Difference between revisions of "2005 AMC 10B Problems/Problem 18"
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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> | 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> | ||
− | Alternatively, we could just choose <math>1</math> out of the <math>8</math> numbers to | + | 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. |
== 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 22:33, 26 November 2013
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.
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 |
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