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

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== Solution ==
 
== Solution ==
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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>
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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>
  
 
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.

Latest revision as of 14:33, 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 (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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