Difference between revisions of "2022 AMC 10A Problems/Problem 22"
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<math>\textbf{(A) }4082\qquad\textbf{(B) }4095\qquad\textbf{(C) }4096\qquad\textbf{(D) }8178\qquad\textbf{(E) }8191</math> | <math>\textbf{(A) }4082\qquad\textbf{(B) }4095\qquad\textbf{(C) }4096\qquad\textbf{(D) }8178\qquad\textbf{(E) }8191</math> | ||
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+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} | ||
==Solution by Omega Learn Using Combinatorial Identities and Overcounting== | ==Solution by Omega Learn Using Combinatorial Identities and Overcounting== | ||
https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6 | https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6 |
Revision as of 04:11, 12 November 2022
Problem
Suppose that 13 cards numbered 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
possible orderings of the cards will the
cards be picked up in exactly two passes?
[asy]size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy]
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.
Solution by Omega Learn Using Combinatorial Identities and Overcounting
https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6