In a race with 10 competitors, the first-place finisher receives a gol

Here the order matters. Why?

Say there are 3 competitors, A, B, C for 2 medals: Gold(G) & Silver(S)

Winners can be:
G S
---
A B
B A
A C
C A
B C
C B

"A" winning the Gold medal & B winning the silver is different from "B" winning the gold & "A" winning the silver. Thus both scenario should be considered.
A B
B A

So, you choose 2 people out of 3 and rearrange them in 2! way.

\(C^3_2*2!=6\)

Likewise, if there are 10 people; you choose 3 winners first i.e. \(C^{10}_3\)
And then re-arrange those 3!.
Total = \(C^{10}_3*3!=720\)

*************************************
If someone says make 2 teams from 3 people A, B, C, then \(C^3_2\) suffices.

Note:
Having a team that contains A and B
IS NO DIFFERENT FROM
Having a team that contains B and A

Thus, order won't matter there.
********************************

we have 10 people and three medals.

so these three medals can be considered as three slots to fill.

one slot is Gold,one slot is Silver and last slot is Bronze.


Order matters here why,
if we seat one person in one place that gives one arrangement and if we re arrange this it gives a different arrangement (eg - Person1 sitting in Gold slot vs Person sitting in Bronze slot - different arrangements. hence order matters).

So its permutations and number of arrangements = 10*9*8 = 720

I too get confused many times with order. good Question

Total Competitors =10
ways of selecting the Gold medal winner = 10 ways (we can select anyone)
ways of selecting the Silver medal winner = 9 ways (we can select anyone but not the Gold medal winner)
ways of selecting the Bronze medal winner = 8 ways (we can select anyone but not the 2 winners already choosen)

so total no of ways will be = 10 x 9 x 8 = 720

hope this helps......

When I first read the topic, I thought GMAT Club was having some competition about solving MGMAT ps questions for 10 people and they were awarding medals. Then I read the post and thought, "why do these competition rules have numbers at the bottom?" *smacks head.

I thought the same.

don't fall into trap of using anagram grid:
10!/3!7!=120. this is wrong, because we are not looking in how many orders can the medals be distributed
but we are looking for number of ways winners can get their medals, so simple permutation is fine

10!/7!=8*9*10=720

This problem can be solved in two ways.

First realize that the number of ways to award a first place medal is 10 because there are 10 entrants.

The total number of possibilities to award a second place medal is 9 because there are 9 remaining participants.

Lastly, the total number of ways to award a third place medal is 8 because there are 8 remaining runners after the first two medals have been awarded.

10*9*8 = 720.

Alternatively, using combinatorics, you can realize that there are 10! ways to arrange the 10 racers. But you've over-counted, so you need to remove the 7! racers who are all collectively "losers." Their order doesn't matter since they are in the group together.

10!/7! = 720.

the answer must be C , actually here the order does matter means whoever comes first will be awarded the gold and so on , so order matter . now we have 10*9*8=720 ways to choose for subsequent outcomes
hence C

Since there are 10 competitors and three positions in which they can place, the number of ways the medals can be awarded is 10P3 = 10 x 9 x 8 = 720.

Answer: C

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