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Difference between revisions of "2025 AIME II Problems/Problem 10"

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~Mitsuihisashi14
 
~Mitsuihisashi14
 
~Small latex fixes by Marcus :)
 
~Small latex fixes by Marcus :)
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== Solution 2: Stars-and-Bars ==
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Let <math> n </math> be the number of groups of adjacent people.
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Clearly, there will be <math> 8-n </math> groups with people, and there are <math> \binom{n}{8-n} </math> ways to select these groups.
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Now, we use the [https://artofproblemsolving.com/wiki/index.php/Ball-and-urn stars-and-bars technique] to find the number of ways to distribute the <math> 8 </math> remaining chairs around the <math> 8 </math> people. Since the <math> n </math> groups are separated, we must choose <math> n-1 </math> chairs to place in between beforehand, leaving <math> 9-n </math> chairs to distribute among <math> n+1 </math> spots. Using stars-and-bars, we find that the number of ways to do this is <math> \binom{9}{9-n} </math>.
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Thus, the total number of arrangements is <math> \binom{n}{8-n} \binom{9}{9-n} </math>. Summing this from <math> n=4 </math> to <math> n=8 </math> yields <math> \sum_{n=4}^{8} \binom{n}{8-n} \binom{9}{9-n} = 2907 \equiv \boxed{907} \ (\text{mod} \ 1000)</math>.
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~ [https://artofproblemsolving.com/community/user/1096232 Pengu14]
  
 
==See also==
 
==See also==

Revision as of 23:56, 17 February 2025

Problem

Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let $N$ be the number of subsets of $16$ chairs that could be selected. Find the remainder when $N$ is divided by $1000$.

Solution 1: Casework

Let's split the problem into a few cases:

Case 1: All $8$ people are sitting isolated (no person sits next to any of them): $^8C_0 \cdot ^9C_1 = 9$

Case 2: $6$ people are isolated, $2$ people sit next to each other (such that each person sits next to either $0$ or $1$ other person): $^7C_1 \cdot ^9C_2 = 7 \cdot 36 = 252$

Case 3: $4$ people are isolated, $2$ people sit next to each other and $2$ other people sit next to each other with the $2$ groups of $2$ people not sitting next to each other (so each person still sits next to either $0$ or $1$ other person): $^6C_2 \cdot ^9C_3 = 1260$

Case 4: $2$ people are isolated, $6$ people are split into $3$ groups of $2$ people, and no $2$ groups sit next to each other: $^5C_3 \cdot ^9C_4 = 10 \cdot 126 = 1260$

Case 5: $4$ groups of $2$, no groups are sitting next to each other: $^4C_4 \cdot ^9C_5 = 126$

Answer: $9 + 252 + 1260 + 1260 + 126 = 2907$

$2907 \equiv 907 \pmod{1000} \Longrightarrow \boxed{907}$.

(Feel free to correct any format/latex problems)

~Mitsuihisashi14 ~Small latex fixes by Marcus :)

Solution 2: Stars-and-Bars

Let $n$ be the number of groups of adjacent people.

Clearly, there will be $8-n$ groups with people, and there are $\binom{n}{8-n}$ ways to select these groups.

Now, we use the stars-and-bars technique to find the number of ways to distribute the $8$ remaining chairs around the $8$ people. Since the $n$ groups are separated, we must choose $n-1$ chairs to place in between beforehand, leaving $9-n$ chairs to distribute among $n+1$ spots. Using stars-and-bars, we find that the number of ways to do this is $\binom{9}{9-n}$.

Thus, the total number of arrangements is $\binom{n}{8-n} \binom{9}{9-n}$. Summing this from $n=4$ to $n=8$ yields $\sum_{n=4}^{8} \binom{n}{8-n} \binom{9}{9-n} = 2907 \equiv \boxed{907} \ (\text{mod} \ 1000)$.

~ Pengu14

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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