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AMC 8 · 2016 · #17

Grade 4 counting
Archetype Overlap / Complement Counting
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 00 to 99, with repeated digits allowable. If no password may begin with the sequence 9,1,1,9,1,1, then how many passwords are possible?

Pick an answer.

(A)
$mbox{ }30$
(B)
$mbox{ }7290$
(C)
$mbox{ }9000$
(D)
$mbox{ }9990$
(E)
$mbox{ }9999$
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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

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) $9999$

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.

⭐ 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.

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Same archetype — closest grade level first.