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

We can split this into a little bit of casework which is easy to do in our head.

Case 1: The first $1$ is ahead of the first $2.$ Then the second $1$ has $5$ places.

Case 2: The first $1$ 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 $\boxed{\textbf{(D) } 15}.$

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
All AMC 10 Problems and Solutions

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2002 AMC 10P Problems/Problem 3

Problem

Solution 1

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