2002 AMC 10P Problems/Problem 3

Problem

Mary typed a six-digit number, but the two $1$ s she typed didn't show. What appeared was $2002.$ How many different six-digit numbers could she have typed?

$\text{(A) }4 \qquad \text{(B) }8 \qquad \text{(C) }10 \qquad \text{(D) }15 \qquad \text{(E) }20$

Solution 1

This is equivalent to ${5 \choose 2} + 5$ since we are choosing $5$ spots (three in between $2002$ and two to the left and right $with two$ 1 $s. We also have to take into account "double$ 1 $s," namely,$ 112002, 211002, 201102, 200112, and 200211 $in our$ 5$spots.

Thus, our answer is$ (Error compiling LaTeX. Unknown error_msg){5 \choose 2} + 5 = \boxed{\textbf{(D) } 15}.$== Solution 2 == We can split this into a little bit of casework which is easy to do in our head.

Case 1: The first$ (Error compiling LaTeX. Unknown error_msg)1 $is ahead of the first$ 2. $Then the second$ 1 $has$ 5$places.

Case 2: The first$ (Error compiling LaTeX. Unknown error_msg)1 $is below the first$ 2. $Then the second$ 1 $has$ 4 $places.$ \dots $This continues as the answer comes to$ 5 + 4 + 3 + 2 + 1 = 15. $Thus, our answer is$ \boxed{\textbf{(D) } 15}.$== Solution 3 == We can just count the cases directly since there are so little.

Thus, our answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D) } 15}.$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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
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2002 AMC 10P Problems/Problem 3

Problem

Solution 1

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