The phrasing of the question is technically ambiguous whether or not the 100 chocolates are considered identical or not. In almost every occurrence of a problem phrased in the way you present it, it is intended to be assumed that the chocolates are identical for all intents and purposes and the only thing we care about it the number of chocolates that each person receives.
This is a classic example of a problem solved by Stars and Bars.
The short explanation is that we can represent any manner of distributing the chocolates as a sequence of $\star$'s and $\mid$'s.
For a smaller example of only eight chocolates and three people the sequences
$$\star\mid\star\star\star\mid\star\star\star\star$$
corresponds to the first person getting one, the second person getting three, and the third person getting four
and
$$\star\star\star\star\star\star\mid\star\mid\star$$
corresponds to the first person getting six and the second and third persons each getting one.
Since each person gets at least one chocolate, we try to pick how many ways in which we can pick the locations of the bars in our stars-and-bars argument. Convince yourself that in any valid arrangement of stars and bars where it corresponds to everyone getting at least one star that no two bars can be touching and no bar can be on the very edge.
Convince yourself also that any way of distributing chocolates can be uniquely described by such a stars and bars configuration and that any such stars and bars configuration corresponds uniquely to a way of distributing the chocolates. (I.e. there is a bijection between the two problems, implying the count must be the same for both)
If there are $100$ stars, how many valid locations are there for bars to go?
$99$ available locations
How many of these locations must we choose (simultaneously)?
$2$
How many ways are there then to distribute?
$\binom{99}{2}$
In the related but technically different question of when the chocolates are all distinct, you should approach via inclusion exclusion.
Letting $A_1,A_2,A_3$ be the sets of arrangements where the first person, the second person, and the third person don't receive any chocolates respectively, and $X$ is the set of arrangements of chocolates regardless, we are tasked with finding $|X\setminus (A_1\cup A_2\cup A_3)|$
By inclusion-exclusion principle, this becomes $|X|-|A_1\cup A_2\cup A_3|=|X|-|A_1|-|A_2|-|A_3|+|A_1\cap A_2|+|A_1\cap A_3|+|A_2\cap A_3|-|A_1\cap A_2\cap A_3|$
$|X|=3^{100}$ can be seen from a straightforward multiplication principle argument as each of the 100 distinct chocolates have three options for who will receive it.
$|A_i|=2^{100}$ again can be seen as if person $i$ doesn't receive any each chocolate has two choices on where to go.
$|A_i\cap A_j|=1^{100}=1$ as if person $i$ and person $j$ (with $i\neq j$) both receive no chocolates, all remaining chocolates must go to the remaining person.
$|A_1\cap A_2\cap A_3|=0$ trivially since it is impossible to distribute the chocolates to the three people and simultaneously none of the three get any chocolates.
This yields a final count of $3^{100}-3\cdot 2^{100}+3$ ways to distribute 100 distinct chocolates among three distinct people.
