Difference between revisions of "2005 AMC 10B Problems/Problem 20"

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== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}

Revision as of 21:49, 23 August 2011

Problem

What is the average (mean) of all -digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?


$\mathrm{(A)}48000\qquad\mathrm{(B)}49999.5\qquad\mathrm{(C)}53332.8\qquad\mathrm{(D)}55555\qquad\mathrm{(E)}56432.8$

Solution

We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936$ Since there are $5! = 120$ such numbers, we divide $6399936 \div 120$ to get $\boxed{\mathrm{(C)}53332.8}$

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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