Difference between revisions of "2005 AMC 10B Problems/Problem 18"
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== Problem == | == Problem == | ||
| + | All of David's telephone numbers have the form <math>555-abc-defg</math>, where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, <math>f</math>, and <math>g</math> are distinct digits and in increasing order, and none is either <math>0</math> or <math>1</math>. How many different telephone numbers can David have? | ||
| + | |||
| + | <math>\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 7 \qquad \mathrm{(D)} 8 \qquad \mathrm{(E)} 9 </math> | ||
== Solution == | == Solution == | ||
| + | The only digits available to use in the phone number are <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>. There are only <math>7</math> spots left among the <math>8</math> numbers, so we need to find the number of ways to choose <math>7</math> numbers from <math>8</math>. The answer is just <math>\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\mathrm{(D)}\ 8}</math> | ||
== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=17|num-a=19}} | |
Revision as of 15:44, 19 July 2011
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 
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 | ||
