Distributing $3$ identical prizes to $10$ players

Question

I have a statement that says:

In a competition, $3$ prizes will be distributed among 10 players.

How many ways can it be done if:

I) All prizes are different

II) All prizes are equal

I) The first prize can be dealt to $10$ players, the second to $9$ players and the third to $8$ players, so the answer is $10 \cdot 9 \cdot 8 = 720$ and, according to my text, is correct.

II) Here the prizes are indistinguishable, but here, I do not know what to do. I tried using the same logic of the first exercise, but that did not work, and I do not know why. I know I can use combinatorics, but that would be solving it without understanding it. Do not use any formula to explain it to me, please, just logic.

Answer 1

Let the first three players with the winner. Their possible rankings are

$$\{(1,2,3), (1,3,2), (2,1,3),(2,3,1),(3,1,2),(3,2,1)\}$$

However the order doesn't matter if the prizes are indistinguishable, hence those $6$ configurations are indistinguishable to us, hence we need to divide by $3!$.

$$\frac{720}{3!}=120.$$

Answer 2

Mattiu, to solve your question:

In the first case, as you did $10*9*8$ what you are really doing is this: If you have just 4 people, say $A,B,C,D$ then your argument would say that we have $4*3*2$ this first $4$ let us choose $A,B,C,D$ then let's pick $A$ then let's pick $B$ out of the $3$ and then pick $C$ out of the $2,$ then the prices are given (in order) to $A,B,C.$
Now, we can pick the same people in a different order, say first $B,$ then $A,$ and for last $C$ hence $B,A,C.$ the problem is that all of this possibilities are the same in the second problem, and hence you have $3!$ possibilities to pick the same people.

So each solution that you have without order say $\{a_1,a_2,a_3\}$ you can associate it with a function that takes care of the order, and so an ordered pick is a tuple $(\{a_1,a_2,a_3\},f)$ with $f$ a bijection from the set $\{1,2,3\}$ to the set $\{1,2,3\}$ and there are $3!$ of those and hence

$$Ordered = Unordered\times \hspace{1mm}Bijection$$ and so $4*3*2 = Unordered*\hspace{1mm}3!$ and hence $\frac{4*3*2}{3*2}=Unordered.$
Just for completion, the Ordered case is called a permutation and the unordered is called a Combination and can be computed using binomial numbers $n\choose k$ (way to choose $k$ elements out of $n$ elements without order).