What does the multinomial formula state? (Formula provided)

Question

The statement of the formula is kind of cryptic to me. In particular, I don't understand how the formula relates the probability of an object, designated by the index i, occurring $n_i$ times since the formula isn't in terms of the index i at all.

Perhaps it's best to demonstrate how much I know about the formula through a usage case:
Let $S = \{1,2,3,4,5,6\}$ and suppose you're rolling an unfair die such that
$P(1)=1/6$
$P(2)=2/6$
$P(3)=3/6$
$P(4)=P(5)=P(6)=0$

You roll the die 10 times in "trials", with different number of rolls in each trial:
Trial 1: 1
Trial 2: 2
Trial 3: 3
Trial 4: 4

Would the formula allow me to find the probability of rolling the number 2, say 7 times?

SOURCES: 2 (at the very bottom)

Answer 1

1. First let me explain the multinomial theorem.Let say you have $n$ independent events and each of the event can have any one of the $r$ outcomes $\{s_1,s_2....s_r\}$ with probability $\{p_1,p_2....p_r \}$ respectively.Now you want the probability that the outcome was $s_i$, $n_i$ times. (ie. $n_1+n_2....n_r=n$). So basically You have total $n$ things of which $n_1$ are of one type , $n_2$ of another .... so on. So number of ways of ordering / permuting such things ( as you might be knowing this formula ) $C=\frac{n!}{n_1!n_2!..n_r!}$. What ever the order might be the probability associated with it will be $P={p_1}^{n_1}{p_2}^{n_2}...{p_r}^{n_r}$. So answer becomes $C.P$. Now you may ask why the name multinomial its so because the co-efficient of ${x_1}^{n_1}{x_2}^{n_2}...{x_r}^{n_r}$ in $(x_1+x_2+...x_r)^n$ where $n=n_1+..n_r$ is $\frac{n!}{n_1!n_2!..n_r!}$. ( when r=2 it reduces to binomial formula with which you might be familiar).
2. Now to the question there are 10 events on any exact 7 events 2 has to occur and on rest 3 any number other than 2 can occur.( This is a special case of multinomial, the binomial, as $r=2$ ).Choose any 7 out of 10 events give them 2 as outcome with probability $\frac{2}{6}$ each and rest 3 get any other number with probability $1-\frac{2}{6}$ each. So answer is $\frac{10!}{7!3!}{(\frac{2}{6})}^7{(1-\frac{2}{6})}^3$.(The events were independent so probability got multiplied). Now why $r=2$ here because you have divided the outcome space into two parts: (1) outcome is 2 (2) outcome is not 2.
   
   Hope it answers your question.