Difference between revisions of "2006 AMC 10A Problems/Problem 21"
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Total # of 4-digit integers w/o 2 or 3: <math>7 * 8 * 8 * 8 = 3584</math> | Total # of 4-digit integers w/o 2 or 3: <math>7 * 8 * 8 * 8 = 3584</math> | ||
| − | Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: 9000-3584=5416 | + | Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: <math>9000-3584=5416 \Rightarrow E </math> |
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== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
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| + | *[[2006 AMC 10A Problems/Problem 20|Previous Problem]] | ||
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| + | *[[2006 AMC 10A Problems/Problem 22|Next Problem]] | ||
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* [[Complementary counting]] | * [[Complementary counting]] | ||
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| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 15:01, 4 August 2006
Problem
How many four-digit positive integers have at least one digit that is a 2 or a 3?
Solution
Since we are asked for the number of positive 4-digit integers with AT LEAST ONE 2 or 3 in it, we can find this by finding the number of 4-digit + integers that DO NOT contain any 2 or 3.
Total # of 4-digit integers: 
Total # of 4-digit integers w/o 2 or 3: 
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: 
