2024 AMC 10B Problems/Problem 17: Difference between revisions
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~lprado | ~lprado | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | |||
https://youtu.be/c6nhclB5V1w?feature=shared | |||
~ Pi Academy | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2024|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:04, 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
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/c6nhclB5V1w?feature=shared
~ Pi Academy
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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