Difference between revisions of "2024 AMC 10B Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | We perform casework based on how many | + | 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>. |
− | <math>5</math> | + | <math>5</math> snails tied: All <math>5</math> snails tied for <math>1</math>st place, so only <math>1</math> way. |
− | <math>4</math> | + | <math>4</math> snails tied: <math>A,B,C,D</math> all tied, and <math>E</math> either got <math>1</math>st or last. <math>{5}\choose{1}</math> ways to choose who isn't involved in the tie and <math>2</math> ways to choose if that snail gets first or last, so <math>10</math> ways. |
− | <math>3</math> | + | <math>3</math> snails tied: We have <math>ABC, D, E</math>. There are <math>3! = 6</math> ways to determine the ranking of the <math>3</math> groups. There are <math>5\choose2</math> ways to determine the two snails not involved in the tie. So <math>6 \cdot 10 = 60</math> ways. |
− | <math>2</math> | + | <math>2</math> snails tied: We have <math>AB, C, D, E</math>. There are <math>4! = 24</math> ways to determine the ranking of the <math>4</math> groups. There are <math>5\choose{3}</math> ways to determine the three snail not involved in the tie. So <math>24 \cdot 10 = 240</math> ways. |
− | It's impossible to have "1 | + | It's impossible to have "1 snail tie", so that case has <math>0</math> ways. |
− | Finally, there are no ties. We just arrange the <math>5</math> | + | Finally, there are no ties. We just arrange the <math>5</math> snail, so <math>5! = 120</math> ways. |
The answer is <math>1+10+60+240+0+120 = \boxed{431}</math>. | The answer is <math>1+10+60+240+0+120 = \boxed{431}</math>. | ||
~lprado | ~lprado |
Revision as of 01:17, 14 November 2024
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