2024 AMC 10B Problems/Problem 17

Solution 1

We perform casework based on how many snails tie. Let's say we're dealing with the following snails: $A,B,C,D,E$ .

$5$ snails tied: All $5$ snails tied for $1$ st place, so only $1$ way.

$4$ snails tied: $A,B,C,D$ all tied, and $E$ either got $1$ st or last. ${5}\choose{1}$ ways to choose who isn't involved in the tie and $2$ ways to choose if that snail gets first or last, so $10$ ways.

$3$ snails tied: We have $ABC, D, E$ . There are $3! = 6$ ways to determine the ranking of the $3$ groups. There are $5\choose2$ ways to determine the two snails not involved in the tie. So $6 \cdot 10 = 60$ ways.

$2$ snails tied: We have $AB, C, D, E$ . There are $4! = 24$ ways to determine the ranking of the $4$ groups. There are $5\choose{3}$ ways to determine the three snail not involved in the tie. So $24 \cdot 10 = 240$ ways.

It's impossible to have "1 snail tie", so that case has $0$ ways.

Finally, there are no ties. We just arrange the $5$ snail, so $5! = 120$ ways.

The answer is $1+10+60+240+0+120 = \boxed{431}$ .

~lprado

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2024 AMC 10B Problems/Problem 17

Solution 1