百码冲刺

把“颁发方式”按最终出现的奖牌集合分成 4 类：

1）只颁发金牌。 2）颁发金牌和银牌。 3）颁发金牌和铜牌。 4）颁发金牌、银牌和铜牌。

分别计数：

1）只颁发金牌意味着第一名至少有 3 人并列（否则会出现银牌或铜牌）。 $$\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6}=42.$$

2）颁发金牌和银牌：必须只有 1 人获得金牌，其余人并列第二；并且并列第二的人数至少为 2（否则会出现铜牌）。 $$\binom{6}{1}\left(\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}\right)=156.$$

3）颁发金牌和铜牌：必须恰好 2 人并列第一（两枚金牌），并且至少 1 人获得铜牌（可以多人并列）。 $$\binom{6}{2}\left(\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}\right)=225.$$

4）颁发金银铜牌：必须恰好 1 人金牌、恰好 1 人银牌，并且至少 1 人铜牌（可并列）。 $$\binom{6}{1}\binom{5}{1}\left(\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}\right)=450.$$

四类互斥相加： $$42+156+225+450=\boxed{873}.$$

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Original Explanation

This question reduces to counting the number of ways in which final assignments can take place. Let's begin by enumerating the possible medals awarded.

1. We could have the most simple case, where only gold medals are awarded.
2. We could have the case where only gold and silver medals are awarded.
3. We could have the case where gold and bronze medals are awarded.
4. Lastly, we could have the case where all gold, silver, and bronze medals are awarded.

Let's now begin to count the number of ways in which each case could occur.

1: If only gold medals are awarded, then there must have been a tie for first place. Furthermore, since no silver or bronze medals are awarded, this further stipulates that the number of people who tied for first must either be 3, 4, 5, or 6 (this would ensure no other medals are awarded). We can count the number of ways to choose these many people from the total by the following sum: $${6 \choose{3}} + {6 \choose{4}} + {6 \choose5} + {6 \choose6} = 42$$

2: We next have the case where gold and silver medals are awarded. In order for this case to occur, however, there can only be $one$ gold medalist, and the remaining people must tie for second (otherwise, if we had multiple gold medalists, it would be impossible to have a silver medalist). Furthermore, amongst the remaining people who tie for second, we $must$ have at least two of them (otherwise, if there's only one, we'd have a bronze medalist as well). We can count the number of ways using the product rule for each subcase: $${6 \choose1}*[{5 \choose2} + {5 \choose3} + {5 \choose4} + {5 \choose5}] = 156$$

3: We now have the case where gold and bronze medals are awarded. Note that in order for this to happen, exactly two gold medals must be awarded, and at least one person must be awarded a bronze (we can have multiple people tie for bronze). This can happen as such: $${6 \choose2}*[{4 \choose1} + {4 \choose2} + {4 \choose3} + {4 \choose4}] = 225$$

4: Lastly, we have the case where gold, silver, and bronze medals are awarded. This can happen if exactly one person is awarded the gold, exactly one silver is awarded, and at least one person gets bronze (just as above, we can have multiple people tie for bronze). $${6 \choose1}{5 \choose1}*[{4 \choose1} + {4 \choose2} + {4 \choose3} + {4 \choose4}] = 450$$

Each of these cases is distinct, they can't occur at the same time. So summing over the possible ways each can occur gives us the total number of ways the medals can be awarded.

$$42 + 156 + 225 + 450 = \boxed{873}$$