N choose k choose j The Android mobile game "Pocket Tanks" has 295 unique weapons. For each match, 20 weapons are inserted into a list. Of those 20, players alternately draw weapons until all are exhausted, leaving each player with 10 weapons.
I'm trying to think of how many combinations of weapons exist. Is this "(295 choose 20) choose 10"? If so this evaluates to an extremely large number. It's large enough that "(295 choose 20) choose 10" in WolframAlpha can't evaluate. You have to calculate the first parenthetical, then evaluate the remaining piece.
 A: As you thought, it would actually be $\binom{295}{20} \binom{20}{10}$. This is because you're selecting every combination of 20 and then every configuration of which 10 weapons the first player received.
This can alternatively be represented as a multinomial with three categories.
$$
\binom{n}{k_1, k_2, k_3}
$$
$n = $ Total number of weapons $ = 295$
$k_1 = $ Player 1's Weapons $= 10$
$k_2 = $ Player 2's Weapons  $= 10$
$k_3 = $ Unused Weapons  $= 275$  
So, the total number of combinations is
$$
\binom{295}{10, 10, 275} = \frac{295!}{10!\ 10!\ 275!}
$$
This is still ludicrous to evaluate, but since it's a division, you can split up some of the work so that Wolfram Alpha doesn't pass out trying to evaluate it.
Edit:
In fact, representing it like this allows Wolfram Alpha to tackle it directly without passing out and gives you a result.
$$
\binom{295}{10, 10, 275} = 979,092,999,029,074,303,631,255,812,346,789,256
$$
Source (Wolfram Alpha link)
A: The number of combinations is $$\binom{295}{20}\binom{20}{10}=\frac{295!}{10!10!275!}$$ 
In the first expression the first factor represents the number of possibilities of choosing $20$ out of $295$. The second factor represents the number of possibilities of choosing $10$ out of $20$.
Note that in the second expression number $20$ is not present (except as summation of $10$ and $10$). This because the 'step in between' of inserting them in a list is somehow irrelevant.
A: The number of ways to choose 20 weapons from 295 is $295\choose20$. The number of ways to choose 10 weapons from 20 is $20\choose10$, which is the same number as possible matches given a set of 20 weapons, since 10 weapons go to both players. So the total number of possible matches is ${295\choose20}{20\choose10}$.
