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 ${5 \choose 2} + 5 = \boxed{\textbf{(D) } 15}.$

Solution 2

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

Case 1: The first $1$ 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

$5 + 4 + 3 + 2 + 1 = \frac{5(6)}{2}=3(5)=15.$

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 $\boxed{\textbf{(D) } 15}.$

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
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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