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

(Solution)
Line 10: Line 10:
 
== 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}}

Revision as of 12:15, 4 July 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS