Multinomial Coefficients Confusion As far as I know for binomial coefficients, we can express one as either $\binom {n} {k}$ or $\binom {n} {k,\ n-k}$. If I'm not wrong they both mean the same thing: $\frac{n!}{k!(n-k)!}$
What about multinomial coefficients? If I have an expression $\binom {n} {i,\ j}$, is it the same as $\binom {n} {i,\ j,\ n-i-j}$? Which one of the two is $\frac{n!}{i!j!n-i-j!}$ and which one's $\frac{n!}{i!j!}$? I'm confused.
Thanks.
 A: You will get a more intuitive understanding by considering dividing people into labelled groups.
If $10$ people are to be divided into two labelled groups $A$ and $B$ of, say, $7$ and $3$,
we can use the binomial coefficient to write $\binom{10}7$ or the multinomial coefficient to write $\binom{10}{7,3}$
If we are to divide them into three groups of $5$, $3$, and $2$,
we can write  $\binom{10}{5,3}$ or $\binom{10}{5,3,2}$ because the last group gets formed automatically with the residue.
The point is that the last part can always be left out of the expression,
and conventionally it is left out when there are two groups, but is left to taste when more than two groups are there.
In $\binom{n}{i,j} \equiv \binom{n}{i,j,n-i-j}$
the $RHS$ is the full expression for $3$ groups with the last group having $0$ members,
and the LHS is the abridged form for the same grouping.
A: As hardmath, I also recommend to read what is stated as definition. Anyway the following can be said (I cite Wikipedia:Multinomial Threorem):

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:
  $$
    (x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\,,
$$
  where
  $
    {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}
$
  is a multinomial coefficient.

So $\binom {n} {i,\ j}$, is not it the same as $\binom {n} {i,\ j,\ n-i-j}$, since for the first $j=n-i$ holds. 
An example: You get $\binom{10}{2,3,5}$, but $\binom{10}{2,3}$ doesn't make sense... 
