Combinatorics of choosing cars Question and the solutions are available here.
The summary of the question:
A van has 4 optional extras and you can choose to have any number.
A sedan has 6 optional extras but you can choose at most 3.
Every vehicle has 4 different colours.
So the combinations of the van are 64 and the sedan is 168.
What I am confused about is the part where you can choose 4 vehicles, and one of them is at least a van. The solution reads ${235 \choose 4} - {171 \choose 4}$.
I was thinking of choosing 3 (with replacement) out of the 232 combinations first and then choosing 1 van out of the 64, so ${234 \choose 3} + {64 \choose 1}$, but my answer is way too big compared to the answer given.
What's wrong with my method?
 A: Note that $\binom{234}{3} + \binom{64}{1} < \binom{235}{4} - \binom{171}{4}$.  If you meant $\binom{234}{3}\binom{64}{1}$, your error was that you count each selection with more than one type of van multiple times, once for each type of van you could have designated as the type of van you selected.  For example, suppose you selected a red van, a blue van, and two sedans.  If you meant $\binom{234}{63}\binom{64}{1}$, you count this selection twice, once when you designate the red van as the van you selected and once when you designate the blue van as the van you selected.
Let $s_i$, $1 \le i \le 168$, be the number of sedans of type $i$ that are selected; let $v_j$, $1 \le j \leq 64$, be the number of vans that are selected.  Since at least one van is selected, if you want to count directly, then you need to solve the system of equations
\begin{align*}
s_1 + s_2 + s_3 + \cdots + s_{168} & = 4 - k \tag{1}\\
v_1 + v_2 + v_3 + \cdots + v_{64} & = k \tag{2}
\end{align*}
in the nonnegative integers as $k$ varies from $1$ to $4$.
Equation $1$ has
$$\binom{4 - k + 168 - 1}{4 - k} = \binom{171 - k}{4 - k}$$
solutions in the nonnegative integers and equation $2$ has
$$\binom{k + 64 - 1}{k} = \binom{63 + k}{k}$$
solutions in the nonnegative integers.   Once you choose how many sedans and how many vans you will select, the choices of sedans and vans are independent, so there are
$$\binom{171 - k}{4 - k}\binom{63 + k}{k}$$
ways to select exactly $k$ vans and $4 - k$ sedans.  Hence, the number of ways to select at least one van is
$$\binom{170}{3}\binom{64}{1} + \binom{169}{2}\binom{65}{2} + \binom{168}{1}\binom{66}{3} + \binom{167}{0}\binom{67}{4} = \binom{235}{4} - \binom{171}{4}$$
