefore the correct choice of $y$ is the smallest $y$ that allows you to find five factors as shown.
I think you could use some calculus to make this a known fact rather than a guess. However, the number of possible choices of three consecutive factors is so few that you don't really need this extra level of sophistication.
Even with the knowledge that smaller $y$ is better, you still have to choose a $y$ that actually works. You could try the first set of three consecutive factors in the list (listed in increasing value), but you would then still have to see whether there is an integer solution for $x.$ That's a trial and possibly an error.
A: I also think that a bit of trial and error is necessary. Let us try to make this efficient. We know which numbers are of the form $x(x-1)$: exactly the numbers $n$ such that $4n+1$ is a square. So we test for square-ness
$$ 4\frac{14400}{y(y-1)(y-2)}+1 $$
provided that $y(y-1)(y-2)$ is a divisor of $14400=2^6 3^2 5^2$. So we may consider the numbers $y\in\{3,4,\ldots,24,25\}$ and cross out the ones such that $y,y-1$ or $y-2$ has a prime divisor greater than $5$. This leaves us with
$$ y\in\{3,4,5,6,10\} $$
and associated to these $y$s we have the following values of $ 4\frac{14400}{y(y-1)(y-2)}+1 $:
$$ \{9601,2401,961,481,81\} $$
The squares $2401=49^2, 961=31^2$ and $81=9^2$ give three solutions: $\color{red}{(25,4),(16,5),(5,10)}$.
