AIME Counting After AMC 12: Use Gaps to Control a No-Adjacent Restriction

In a counting transition from AMC 12 to AIME practice, a useful goal is learning to build the restriction into the representation. Counting everything and subtracting a few obvious bad cases may leave hidden overlaps.

The original exercise below is a foundation example, not a released AIME question or an official difficulty benchmark.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 AIME COUNTING and tell us your school year and question.

Define the objects and the restriction

Count binary strings of length eight containing exactly three ones, with no two ones adjacent. Positions are labelled from left to right, so rotations are not identified.

Without the adjacency condition, there would be C(8,3) = 56 strings. We need a representation that keeps each pair of ones separated.

Method 1: compress the positions

Let the positions of the ones be p1 < p2 < p3, with p2 at least p1 + 2 and p3 at least p2 + 2. Set q1 = p1, q2 = p2 - 1 and q3 = p3 - 2.

Then 1 ≤ q1 < q2 < q3 ≤ 6. Choosing any three distinct positions from 1 through 6 determines one permitted original string by reversing the shifts. The count is C(6,3) = 20.

Method 2: distribute the remaining zeros

There are five zeros. Reserve one zero in each of the two internal gaps between the three ones. Three zeros remain to distribute among the four gaps: before, between, between and after.

The number of nonnegative gap allocations is C(6,3) = 20. Explain why each allocation creates exactly one string, rather than treating the formula as a substitute for the correspondence.

Test a circular variation

If the first and last positions are also considered adjacent, the original count includes four forbidden strings with ones at both endpoints. The third one can then occupy position 3, 4, 5 or 6.

Removing those gives 16 for this labelled-position circular-adjacency variation. This is not yet a count up to rotation; identifying rotated strings would be a separate problem.

Make the bridge skill explicit

The transferable lesson is to choose a representation in which constraints are automatic, and then prove the mapping is reversible. That prevents both omissions and duplicate counting.

After a delayed redo, change the length or number of ones and derive the new bounds. For a discussion of counting difficulties, send AMC12 AIME COUNTING with an attempt showing how you handled the restriction. Official competition rules should be checked separately.

Continue reading: From AMC 12 Geometry to AIME Practice: Turn a Sketch into an Exact Model.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 AIME COUNTING. Include your school year, country or competition region, experience and the question you want to discuss.

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Official information and further reading: Source 1 Source 2

AMC 12 Algebra: Find One Binomial Coefficient Without Expanding Everything

A coefficient question asks for one part of a polynomial, not its value at a convenient input. Thinking about which factors supply x helps explain the binomial coefficient directly.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 BINOMIAL 1019 and tell us your school year and question.

Choose the factors supplying x

Consider (1 + 2x)5 as five multiplied factors. To obtain an x3 term, choose 2x from exactly three factors and 1 from the remaining two.

There are C(5,3) = 10 ways to choose those three factors. Each choice contributes (2x)3 = 8x3, so the total term is 80x3 and the coefficient is 80.

Separate three different quantities

The requested coefficient is 80. The corresponding term is 80x3. The value of the whole polynomial at x = 1 is 35 = 243, which includes all coefficients.

Substituting x = 1 is a useful overall expansion check, but it cannot by itself isolate the coefficient of x3.

Adapt the reasoning to a sign change

For (1 - 2x)5, the same choice count applies, but each selection contributes (-2)3x3. The coefficient is therefore -80.

Record the exponent, the number of choices and the numerical contribution separately. This prevents losing a power of 2 or an odd negative sign. The example is original teaching material for AMC 12 preparation; current MAA information should be used for the competition itself.

Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 BINOMIAL 1019. Include your school year, country or competition region, experience and the question you want to discuss.

Scan to contact Hanlin on WeChat. Send AMC12 BINOMIAL 1019.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1

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