How many ways to divide 3 medals (only) in a group of ten people.

Question

3 medals: Gold, Silver and Bronze.

Gold- $1^{st}$ place
Silver- $2^{nd}$ place
Bronze- $3^{rd}$ place

How many lists of winners are possible?

I did it like this: $(10\cdot9\cdot8)\cdot3$

but the answer for some reason in only ($10\cdot9\cdot8$)

I don't understand why. There are $10$ options to get a bronze medal then $9$ silver then $8$ bronze.
It is possible to change the order of medal received $3$ times.
Hence I multiplied it by $3$. But the answer is only ($10\cdot9\cdot8$)

Answer 1

Let's pick the person with gold medal, there are $10$ options.

Let's pick the person with the silver medal, the person who won the gold medal can no longer receive the prize, there are $9$ options.

Last, assign the bronze medal, there are $8$ options.

We have completely who receive which medal, by multiplication principle, there are $10\cdot 9 \cdot 8$ ways.

Answer 2

There are 10 options to get a bronze medal then 9 silver then 8 bronze

The fact that you're saying "X then Y then Z" means that you're already assigning an order to the medals, so you don't need to multiply your count again to account for order.

If we counted this in an order-agnostic manner first, then we would say there are $\binom{10}{3}$ ways to give the 3 medals to any $3$ of the $10$ people, and then $3!$ ways to reorder these medals, giving $$\binom{10}{3}\cdot 3! = 720 = 10 \cdot 9 \cdot 8$$ ways. Both approaches give the same result.

Answer 3

hope this might help (bit long for a comment). I think the key thing to understand is that when, in effect, you are using the multiplication principle for how many choices you have in a sequence of independent decisions, you are counting all possible orderings. Consider for example the set of all possible ordered $2$-tuples you can make from the set $\{a_1,a_2,a_3\}$:

$$\begin{matrix} (a_1,a_1) & (a_1,a_2) & (a_1,a_3) \\ (a_2,a_1) & (a_2,a_2) & (a_2,a_3) \\ (a_3,a_1) & (a_3,a_2) & (a_3,a_3) \\ \end{matrix}$$

Here you can see that $(a_2,b_1)$ is being counted separately to $(a_1,b_2)$, (however there is only one entry each time an element is selected twice). From this example, if you delete the repeated elements you can see we have $3 \times 2$ arrangements (the off-diagonals), in which we have fully accounted for all possible re-orderings.

Now in your case you are making ordered $3$-tuples from $10$ elements, without replacement, which, if represented like the above, would give rise to a $10 \times 10 \times 10$ cube, from which we would have to delete all the ordered $3$-tuples where elements have been repeated. My mathjax isn't good enough to represent that :) but the $10 \times 9 \times 8$ calculation will fully count all the possible orderings you need excluding the repeats. This has been stated of course in the answers but just trying to give a visualisation of this important feature of the multiplcation principle - it counts ordered selections, with or without replacement, and every ordering of the sample contributes!

Addendum: And just to be explicit

• The people are labelled / distinguishable
• The positions into which they are being placed are labelled / distinguishable
• The physical process of when they receive their medals is immaterial, all that matters is what kind of medal they receive.

The multiplication principle deals with everything for you given the above context.