Difference between revisions of "2020 AIME I Problems/Problem 5"

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== Problem ==
 
== Problem ==
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Six cards numbered <math>1</math> through <math>6</math> are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
  
 
== Solution ==
 
== Solution ==

Revision as of 16:18, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.

Solution

Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.

If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\boxed{052}$.

-molocyxu

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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