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

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.

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

Then we have

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

Bingo!

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