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

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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?
 
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>
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 </math>
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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>
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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 then <math>\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\textbf{(D) } 8}</math>
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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.
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Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly
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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}}
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{{MAA Notice}}

Latest revision as of 15:33, 16 December 2021

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?

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

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 then $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\textbf{(D) } 8}$

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

Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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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