Difference between revisions of "2005 AMC 10B Problems/Problem 18"
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Alternatively, we could just choose <math>1</math> out of the <math>8</math> numbers not to be used. There are obviously <math>\boxed{8}</math> ways to do so. | Alternatively, we could just choose <math>1</math> out of the <math>8</math> numbers not to be used. There are obviously <math>\boxed{8}</math> ways to do so. | ||
| + | |||
| + | Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2005|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:29, 29 November 2019
Problem
All of David's telephone numbers have the form 
, where 
, 
, 
, 
, 
, 
, and 
are distinct digits and in increasing order, and none is either 
or 
. How many different telephone numbers can David have?
Solution
The only digits available to use in the phone number are 
, 
, 
, 
, 
, 
, 
, and 
. There are only spots left among the numbers, so we need to find the number of ways to choose numbers from . The answer is just 
Alternatively, we could just choose out of the numbers not to be used. There are obviously 
ways to do so.
Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
