Difference between revisions of "2006 AMC 10A Problems/Problem 21"
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== Problem == | == Problem == | ||
− | How many four- | + | How many four-digit positive integers have at least one digit that is a <math>2</math> or a <math>3</math>? |
<math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math> | <math>\textbf{(A) } 2439\qquad\textbf{(B) } 4096\qquad\textbf{(C) } 4903\qquad\textbf{(D) } 4904\qquad\textbf{(E) } 5416</math> | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | == Solution (Complementary Counting) == | + | == Solution 1 (Complementary Counting) == |
− | Since we are asked for the number of positive 4-digit | + | Since we are asked for the number of positive <math>4</math>-digit integers with at least <math>2</math> or <math>3</math> in it, we can find this by finding the total number of <math>4</math>-digit integers and subtracting off those which do not have any <math>2</math>s or <math>3</math>s as digits. |
− | The total number of 4-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have 10 choices for each digit except the first (which can't be 0). | + | The total number of <math>4</math>-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have <math>10</math> choices for each digit except the first (which can't be <math>0</math>). |
− | Similarly, the total number of 4-digit integers without any 2 or 3 is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>. | + | Similarly, the total number of <math>4</math>-digit integers without any <math>2</math> or <math>3</math> is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>. |
− | Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 | + | Therefore, the total number of positive <math>4</math>-digit integers that have at least one <math>2</math> or <math>3</math> is <math>9000-3584=\boxed{\textbf{(E) }5416}.</math> |
+ | |||
+ | == Solution 2 (Casework)== | ||
+ | We proceed to every case. | ||
+ | |||
+ | Case <math>1</math>: There is ONLY one <math>2</math> or <math>3</math>. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>512</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>7 \cdot 8^2</math> = <math>448</math> arrangements. Therefore, we have <math>2(448 \cdot 3 + 512) = 3712</math> arrangements. | ||
+ | |||
+ | Case <math>2</math> : There are two <math>2</math>s OR two <math>3</math>s. If the <math>2</math> or <math>3</math> is occupying the first digit, we have <math>64</math> arrangements. If the <math>2</math> or <math>3</math> is not occupying the first digit, there are <math>56</math> arrangements. There are <math>3</math> ways for the <math>2</math> or the <math>3</math> to be occupying the first digit and <math>3</math> ways for the first digit to be unoccupied. There are <math>2(3 \cdot (56+64))</math> = <math>720</math> arrangements. | ||
+ | |||
+ | Case <math>3</math> : There is ONLY one <math>2</math> and one <math>3</math>. If the <math>2</math> or the <math>3</math> is occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>64</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 64</math> = <math>384</math> arrangements. If the <math>2</math> or the <math>3</math> isn't occupying the first digit, we have <math>6</math> types of arrangements of where the <math>2</math> or <math>3</math> is. We also have <math>56</math> different arrangements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <math>336 + 384</math> = <math>720</math> total arrangements for this case. | ||
+ | |||
+ | Notice that we already counted <math>3712 + 720 + 720 = 5152</math> cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is <math>\boxed{\textbf{(E) }5416}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
== See also == | == See also == |
Latest revision as of 20:25, 30 April 2024
Contents
Problem
How many four-digit positive integers have at least one digit that is a or a
?
Video Solution
https://youtu.be/0W3VmFp55cM?t=3291
~ pi_is_3.14
Solution 1 (Complementary Counting)
Since we are asked for the number of positive -digit integers with at least
or
in it, we can find this by finding the total number of
-digit integers and subtracting off those which do not have any
s or
s as digits.
The total number of -digit integers is
, since we have
choices for each digit except the first (which can't be
).
Similarly, the total number of -digit integers without any
or
is
.
Therefore, the total number of positive -digit integers that have at least one
or
is
Solution 2 (Casework)
We proceed to every case.
Case : There is ONLY one
or
. If the
or
is occupying the first digit, we have
arrangements. If the
or
is not occupying the first digit, there are
=
arrangements. Therefore, we have
arrangements.
Case : There are two
s OR two
s. If the
or
is occupying the first digit, we have
arrangements. If the
or
is not occupying the first digit, there are
arrangements. There are
ways for the
or the
to be occupying the first digit and
ways for the first digit to be unoccupied. There are
=
arrangements.
Case : There is ONLY one
and one
. If the
or the
is occupying the first digit, we have
types of arrangements of where the
or
is. We also have
different arrangements for the non-
or
digits. We have
=
arrangements. If the
or the
isn't occupying the first digit, we have
types of arrangements of where the
or
is. We also have
different arrangements for the non-
or
digits. We have
=
arrangements for this case. We have
=
total arrangements for this case.
Notice that we already counted cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is
.
~Arcticturn
See also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
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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