Expected Number of Successes in a Sample 
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.

Solution: 
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?
For example:
\begin{align}
{10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\
{10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\
\vdots & \\
{10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1  \cdot 10^{-10}. \tag{10}
\end{align}
Adding them together:
\begin{align}
E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10.
\end{align}
Or is it enough to calculate
\begin{align}
E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right).
\end{align}
Thanks.

Your shorter formula doesn't work. Suppose the question asked you to select two calculators instead of ten. To compute the expected value "by hand", there are three cases to consider (we don't care about the case when no calculators are broken).
First Calculator Broken: This event happens with probability $(1/10)(9/10) = 9/100$.
Second Calculator Broken: This event happens with probability $(9/10)(1/10) = 9/100$.
Both Calculators Broken: This event happens with probability $(1/10)(1/10) = 1/100$.
The expected number of broken calculators (in my modified example) is therefore
$$
(1 \cdot 9/100) + (1 \cdot 9/100) + (2 \cdot 1/100) = 20/100 = 0.2.
$$
The fact that we had two instances of "one calculator broken" would normally be accounted for by the factor of $C(2,1) = 2$.
To address your concern more directly, your shorter formula does not reflect the fact that "exactly one broken calculator" really means "one is broken and nine are functional", which is why that formula is giving a higher expected value than is correct.
