2024 AMC 10B Problems/Problem 17: Difference between revisions

Revision as of 12:06, 14 November 2024

Problem

In a race among $5$ 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?

$\textbf{(A) } 180 \qquad\textbf{(B) } 361 \qquad\textbf{(C) } 420 \qquad\textbf{(D) } 431 \qquad\textbf{(E) } 720$

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

Solution 2 (Solution 1 but less words)

Split the problem into cases. A tie of $n$ snails has $\dbinom{5}{n}$ ways to choose the snails that are tied, $6-n$ ways to choose which place they tie for, and $(5-n)!$ to place the remaining snails.

1. No tie $\implies5!=120$ 2. Tie of 2 snails $\implies\dbinom{5}{2}\cdot4\cdot3!=240$ 3. Tie of 3 snails $\implies\dbinom{5}{3}\cdot3\cdot2!=60$ 4. Tie of 4 snails $\implies\dbinom{5}{4}\cdot2=10$ 5. Tie of all 5 snails $\implies1$

The answer is $120+240+60+10+1=\boxed{\text{(D) }431}$ ~Tacos_are_yummy_1

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/c6nhclB5V1w?feature=shared

~ Pi Academy

@@ Line 22: / Line 22: @@
 ~lprado
+==Solution 2 (Solution 1 but less words)==
+Split the problem into cases. A tie of <math>n</math> snails has <math>\dbinom{5}{n}</math> ways to choose the snails that are tied, <math>6-n</math> ways to choose which place they tie for, and <math>(5-n)!</math> to place the remaining snails.
+. No tie <math>\implies5!=120</math>
+. Tie of 2 snails <math>\implies\dbinom{5}{2}\cdot4\cdot3!=240</math>
+. Tie of 3 snails <math>\implies\dbinom{5}{3}\cdot3\cdot2!=60</math>
+. Tie of 4 snails <math>\implies\dbinom{5}{4}\cdot2=10</math>
+. Tie of all 5 snails <math>\implies1</math>
+The answer is <math>120+240+60+10+1=\boxed{\text{(D) }431}</math> ~Tacos_are_yummy_1
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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