AMC 8 · 2016 · #17

Grade 4 counting

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

permutations-basiccomplementary-countingsystematic-enumeration complementary-countingsystematic-enumeration ↑ Prerequisites: permutations-basicmulti-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

Pick an answer.

(A)

$mbox{ }30$

(B)

$mbox{ }7290$

(C)

$mbox{ }9000$

(D)

$mbox{ }9990$

(E)

$mbox{ }9999$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A password is $4$ digits, and each digit is anything from $0$ to $9$ (repeats are fine). The only rule is that the password is NOT allowed to start with the three digits $9, 1, 1$ (in that order). How many passwords are still allowed?

Givens: Password length = $4$ digits; Each digit can be any of $0, 1, 2, \ldots, 9$ ($10$ choices); Repeats are allowed; Forbidden: the first three digits are $9, 1, 1$ (any $4$th digit); Answer choices: (A) $30$, (B) $7290$, (C) $9000$, (D) $9990$, (E) $9999$

Unknowns: The number of $4$-digit passwords that do NOT begin with $9, 1, 1$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #2 Make a Systematic List

Counting allowed passwords directly is messy — we would have to handle every first-digit case carefully. The forbidden set is much smaller, so Tool #16 (Complement) wins: count ALL $4$-digit strings, count just the forbidden ones starting with $9,1,1$, and subtract. Tool #2 (Systematic List) is the natural way to confirm the forbidden count — list $9110, 9111, \ldots, 9119$ and check we have exactly $10$ items.

Execute — Answer: D

#16 Change Focus / Count the Complement 4.OA.A.1 Step 1

Count the universe: every $4$-digit password with no rules.
Each of the $4$ positions has $10$ choices ($0$ through $9$), and choices multiply.

$$10 \times 10 \times 10 \times 10 = 10{,}000$$

💡 Multiplying choices across independent positions is the basic counting move from Grade 4 multiplicative reasoning.

#16 Change Focus / Count the Complement 4.OA.A.1 Step 2

Count the forbidden passwords — the ones that start with $9, 1, 1$.
The first three digits are pinned ($1$ choice each), and the $4$th digit is free ($10$ choices).

$$1 \times 1 \times 1 \times 10 = 10$$

💡 Fixing positions reduces the count: a forced slot contributes a factor of $1$, an open slot contributes its full count.

#2 Make a Systematic List 3.OA.A.1 Step 3

Verify the forbidden count by listing them in order (Tool #2).
The only thing that varies is the last digit.

$$9110, 9111, 9112, 9113, 9114, 9115, 9116, 9117, 9118, 9119 \;\Rightarrow\; 10 \text{ items}$$

💡 Listing in order is the safest way to confirm a small count and avoid off-by-one errors.

#16 Change Focus / Count the Complement 4.NBT.B.4 Step 4

Subtract the forbidden count from the universe (the complement step).

$$10{,}000 - 10 = 9{,}990 \;\Rightarrow\; \textbf{(D)}$$

💡 "Allowed = total minus forbidden" is the complement principle in one line.

[1] #16 4.OA.A.1 Count the universe: every $4$-digit password with no rules. Each of the $4$ posi

[2] #16 4.OA.A.1 Count the forbidden passwords — the ones that start with $9, 1, 1$. The first th

[3] #2 3.OA.A.1 Verify the forbidden count by listing them in order (Tool #2). The only thing th

[4] #16 4.NBT.B.4 Subtract the forbidden count from the universe (the complement step).

Review

Reasonableness: Out of $10{,}000$ possible passwords, only $10$ are banned — a tiny fraction ($0.1\%$). So the allowed count should be just slightly below $10{,}000$. Choice (D) $9{,}990$ matches that exactly. Choices (A) $30$ and (B) $7290$ are far too small, and (E) $9999$ would mean only $1$ forbidden password — but the $4$th digit gives $10$ forbidden options, not $1$. (C) $9000$ would correspond to banning every password whose first digit is $9$, which is not the rule.

Alternative: Direct count (without complement): the password is OK as long as the first three digits are NOT $(9,1,1)$. Of the $10 \times 10 \times 10 = 1000$ possible $3$-digit prefixes, exactly $1$ is bad, so $999$ prefixes are OK. Each of those has $10$ choices for the last digit: $999 \times 10 = 9{,}990$. Same answer, more work — which is exactly why Tool #16 (complement) is the cleaner path.

CCSS standards used (min grade 4)

3.OA.A.1 Interpret products of whole numbers as repeated equal groups (Listing the $10$ forbidden passwords $9110, 9111, \ldots, 9119$ — a Grade 3 "count by ones" check.)
4.OA.A.1 Interpret a multiplication equation as a comparison; multiplicative reasoning (Counting $10 \times 10 \times 10 \times 10 = 10{,}000$ total passwords and $1 \times 1 \times 1 \times 10 = 10$ forbidden ones — multiplying independent choices.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing $10{,}000 - 10 = 9{,}990$ as the final complement subtraction.)

⭐ When the "not allowed" cases are few, count those instead and subtract — that's the complement trick, and it only needs Grade 4 multiplication and subtraction.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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