How i can solve the combinatorial problem, using only the multiplicative and additive principle?

Question

I have a statement:

In a class of 10 students, $3$ prizes will be awarded.

What is the total combination of awards, if:

$i)$ Each player receives a SINGLE prize and the prizes are different.

$ii)$ Each player receives a SINGLE prize and the prizes are the same.

$iii)$ A player can repeat all the prizes, and the prizes are the same.

Well, my development was:

Let the prizes, be called $A, B, C$

$i)$ The price $A$ have 10 players to be distributed, $B$ have 9 players, $C$ have 8 players, so the total cases are: $10*9*8 = 720$

$ii)$ The same argument for $i)$, but now the order doesn't matter, so I can assume that: $ABC = BCA = CBA$, because $A = B = C $....

Therefore, to undo the repeated combinations i need the permutation of $3$, so the total combinations will be: $\frac{10*9*8}{3!} = 120$

$iii)$ Well, with the formula of $C_{repetition}(n, r) = \frac{(n+r-1)!}{(n-1)!r!}$

The result will be: $C_{rep}(10, 3) = \frac{12!}{9!3!} = 220$

The three results are good, but with the exercise $iii)$ i have some problems of understanding.

$A)$ ¿How i can get the correct result of the third exercise, using only the multiplicative and additive principle, in a intuitive way?

$B)$ ¿What is the intuitive and the formal proof of the formula: $\frac{(n+r-1)!}{(n-1)!r!}$ ?

Answer 1

One way is to break the $iii)$ into three scenarios: 1) Three players get a prize. 2) Two players get a prize (one gets two). 3) One player gets three prizes. Case one is the same as $ii)$. Case two is $10*9$. Case three is simply $10$.

Answer 2

This is a tricky question to answer in general. Consider the following related problem:

There are $n$ distinct people and $r$ distinct hats. Each hat must be placed on a person. It is allowed to stack multiple hats on a single person; in the case, the order of the stack matters.

How many ways are there to place all the hats?

We can solve this using the multiplication principle. Place the hats one at a time, in the order hat 1, hat 2, ... , hat $r$.

1. Hat $1$ can be placed in $n$ ways, on any of the $n$ people.
2. Hat $2$ can either be placed directly on top of any of the $n$ heads, or on top of hat $1$. This is $n+1$ choices.
3. Hat $3$ can be placed on any of of the $n$ heads, or on hat $1$, or on top of hat $2$. This is $n+2$ choices.

You see the pattern. The $i^\text{th}$ hat has $n+i-1$ choices, so the number of ways is $$ n(n+1)(n+2)\cdots (n+r-1) $$

Now, back to the problem of giving $r$ identical prizes to $n$ distinct people. This is just like the previous problem, but the $r$ items begin given out are no longer distinct. Therefore, we must divide by $r!$ in order to get rid of this over counting, resulting in $$ \frac{n(n+1)(n+2)\cdots (n+r-1)}{r!}=\binom{n+r-1}{r} $$

Note the analogy between cases (i) and (ii); the result for (ii) was obtained by taking the result for (i) and dividing by $3!$ to account for the fact that we do no care about order.