Difference between revisions of "2002 AMC 10P Problems/Problem 3"
(→Solution 1) |
(→Solution 2) |
||
Line 32: | Line 32: | ||
This continues as the answer comes to | This continues as the answer comes to | ||
− | <math> 5 + 4 + 3 + 2 + 1 = 15.</math> | + | <math> 5 + 4 + 3 + 2 + 1 = \frac{5(6)}{2}=3(5)=15.</math> |
Thus, our answer is <math>\boxed{\textbf{(D) } 15}.</math> | Thus, our answer is <math>\boxed{\textbf{(D) } 15}.</math> |
Latest revision as of 21:00, 15 July 2024
Problem
Mary typed a six-digit number, but the two s she typed didn't show. What appeared was
How many different six-digit numbers could she have typed?
Solution 1
This is equivalent to since we are choosing
spots (three in between
and two to the left and right with two
s. We also have to take into account "double
s," namely,
and
in our
spots.
Thus, our answer is
Solution 2
We can split this into a little bit of casework which is easy to do in our head.
Case 1: The first is ahead of the first
Then the second
has
places.
Case 2: The first is below the first
Then the second
has
places.
This continues as the answer comes to
Thus, our answer is
Solution 3
We can just count the cases directly since there are so little.
Thus, our answer is
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.