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

Revision as of 16:07, 22 January 2014

Problem

What is the average (mean) of all 5-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 $\boxed{\mathrm{(C)}53332.8}$

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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Revision as of 11:15, 4 July 2013 (view source) Nathan wailes (talk \| contribs) ← Older edit		Revision as of 16:07, 22 January 2014 (view source) Smart99 (talk \| contribs) m (→‎Problem) Newer edit →
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	== Problem ==		== 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?	+	What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?

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

Revision as of 16:07, 22 January 2014

Problem

Solution

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