2002 AMC 10P Problems/Problem 3

Revision as of 19:44, 14 July 2024 by Wes (talk | contribs) (Solution 1)

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 $$ (Error compiling LaTeX. Unknown error_msg)\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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png