Combinatorics - Distribution of Coins I am preparing for a competitive exam. Sorry if the question sounds naive.
The Below example was listed when I was studying Fundamental Principle of Counting which is

If you must make a number of separate decisions, then MULTIPLY the
  number of ways to make each individual decision to find the number of
  ways to make all the decisions

The below question was not solved using the above technique,instead they listed all the possibilities.
My question is why the below problem cannot be solved using the above technique?
If yes, then how?
Question:
In how many ways 6 identical coins can be distributed among Alex,Bea and Chad?
NOTE: Some people may recieve zero coins?
Answer is 28
 A: The fundamental principle of counting works when decisions are independent, in other words if taking one decision does not affect the choices for the others. In this case this does not work. If we give $6$ coins to Alex we are forced to give $0$ coins to everyone else.
This problem in particular can be solved through the method of stars and bars, which itself relies on the "binomial coefficients" (which can be calculated with help of the fundamental principle of counting).
The idea behind stars and bars is that we have some objects (stars, although coins in this case) that we need to distribute amongst some people. Suppose we have six stars and three people.
Then we can represent each distribution symbolically as in the following example:
$***|**|*$
In this example Alex gets three stars, Bea $2$ and Chad $1$.
$*|**|***$
In this example Alex gets one star, Bea $2$ and Chad $3$.
As a final example consider the following:
$******||$
In this example Alex gets $6$ stars, Bea $0$ and Chad $0$.
So we have  transformed the problem of distributing $n$ stars between $k$ people to the problem of placing $n$ stars and $k-1$ bars in a line.
Since we have $n+k-1$ elements in total and $k-1$ bars there are $\binom{n+k-1}{k-1}$ ways to do this.
in your case there are $6$ stars(coins) and $3$ persons so the answer is $\binom{6+3-1}{3-1}=\binom{8}{2}=\frac{8\cdot7}{2\cdot 1}=28$

Note that $\binom{n}{k}$ represents a binomial coefficient, you can find out more about binomial coefficients and how to calculate them here
A: I think the answer is $n+k-1\choose n $
where n= 6 , and k=3
since, we have no limit about how much coin we can gave to each person !
A: Here's a solution using the FPC:
Imagine Alex gets $a$ coins. The problem is then how many ways can we divide $6-a$ coins between Bea and Chad, which is a simple partition.
Imagine the $o$'s are coins:
$$o\;o\;o\;o\;o\;o$$
There are $7$ possible partition places, so for an example if $a=0$ and we have $oo|oooo$, then the distribution of coins is Alex:$0$, Bea:$2$ and Chad:$4$.
Next we observe that if Alex has $a$ coins, then there are $7-a$ possible partitions of the $6-a$ remaining coins. This allows us to group the number of coins Alex gets into $\{0\}, \{1,6\}, \{2,5\}, \{3,4\}$, with a total of $7$ possible partitions in each, hence the answer, by the FPC, is $4\times7=28$.
