There are $10$ drama episodes, $8$ comedy, $3$ news bulletins, and $6$ animations. A user is purchasing $5$ episodes. How many sets of $5$ episodes are there in which all four categories are purchased?
I have done the following :
$c(10,2)c(8,1)c(3,1)c(6,1) + c(10,1)c(8.2)c(3,1)c(6,1) + c(10,1)c(8,1)c(3,2)c(6,1) + c(10,1)c(8.1)c(3,1)c(6,2)$
which I calculate as a total of $16560$.
But thinking about it shouldn't it also be:
$c(10,1)c(8,1)c(3,1)c(6,1)c(23,1)$ which I calculate as $33120$.
So what am I doing wrong? Both answers make sense to me!
The first solution you wrote out is the correct one.
In the second solution, you are actually double-counting every possible outcome, as I'm sure you noticed that $33120$ was $2(16560)$.
Let's see why this is happening. Using the process of the second solution, notice that we are choosing the last episode out of the remaining $23$. But if we fail to choose any particular episode the first time through (as in, choosing it before we choose the last episode from the $23$ remaining) we have the opportunity to pick it again as the fifth one. Just for a quick example, say we picked comedy episode $A$ in the first four, then comedy episode $B$ as the fifth one. Another way this method is counting is picking $B$ in the first four then $A$ as the fifth one, which happens to be the same combination of episodes. That means using this method we are double counting the number of outcomes.