Difference between revisions of "2022 AMC 12A Problems/Problem 19"
Pi is 3.14 (talk | contribs) (→See Also) |
Mathischess (talk | contribs) |
||
| Line 6: | Line 6: | ||
two passes? | two passes? | ||
| + | == Video Solution By ThePuzzlr == | ||
| + | https://youtu.be/p9xNduqTKLM | ||
| + | |||
| + | ~ MathIsChess | ||
==Solution by OmegaLearn Using Combinatorial Identities and Overcounting== | ==Solution by OmegaLearn Using Combinatorial Identities and Overcounting== | ||
Revision as of 10:44, 12 November 2022
Contents
Problem
Suppose that 13 cards numbered 1, 2, 3, . . . , 13 are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second pass, 6 on the third pass, 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass.
For how many of the 13! possible orderings of the cards will the 13 cards be picked up in exactly two passes?
Video Solution By ThePuzzlr
~ MathIsChess
Solution by OmegaLearn Using Combinatorial Identities and Overcounting
~ pi_is_3.14
Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
