Difference between revisions of "2005 AMC 10B Problems/Problem 18"

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== 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>
 
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>
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Alternatively, we could just choose <math>1</math> out of the <math>8</math> numbers to not be used. There are obviously <math>\boxed{8}</math> ways to do so.
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== 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}}

Revision as of 23:32, 24 May 2013

Problem

All of David's telephone numbers have the form $555-abc-defg$, where $a$, $b$, $c$, $d$, $e$, $f$, and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$. How many different telephone numbers can David have?

$\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 7 \qquad \mathrm{(D)} 8 \qquad \mathrm{(E)} 9$

Solution

The only digits available to use in the phone number are $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$. There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$. The answer is just $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\mathrm{(D)}\ 8}$

Alternatively, we could just choose $1$ out of the $8$ numbers to not be used. There are obviously $\boxed{8}$ ways to do so.

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions
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