Difference between revisions of "2006 AMC 10A Problems/Problem 21"

(wikified and linked to "complementary counting")
m (added category and link to previous and next problem)
Line 11: Line 11:
 
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 (E)
+
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>
 
 
  
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]
 +
 +
*[[2006 AMC 10A Problems/Problem 20|Previous Problem]]
 +
 +
*[[2006 AMC 10A Problems/Problem 22|Next Problem]]
 +
 
* [[Complementary counting]]
 
* [[Complementary counting]]
 +
 +
[[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?

$\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad$

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: $9 * 10 * 10 * 10 = 9000$

Total # of 4-digit integers w/o 2 or 3: $7 * 8 * 8 * 8 = 3584$

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in it equals: $9000-3584=5416 \Rightarrow E$

See Also

Invalid username
Login to AoPS