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Difference between revisions of "1994 AHSME Problems/Problem 22"

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<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630 </math>
 
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630 </math>
 
==Solution==
 
==Solution==
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Let the seats be numbered in order clockwise around the table <math>1,2,3,\dots 9</math>. Let <math>a,b,c</math> be the seats occupied by Alpha, Beta, and Gamma, respectively. Without loss of generality, fix Alpha at seat <math>1</math> to account for rotations. Then there are ten choices for the set <math>\{b,c\}</math>, namely <math>\{b,c\}=\{3,5\},\{3,6\},\{3,7\},\{3,8\},\{4,6\},\{4,7\},\{4,8\},\{5,7\},\{5,8\},\{6,8\}</math>. After multiplying by <math>3!=6</math> (number of ways to permute Alpha, Beta, and Gamma), we get the final answer of <math>\boxed{\textbf{(C)}\ 60}</math>
+
Since each professor must sit between two students, they cannot be seated in seats <math>1</math> or <math>9</math> (the seats at either end of the row). Hence, each professor has <math>7</math> seats they can choose from and must be at least <math>1</math> seat apart.
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This question is equivalent to choosing <math>3</math> seats from a row of <math>5</math> seats with no restrictions because we can simply generate a valid arrangement by inserting a seat right after the first and second seat chosen.
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Hence, the answer is <math>\binom{5}{3} \cdot 3! =</math> <math>60</math> <math>\textbf{(C)}</math>.
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==See Also==
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{{AHSME box|year=1994|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 02:24, 28 May 2021

Problem

Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?

$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630$

Solution

Since each professor must sit between two students, they cannot be seated in seats $1$ or $9$ (the seats at either end of the row). Hence, each professor has $7$ seats they can choose from and must be at least $1$ seat apart.

This question is equivalent to choosing $3$ seats from a row of $5$ seats with no restrictions because we can simply generate a valid arrangement by inserting a seat right after the first and second seat chosen.

Hence, the answer is $\binom{5}{3} \cdot 3! =$ $60$ $\textbf{(C)}$.

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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