Expected Number of Outcomes for Tossing a Dice Four Times Such that at least one $i$ occurs

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There is a fair dice being tossed four times. Let $X_i$ for $1 \leq i \leq 6$ takes value of $1$ if there is at least one occurrence of $i$ and $0$ otherwise. Let $X$ be the number of different outcomes (if the four tosses are $4,4,6,1$, then $X$ equals $3$ since there are $3$ different outcomes.) What is the expectation of $X$?

I understand $X_i$ is Bernoulli variable with $\Pr(X_i=1) = 1-(5/6)^4$. But the answer says $X=\sum_1^6 X_i$ so $\mathbb E [X] = \mathbb E \left[\sum_1^6 X_i \right] = \sum_1^6 \mathbb E[X_i]$. I only do not understand why $X=\sum_1^6 X_i$? What prevents $X$ from equalling to $5$ or 6? Maybe $X_i$'s are not independent of each other but that shouldn't prevent their sum from adding up to $5$ or $6$?

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You are summing what's called indicator variables, thus $$X_i,1\leq i\leq 6$$ are either one or zero.

This sum is maximized when all the tosses are distinct, but there are only 4 tosses, thus the maximum possible values of this sum is 4. Stated another way, there are at least two zeroes among the 6 variables $X_i$ since you cannot have more than 4 distinct numbers appear in 4 tosses.

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Consider $6$ buttons labeled from $1$ to $6$. They light up if the corresponding number has been tossed and once they light up, they remains light up.

After the first toss, one of the button lights up and it stays lights up forever.

For the second toss, there are two possibilities, if the number appears during this toss appeared previously, no new button lights up. If a new number appears, a new button lights up. At this moment, at most two buttons can lights up.

Similar for the third toss, at most one more button can lights up.

The same thing applies for the fourth toss.

After the $4$-th toss, at most $4$ buttons would light out as during each round, either $0$ or $1$ new buttons lights up.

These buttons corresponds to $X_i$, $X_i=1$ if it lights up and equal to $0$ otherwise.

Hence $\sum_{i=1}^6 X_i \leq 4$.