Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #20
Grade 7 probabilitycountingPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Probability problems split cleanly into three subproblems (Tool #7): (a) count the sample space, (b) count the favorable outcomes, (c) form the ratio and reduce. For (b) we count digit by digit using the multiplication principle — a structured systematic list (Tool #2) where the ORDER of filling matters: pick the most restricted slot (units, must be odd) first, then the next-most restricted (thousands, cannot be 0 and cannot repeat units), then the remaining slots. After computing, Tool #3 (Eliminate) confirms the simplified fraction matches exactly one answer choice.
Count all four-digit numbers
Subproblem A — the sample space. The integers 1000 to 9999 are exactly the 4-digit numbers, so there are 9000 of them.
Knowing that 4-digit whole numbers run from 1000 to 9999 and subtracting to count them is a Grade 4 multi-digit place-value skill.
4.NBT.A.2Identify SubproblemsPlan the order of digits
Subproblem B — favorable count. Fill the slots most-restricted-first: choose the odd units digit before the thousands digit.
Making a systematic list of digit positions, filled in a fixed order, is a Grade 4 multi-step word-problem strategy using the multiplication principle.
4.OA.A.3Make A Systematic ListCount choices for each digit
Per slot: units d₄ has 5 odd choices; thousands d₁ excludes 0 and d₄ for 8; hundreds d₂ has 8; tens d₃ has 7.
Counting how many digits remain after each restriction is repeated subtraction 10 - k — well within Grade 4 multi-step arithmetic.
4.OA.A.3Make A Systematic ListMultiply the choices
Multiply the per-slot choices: 5 × 8 × 8 × 7 = 2240 favorable integers.
Multi-digit multiplication like 40 × 56 = 2240 is the Grade 5 fluency standard for multiplying multi-digit whole numbers.
The number of favorable four-digit integers — odd, with all four digits different — is the product 5 × 8 × 8 × 7 = 2240.
▸ Why?
A favorable number is built by four choices made one after another (units, then thousands, then hundreds, then tens), and the number of ways to make such a run of choices is the product of how many options each step has.
▸ Why?
Each option for one slot leads to a fresh full set of options for the next slot, so the outcomes stack up as equal groups sitting inside equal groups — and multiplying the option-counts is exactly what counts equal groups of equal groups.
▸ Why?
The four option-counts are 5, 8, 8, 7, each found by starting from the ten digits 0–9 and taking away the ones that slot is not allowed to use.
▸ Why?
For any slot the ten digits split with no overlap into 'forbidden here' and 'still allowed', so the allowed count is 10 minus the forbidden ones — units keeps only the five odd digits, thousands drops both 0 and the units digit to leave 8, and each later slot drops the digits already used.
Form and reduce the fraction
Subproblem C — reduce. gcd(2240, 9000) = 40, so = .
Using prime factorization to find the GCF and reduce a fraction is the Grade 6 GCF/LCM standard.
6.NS.B.4Identify SubproblemsMatch to the choices
≈ 0.249 is a valid probability, and it matches exactly one choice — (B).
Interpreting a count ratio as a probability between 0 and 1 is the Grade 7 introduction to chance probability.
7.SP.C.5Eliminate PossibilitiesThis AMC 8 problem only needs Grade 7 probability — favorable cases divided by total cases — that you already know!
- Count all four-digit numbers
- Plan the order of digits
- Count choices for each digit
- Multiply the choices
- Form and reduce the fraction
- Match to the choices
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