2006 AMC 10A Problems/Problem 21

Problem

How many four-digit positive integers have at least one digit that is a $2$ or a $3$ ?

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

Video Solution

~ pi_is_3.14

Solution (Complementary Counting)

Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.

The total number of 4-digit integers is $9 \cdot 10 \cdot 10 \cdot 10 = 9000$ , since we have 10 choices for each digit except the first (which can't be 0).

Similarly, the total number of 4-digit integers without any 2 or 3 is $7 \cdot 8 \cdot 8 \cdot 8 ={3584}$ .

Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is $9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)}$

Solution (Casework)

We proceed to the cases.

Case $1$ : There is only one $2$ or $3$ . If the $2$ or $3$ is occupying the first digit, we have $512$ arrangements. If the $2$ or $3$ is not occupying the first digit, there are $7 \cdot 8^2$ = $448$ arrangements. Therefore, we have $2(448 \cdot 3 + 512) =$ 3712$arrangements.

Case$ (Error compiling LaTeX. Unknown error_msg)2 $: There are Two$ 2 $s or two$ 3 $s but not both. If the$ 2 $or$ 3 $is occupying the first digit, we have$ 64 $arrangements. If the$ 2 $or$ 3 $is not occupying the first digit, there are$ 56 $arrangements. There are$ 3 $ways for the$ 2 $or the$ 3 $to be occupying the first digit and$ 3 $ways for the first digit to be unoccupied. There are$ 2(3 \cdot (56+64)) = 720$arrangements.

Case$ (Error compiling LaTeX. Unknown error_msg)3 $: There is one$ 3 $and one$ 2 $but no more. If the$ 2 $or the$ 3 $is occupying the first digit, we have$ 6 $types of arrangements of where the$ 2 $or$ 3 $is. We also have$ 64 $different arrangements for the non-$ 2 $or$ 3 $digits. We have$ 6 \cdot 64 $=$ 384 $arrangements. If the$ 2 $or the$ 3 $isn't occupying the first digit, we have$ 6 $types of arrangements of where the$ 2 $or$ 3 $is. We also have$ 56 $different arrangements for the non-$ 2 $or$ 3 $digits. We have$ 6 \cdot 56 $=$ 336 $arrangements for this case. We have$ 336 + 384 $=$ 720$total arrangements for this case.

Notice that we already counted$ (Error compiling LaTeX. Unknown error_msg)3712 + 720 + 720 = 5152 $cases and we still have a lot of cases left over to count. This is already larger than the second largest answer choice, and therefore, our answer is$ \boxed{\textbf{(E) }

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2006 AMC 10A Problems/Problem 21

Contents

Problem

Video Solution

Solution (Complementary Counting)

Solution (Casework)

See also