Difference between revisions of "2024 AMC 10B Problems/Problem 17"
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| + | ==Problem== | ||
| + | In a race among <math>5</math> snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible? | ||
| + | <math>\textbf{(A) } 180 \qquad\textbf{(B) } 361 \qquad\textbf{(C) } 420 \qquad\textbf{(D) } 431 \qquad\textbf{(E) } 720</math> | ||
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==Solution 1== | ==Solution 1== | ||
We perform casework based on how many snails tie. Let's say we're dealing with the following snails: <math>A,B,C,D,E</math>. | We perform casework based on how many snails tie. Let's say we're dealing with the following snails: <math>A,B,C,D,E</math>. | ||
Revision as of 00:20, 14 November 2024
Problem
In a race among 
snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?
Solution 1
We perform casework based on how many snails tie. Let's say we're dealing with the following snails: 
.
snails tied: All snails tied for 
st place, so only way.
snails tied: 
all tied, and 
either got st or last. 
ways to choose who isn't involved in the tie and 
ways to choose if that snail gets first or last, so 
ways.
snails tied: We have 
. There are 
ways to determine the ranking of the groups. There are 
ways to determine the two snails not involved in the tie. So 
ways.
snails tied: We have 
. There are 
ways to determine the ranking of the groups. There are 
ways to determine the three snail not involved in the tie. So 
ways.
It's impossible to have "1 snail tie", so that case has 
ways.
Finally, there are no ties. We just arrange the snail, so 
ways.
The answer is 
.
~lprado
