Difference between revisions of "1994 AHSME Problems/Problem 22"
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Hence, the answer is <math>\binom{5}{3} \cdot 3! =</math> <math>60</math> <math>\textbf{(C)}</math>. | 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}} | ||
| + | {{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?
Solution
Since each professor must sit between two students, they cannot be seated in seats 
or 
(the seats at either end of the row). Hence, each professor has 
seats they can choose from and must be at least seat apart.
This question is equivalent to choosing 
seats from a row of 
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
