5
$\begingroup$

An Olympic final consists of two runners from each of four different countries. An ordered selection of three of the eight runners win the medals (gold, silver, and bronze). How many different possibilities are there if the medal winners are all from different countries?#

They said its 192 (8x6x4) but I think its 192 * 3 since it's an ordered selection no?

Edit: My confusion came from a different variation of the question. How many different possibilities are there if two of the medal winners are from the same country? The answer for this was 144 and to my knowledge because of 8x1x6 and x3 for the order

$\endgroup$
4
  • 6
    $\begingroup$ The gold winner is chosen from one of the countries ($4$ choices for country, $2$ choices for gold winner, so $8$ choices). Given that, the silver winner must be from one of $3$ countries (so $6$ choices). That leaves $4$ runners to choose the bronze winner from. The order is taken into account already. $\endgroup$
    – lulu
    Commented Jan 15, 2023 at 16:36
  • 2
    $\begingroup$ Thanks for taking the time to state your question so clearly. $\endgroup$
    – MJD
    Commented Jan 15, 2023 at 16:40
  • 1
    $\begingroup$ @MJD is that sarcastic :( $\endgroup$
    – Tingo Hugo
    Commented Jan 15, 2023 at 16:45
  • 2
    $\begingroup$ Not at all sarcastic. $\endgroup$
    – MJD
    Commented Jan 15, 2023 at 17:35

1 Answer 1

7
$\begingroup$

Addendum added to respond to the question of Tingo Hugo.


$8 \times 6 \times 4,~$ does represent an ordered selection. I would equivalently express it as follows:

  • There are $4$ choices for which team wins the gold, then $3$ choices for the silver, and then $2$ choices for the bronze.

  • Within each of the three winning teams, there are two choices for which runner wins the medal.

$$4! \times 2^3 = 8 \times 6 \times 4. \tag1 $$

The RHS of (1) above may be alternatively reasoned by:

  • There are $8$ choices for who wins the gold.

  • Then, since the other runner of the team that won the gold can't win the silver, there are then $6$ choices for who wins the silver.

  • Similarly, there are then $4$ choices for who wins the bronze.


Addendum
Responding to the question of Tingo Hugo.

How many different possibilities are there if two of the medal winners are from the same country? : the answer is $144 ~(8*1*6) * 3~$?

If two of the medal winners are from the same country, I would compute the possibilities as follows:

  • There are $4$ choices for which team wins two medals.

  • There are then $3$ choices for which other team only wins one medal.

  • Then, there are $~3~$ choices for which medal the other team wins.

  • Then, there are $~2~$ choices for how the two medals are distributed among the two runners of the team that won two medals.

  • Then, there are $~2~$ choices for which of the two runners wins the third medal, on the other team, which only won one medal.

Putting this all together, the computation is

$$4 \times 3 \times 3 \times 2^2 = 144.$$

The somewhat offbeat equivalent expression of

$$8 \times 1 \times 6 \times 3$$

can be explained as follows.

The factor of $3$ determines which medal was won by the other team, that only won 1 medal.

Assume that this other medal is the bronze.

Then, there are $8$ choices for which person wins the gold medal.

Then, there is only $1$ choice for which person wins the silver medal. This is because it is being assumed that the gold and silver medal were won by the same team.

Then, there are $6$ choices for who wins the bronze medal.

This explains the computation of

$$8 \times 1 \times 6 \times 3.$$

$\endgroup$
3
  • $\begingroup$ Oh, I don't quite understand because for a different variation of the question, 'How many different possibilities are there if two of the medal winners are from the same country?' the answer is 144 (8*1*6) * 3? $\endgroup$
    – Tingo Hugo
    Commented Jan 15, 2023 at 16:42
  • 1
    $\begingroup$ @TingoHugo See the Addendum that I have just added to the end of my answer. $\endgroup$ Commented Jan 15, 2023 at 16:56
  • $\begingroup$ thank you very much $\endgroup$
    – Tingo Hugo
    Commented Jan 15, 2023 at 17:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .