2022 AMC 10 A Problem 22

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To help you master 2022 AMC 10 A Problem 22, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2022 AMC10 A Problem 22

Solution

Solution 1

Position Selection Method

If the cards are picked up in exactly two passes, there must be some card $k$ where the first pass ends. By choosing the positions for the cards in the first pass, we can count the valid arrangements.

The sum of all binomial coefficients for a given $n$ is $2^{n}$ . This represents the total number of subsets of a set of size $n$ . We use this to quickly evaluate the sum of $(k 13)$ without calculating each term individually.

Step 1: Understand the condition for exactly two passes

If the cards are picked up in exactly two passes, there must be some integer $k$ such that the first group of cards is picked up in the first pass, and the remaining cards are picked up in the second pass. Because the pickup process is deterministic, each valid arrangement has exactly one such $k$ . Thus, we can count the number of valid arrangements for each $k$ and sum them up.

First pass: cards 1, 2, \dots, k

Second pass: cards k + 1, \dots, 13

1 \leq k \leq 12

Step 2: Identify the requirements for a specific $k$

For the first group of cards to be picked up in the first pass, they must appear in left-to-right order. The same applies to the second group. Finally, to ensure the first card of the second group is not picked up early, it must be located to the left of the last card of the first group.

pos (1) < pos (2) < \dots < pos (k)

pos (k + 1) < pos (k + 2) < \dots < pos (13)

pos (k + 1) < pos (k)

Step 3: Count the ways to place the cards for a fixed $k$

An arrangement is completely determined by choosing which positions will hold the first group of cards. Once the positions are chosen, the cards for the first pass must be placed in these positions from left to right. The remaining cards must be placed in the remaining positions from left to right.

Number of ways to choose positions = (k 13)

Step 4: Exclude the invalid arrangements

We must ensure the condition that the first card of the second pass is to the left of the last card of the first pass. This fails only if all chosen positions are to the left of all unchosen positions, which happens exactly when the chosen positions are the first consecutive spots. In this single failing case, the cards are perfectly sorted and would be picked up in one pass. We subtract this 1 invalid case.

Valid arrangements for a fixed k = (k 13) - 1

Step 5: Sum over all possible values of $k$

The number of cards in the first pass can be any integer from 1 to 12. We sum the valid arrangements over all these possibilities. Using the sum of binomial coefficients, we can evaluate the total.

Total = k = 1 \sum 12 ((k 13) - 1)

Total = (k = 1 \sum 12 (k 13)) - 12

k = 1 \sum 12 (k 13) = 2^{13} - (0 13) - (13 13) = 8192 - 1 - 1 = 8190

Total = 8190 - 12 = 8178

Correct answer: D. $8178$
The exact answer is 8178, which corresponds to choice D. We can verify the order of magnitude: the total number of subsets of 13 items is 8192. Since exactly 14 subsets result in a perfectly sorted array (1 pass), the remaining 8178 subsets result in exactly 2 passes. This confirms our subtraction is correct.

Solution 2

Inverse Permutation Method

Track the positions of the cards $1, 2, \dots, 13$ . The number of passes is determined by how many times we have to ‘jump backwards’ to find the next number. We can count the sequences of positions that have exactly one backward jump.

A ‘descent’ in a sequence occurs when a term is strictly greater than the next term. Counting passes in a left-to-right scanning process is equivalent to counting descents in the sequence of the items’ positions.

Step 1: Translate the process into a sequence of positions

Let the position of each card be recorded. We start by finding the first card. We then look for the next card. If its position is greater, we find it in the same pass. If its position is smaller, we must start a new pass. In general, a new pass begins exactly when the position drops.

New pass condition: pos (x + 1) < pos (x)

Total passes = 1 + (number of drops in position)

Step 2: Relate passes to descents

We want the cards to be picked up in exactly two passes. This means the sequence of positions must have exactly one drop, known as a ‘descent’. Since each card ordering has a unique sequence of positions, we just need to count the number of position sequences that have exactly one descent.

Target: Exactly 1 descent in pos (1), pos (2), \dots, pos (13)

Step 3: Construct permutations with exactly 1 descent

A permutation with exactly 1 descent consists of two increasing blocks. We can construct such a sequence by choosing a subset of the numbers to form the first block. The remaining numbers will form the second block. We arrange the chosen subset in increasing order, followed by the remaining numbers in increasing order.

Number of ways to choose the subset S = 2^{13} = 8192

Step 4: Exclude subsets that produce 0 descents

This construction creates a sequence with at most 1 descent. It fails to have exactly 1 descent only if it is perfectly sorted, which happens when the chosen subset consists of the first consecutive integers. There is one such subset for each possible size from 0 to 13.

Number of invalid subsets = 14

Step 5: Conclude the count

Because a descent can only occur at the boundary between the two blocks, the elements of the chosen subset are exactly the numbers appearing before the descent. Thus, distinct valid subsets produce distinct permutations.

Total valid sequences = 8192 - 14 = 8178

Correct answer: D. $8178$
The result is 8178, matching choice D. Both the direct position selection and the inverse permutation descent counting yield the exact same logical structure and final arithmetic, confirming the result.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Sum of Binomial Coefficients:

k = 0 \sum n (k n) = 2^{n}

$n$ is a non-negative integer, and $k$ ranges from $0$ to $n$ . It allows us to quickly sum the number of valid arrangements $(k 13)$ across all possible sizes of the first pass.

Permutation Descents:

A descent in a sequence

a_{1}, a_{2}, \dots, a_{n}

is an index

i

such that

a_{i} > a_{i + 1}

.

Applies to any sequence of numbers. The number of passes required to pick up the cards is exactly $1$ plus the number of descents in the sequence of the cards' positions.

Key Takeaway: Remember This Strategy for Next Time

A process that repeatedly scans an array from left to right to find items in a specific order.

Try: Track the *positions* of the items rather than the items themselves. A new pass is triggered exactly when the position of the next item is smaller than the current one.

Why: This transforms a dynamic algorithmic process into a static property of a sequence (counting descents), which is much easier to model combinatorially.

A sequence that is partitioned into two increasing blocks.

Try: Model the sequence by choosing a subset of values for the first block. The remaining values automatically form the second block.

Why: Since the order within each block is strictly increasing, the subset of elements completely determines the entire sequence. This reduces a permutation problem to a simpler subset-choosing problem.

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2022 AMC 10 A Problem 22

In This Page:

2022 AMC10 A Problem 22