# How many ways are there for three medals to be awarded if ties are possible?

The Question

There are six runners in the 100-yard dash. How many ways are there for three medals to be awarded if ties are possible? (The runner or runners who finish with the fastest time receive gold medals, the runner or runners who finish with exactly one runner ahead receive silver medals, and the runner or runners who finish with exactly two runners ahead receive bronze medals.)

My Attempt

I saw a few cases:

Case I: No Ties

P(6,3) = 120 ways to pick the gold medal

Case II: 2 people tie

First we must pick the two people who tie, this can be done C(6,2) = 15 ways. Now I have to pick a medal for them to win, which opens up more cases because if they tie for first no silver medal is awarded and if they tie for second no bronze medal is awarded.

My Problem

My problem with this method is that it takes way too long to consider all the cases and wouldn't be practical if I was writing a test, or if I was designing an algorithm for this type of question. I'm wondering if there is a more efficient way to solve this?

My book gave the answer $873$ if that helps at all

• And I guess, in theory, you could have 3-way, 4-way, 5-way, and 6-way ties too! If 4 people tied for first, it seems that that would break the question, since more than 3 medals would be needed. I guess the question really becomes: How many ways can 6 things be partitioned? – JLee Nov 7 '14 at 1:05
• So what you're saying is this is basically the question how many ways can I put 6 marbles in 3 distinct jars? – Dunka Nov 7 '14 at 1:12
• I think it's a little different dunka, because if 3 people tie for gold, the silver and bronze medals are not awarded. That's to say, you don't want to count the case in which there are 3 marbles in the first jar, then 2 in the second, then 1 in the third. – Shane Nov 7 '14 at 1:15
• @Shane very true. Should I proceed with cases, knowing that 4 way ties or more would actually break the questions rules? – Dunka Nov 7 '14 at 1:17
• I'm not sure, to be honest. I'm one of those annoying commenters who only points out problems with what you're doing rather than offering solutions! Hopefully somebody smarter than I will intervene at any moment now :P – Shane Nov 7 '14 at 1:18

Let $n$ be the number of gold medals awarded.

$\textbf{1)}\;\;$If $n=2$, then there are $\binom{6}{2}=15$ ways to choose the gold medalists and

$\hspace{.3 in}2^4-1=15$ ways to choose the bronze medalists, so there are $15\cdot15=225$ possibilities.

$\textbf{2)}\;\;$ If $n=1$, there are $\binom{6}{1}$ ways to award the gold medal, and then

there are $\binom{5}{1}=5$ ways to award one silver medal and $2^4-1=15$ ways to award bronze medals, and

there are $\binom{5}{2}+\cdots+\binom{5}{5}=2^5-\binom{5}{0}-\binom{5}{1}=26$ ways to award more than one silver medal;

so in this case there are $6[5\cdot15+26]=606$ possibilities.

$\textbf{3)}\;\;$ If $n\ge3$, there are $\binom{6}{3}+\cdots+\binom{6}{6}=2^6-\binom{6}{2}-\binom{6}{1}-\binom{6}{0}=42$ ways to award the gold.

Therefore there are a total of $225+606+42=873$ ways to award the medals.

• @Shane this was the answer we were looking for! – Dunka Nov 9 '14 at 0:34

Maybe the best way to break it down is by cases likes this:

1. 6 win gold (1 possibility)
2. 5 win gold (6 possibilities)
3. 4 win gold (${6 \choose 2}=15$ possibilities)
4. 3 win gold (${6 \choose 3}=20$ possibilities)
5. 2 win gold, four win bronze (15 possibilities)
6. 2 win gold, three win bronze (60 poss.)
7. 2 win gold, two win bronze (90 poss.)
8. 2 win gold, one wins bronze (60 poss.)
9. 1 wins gold, 5 win silver (6 poss.)
10. 1 wins gold, 4 win silver (30 poss.)
11. 1 wins gold, 3 win silver (60 poss.)
12. 1 wins gold, 2 win silver (60 poss.)
13. 1 wins gold, 1 win silver, 4 win bronze (30 poss.)
14. 1 wins gold, 1 win silver, 3 win bronze (120 poss.)
15. 1 wins gold, 1 win silver, 2 win bronze (180 poss.)
16. 1 wins gold, 1 win silver, 1 win bronze (120 poss.)

Then we have

$$1 + 6 + 15 + 20 + 15 +60+90+60+6+30+60+60+30+120+180+120= 873$$

Bingo!

• This may be right. My book gives the answer of $873$ however, this is still a time consuming method. I'm wondering if there is an easier way than cases to do this. – Dunka Nov 7 '14 at 1:34
• Ooh... now you've given me an answer... let's see if I can get to it. Here come some edits! – Shane Nov 7 '14 at 1:36
• Now the goal is for somebody smarter to show us how to do this faster. Anybody? – Shane Nov 7 '14 at 1:53
• Thanks for the solution. However, I still want to know if there is a better way than cases to do this one. – Dunka Nov 7 '14 at 1:54
• I would be curious to see that also. In particular, if there are $n$ runners rather than $6$, my approach wouldn't be practical. I'm sure you express the above as some rather complicated sum of sums... – Shane Nov 7 '14 at 1:57