Difference between revisions of "2022 AMC 10A Problems/Problem 22"

(Solution Using Combinatorial Identities and Overcounting)
(Problem)
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Suppose that 13 cards numbered <math>1, 2, 3, \cdots, 13</math> 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 <math>13!</math> possible orderings of the cards will the <math>13</math> cards be picked up in exactly two passes?
 
Suppose that 13 cards numbered <math>1, 2, 3, \cdots, 13</math> 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 <math>13!</math> possible orderings of the cards will the <math>13</math> cards be picked up in exactly two passes?
  
[asy]
+
[say]
 +
/import olympiad
 +
/import cse5
 
size(11cm);
 
size(11cm);
 
draw((0,0)--(2,0)--(2,3)--(0,3)--cycle);
 
draw((0,0)--(2,0)--(2,3)--(0,3)--cycle);

Revision as of 03:10, 12 November 2022

Problem

Suppose that 13 cards numbered $1, 2, 3, \cdots, 13$ 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?

[say] /import olympiad /import cse5 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]

$\textbf{(A) }4082\qquad\textbf{(B) }4095\qquad\textbf{(C) }4096\qquad\textbf{(D) }8178\qquad\textbf{(E) }8191$

Solution by Omega Learn Using Combinatorial Identities and Overcounting

https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6