The Vault Code — Teacher Answers

🔐 Student Activity Unlock Codes (case-insensitive)

Phase 1 → 2

STAGES

Phase 2 → 3

SHRINKING

Phase 3 → 4

COMPLEMENT

Phase 4 → Done

CRACKED

🎯 Why this activity, and how it's different

This activity is deliberately layered with four distinct insights, not one idea applied to bigger numbers: independent stages (P1), shrinking choices under a no-repeat rule (P2), complementary counting for "at least one" conditions (P3), and combining both P2 and P3 together (P4). Resist confirming complementary counting is the right move in Phase 3 until a group has genuinely tried the messy direct casework first and felt why it's painful — that discomfort is what makes the shortcut land.

🗣 Oral Launch — Read Aloud to Class

"You've been hired as a security consultant. The vault's owner wants to know exactly how many codes their system allows — get the count wrong, and you'll either miss a real vulnerability or waste money fixing one that isn't there."

Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.

Setting the Baseline

The current system

💬 Hint ladder — give ONE at a time, only if truly stuck

1 "How many choices do you have for the very first character?"

2 "Since repeats are allowed, does your choice for the first character limit what's available for the second?"

3 "If every one of the 4 positions independently has 10 choices, how do you combine four separate counts into one total?"

Task 1a — The current code

$10\times10\times10\times10 = 10^4$

10,000 codes

Task 1b — Scaling up

6-character version: $10^6 = 1{,}000{,}000$

General rule: a code of length $n$, digits 0–9, repeats allowed, gives $10^n$ possible codes.

✓ Gate check before giving STAGES

Require: correct count for 4 and 6 characters, and the general pattern stated as $10^n$ (or described correctly in words) — this phase should move quickly for a strong group; don't let it drag.

No Repeats Allowed

A new security policy

💬 Hint ladder — give ONE at a time, only if truly stuck

1 "Once a digit has been used in the first position, is it still available for the second?"

2 "How many digits remain for the second position? The third? The fourth?"

3 "For 2c: which position has the most restrictive rule attached to it?"

4 "Try counting that most-restricted position first — how many digits remain for the others afterward, remembering 0 becomes available again?"

Task 2a — Tightening security

$10\times9\times8\times7$

5,040 codes

Task 2b — What actually changed

In Phase 1 every position had exactly 10 choices no matter what came before (independent stages). Here, each choice removes an option from the remaining pool, so the number of choices depends on earlier decisions (dependent stages) — that's a genuinely different structure, not just "a smaller number."

Task 2c — One more rule

Order of reasoning matters here: count the most restricted position first.

First digit: 9 choices (1–9, excludes 0)
Second digit: 9 choices remaining (0–9 minus the one digit already used — 0 is available here)
Third digit: 8 choices
Fourth digit: 7 choices
$9\times9\times8\times7$

4,536 codes

🔍 What to look for

The classic error here is writing $9\times9\times8\times7$ correctly but for the wrong reason, or writing $9\times8\times7\times6$ by incorrectly assuming the second position also loses a "generic" option rather than specifically losing whichever digit was used first. Ask a group to justify each factor individually if the answer is right but the reasoning is shaky.

✓ Gate check before giving SHRINKING

Require: correct 5,040 answer with reasoning that references shrinking/dependent choices (not just "because there's one less digit"), and correct 4,536 with the most-restricted-position-first logic articulated.

The Letter Requirement

A smarter security upgrade

💬 Hint ladder — give ONE at a time, only if truly stuck

1 "Try directly counting codes with exactly one letter. How many pieces does that calculation need?"

2 "What's the exact opposite of 'at least one letter'? Is that opposite case easier to count?"

3 "If you know the total number of codes, and you know how many have zero letters, how could you get the number with at least one?"

Task 3a & 3b — At least one letter

Direct approach (what most groups try first): count exactly-1-letter, exactly-2, exactly-3, and exactly-4 codes separately, then add. This works but is genuinely tedious:

exactly 1 letter: $\binom{4}{1}\times26\times10^3 = 104{,}000$
exactly 2 letters: $\binom{4}{2}\times26^2\times10^2 = 405{,}600$
exactly 3 letters: $\binom{4}{3}\times26^3\times10 = 703{,}040$
exactly 4 letters: $26^4 = 456{,}976$
sum $= 1{,}669{,}616$

Complementary approach (the intended discovery): total codes minus all-digit codes.

total: $36^4 = 1{,}679{,}616$
all-digit (no letters at all): $10^4=10{,}000$
at least one letter: $1{,}679{,}616-10{,}000$

1,669,616 codes (both methods agree — I verified this directly)

🔍 Facilitation note

Let a group finish the direct method if they're determined to — arriving at the same answer two different ways is genuinely satisfying and reinforces that complementary counting isn't a trick, it's a shortcut to something they could always have done the long way. Don't rush them out of the direct method if they're making real progress; the "why bother with 4 cases when there's 1" realization lands harder after some real effort.

✓ Gate check before giving COMPLEMENT

Require: correct final count (1,669,616) by any valid method, and — this is the key one — an explicit statement of the complementary idea in their own words, even if they only reach it after finishing the direct method.

Cracking the Vault

Combining every rule at once

💬 Hint ladder — give ONE at a time, only if truly stuck

1 "This problem has two rules layered together — which earlier phase dealt with 'no repeats', and which dealt with 'at least one letter'?"

2 "Can you count the total number of no-repeat codes using 36 characters, the same way you counted no-repeat codes using 10 digits in Phase 2?"

3 "Can you count the all-digit, no-repeats case — haven't you already done exactly this calculation before?"

4 "How do those two totals combine to answer the actual question?"

Task 4a & 4b — Every rule at once

total no-repeat, 36 characters: $36\times35\times34\times33 = 1{,}413{,}720$
all-digit no-repeat (this is exactly Phase 2's answer!): $10\times9\times8\times7 = 5{,}040$
at least one letter, no repeats: $1{,}413{,}720-5{,}040$

1,408,680 codes

🔍 What to look for

Groups should notice — ideally without being told — that the "all-digit no-repeat" count is identical to their Phase 2 answer. Pointing this out explicitly if they miss it is worthwhile: it's good evidence they've built a genuinely reusable technique, not just solved four separate puzzles.

✓ Gate check before giving CRACKED

Require: correct total no-repeat count (1,413,720), correct all-digit no-repeat count (5,040) explicitly connected back to Phase 2, and the correct final answer with the subtraction reasoning stated clearly.

Challenge Extension — first character forced to be a letter

Once the first character is forced to be a letter, an all-digit code is already impossible — so the "at least one letter" condition is automatically satisfied and needs no separate handling.

first character: 26 choices (any letter)
remaining 3 characters, no repeats, from the remaining 35: $35\times34\times33$
$26\times35\times34\times33$

1,021,020 codes

🔍 Facilitation note

The trap here is a group mechanically reapplying the Phase 4 subtraction method (total minus all-digit) without noticing it's now unnecessary — that would still arrive at the right structure by coincidence in some setups, but the reasoning would be wrong. Push for the explicit realization that no subtraction is needed at all this time.

📋 Consolidation sequence (whole class, after all groups finish)

1. Have a group present Phase 2c's ordering trap — why counting the most-restricted position first matters — since this generalizes far beyond this activity.
2. For Phase 3, ask for a show of hands: who tried the direct 4-case method first? Have one of those groups briefly show how many separate calculations it needed, then have another group show the 2-line complementary version, side by side.
3. Explicitly name the technique: "counting the complement" — and ask the class where else they might use "total minus the case I don't want" outside of this activity.
4. If time allows, preview that next class extends this same counting logic to circular arrangements — where the fundamental counting principle alone isn't quite enough.