Probability: Finding the Number of Pears Given Two Scenarios 
You have a bag containing 20 apples, 10 oranges, and an unknown number
  of pears. If the probability that you select 2 apples and 2 oranges is
  equal to the probability that you select 1 apple, 1 orange, and 2
  pears, then what is the number of pears originally in the bag?

Using $n$ as the number of pears, I found the probability of selecting 2 apples and 2 oranges to be: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}$$
seeing as the number of ways to choose two apples and 2 oranges (order shouldn't matter) would be given by $\dbinom{20}{2}*\dbinom{10}{2}$. I put this over the total number of possibilities, which was found by choosing $4$ fruits from a total of $30+n$ fruits.
As the problem stated, this equaled the probability of choosing 1 apple, 1 orange, and 2 pears, which would be:
$$\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
And thus: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}=\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
$$\Rightarrow\dbinom{20}{2}*\dbinom{10}{2}=\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}$$
$$\Rightarrow\dbinom{n}{2}=42.75$$
which doesn't seem to be correct. Can anyone tell me where I went wrong? I believe it is most likely a conceptual mistake; is this not how you calculate the respective probabilities?

It is good that you were alert. I believe everything you are doing is correct. If we continue, then
$$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$
I think it is ok to have a decimal number here.
This gives
\begin{align*}
\frac{n!}{2!(n-2)!} &= \frac{171}{4}\\
\implies \frac{n!}{(n-2)!} &= \frac{171}{2}\\
n(n-1) &= \frac{171}{2}\\
n^2-n+\frac{1}{4} &= \frac{171}{2}+\frac{1}{4}\\
\left(n-\frac{1}{2}\right)^2 &= \frac{343}{4}\\
\implies n &= \sqrt{\frac{343}{4}}+\frac{1}{2}\\
n&=9.76013
\end{align*}
which I feel is an appropriate approach. The numbers are usually chosen so that when you complete the square, you will end up with a whole number $n$, but I think there was an error.
