Difference between revisions of "2005 AMC 10B Problems/Problem 20"
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== Solution == | == 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 <math> 4! = 24 </math> ways to arrange the other numbers, so each number appears in each spot <math> 24 </math> times. Therefore, the sum of all such numbers is <math> 24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936 </math> Since there are <math> 5! = 120 </math> such numbers, we divide <math> 6399936 \div 120 </math> to get <math> \boxed{53332.8} </math> | + | 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 <math> 4! = 24 </math> ways to arrange the other numbers, so each number appears in each spot <math> 24 </math> times. Therefore, the sum of all such numbers is <math> 24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936 </math> Since there are <math> 5! = 120 </math> such numbers, we divide <math> 6399936 \div 120 </math> to get <math> \boxed{\mathrm{(C)}53332.8} </math> |
== See Also == | == See Also == | ||
*[[2005 AMC 10B Problems]] | *[[2005 AMC 10B Problems]] | ||
Revision as of 18:15, 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?
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 
ways to arrange the other numbers, so each number appears in each spot 
times. Therefore, the sum of all such numbers is 
Since there are 
such numbers, we divide 
to get 
