Difference between revisions of "1990 AJHSME Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | p | + | Let <math>p</math> be a person seated and <math>o</math> is an empty seat |
| − | The pattern of seating that results in the fewest occupied seats is opoopoopoo...po | + | The pattern of seating that results in the fewest occupied seats is <math>\text{opoopoopoo...po}</math>. |
| − | + | We can group the seats in 3s like this: <math>\text{opo opo opo ... opo}.</math> | |
| − | opo opo opo ...opo | + | |
| + | There are a total of <math>40=\boxed{B}</math> groups | ||
| − | |||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1990|num-b=18|num-a=20}} | {{AJHSME box|year=1990|num-b=18|num-a=20}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
Latest revision as of 00:59, 25 November 2020
Problem
There are 
seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
Solution
Let 
be a person seated and 
is an empty seat
The pattern of seating that results in the fewest occupied seats is 
.
We can group the seats in 3s like this: 
There are a total of 
groups
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
