AMC 10 · 2018 · #11

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

combinations-basicsystematic-enumeration easier-related-problemsystematic-enumeration ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 2 insights

Problem

Seven fair ordinary 6-sided dice are rolled. The chance their top faces add to 10 is written as (n)/(6⁷). Find the whole number n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are 6⁷ equally likely ordered rolls, so n is simply the count of rolls that total 10 — a 'how many ways' counting question, which is Tool #2 (Make a Systematic List). Tool #9 (Solve an Easier Related Problem): instead of asking each face to land between 1 and 6, give every die its smallest value 1 first; that uses up 7 and leaves only 3 extra dots to hand out, a much smaller problem. Tool #4 (Introduce a Variable): write each die as 1+e_i so the leftover dots e_i become the real unknowns. Tool #7 (Identify Subproblems): the count splits cleanly by how the 3 extra dots are grouped, and the three small counts add to the answer.

1STEP 1

See what n counts

Every ordered roll is equally likely and there are 6⁷ in all, so n is the count of ordered rolls that add to 10.

P=(favorable rolls)/6⁷=n/6⁷ → n=#{(d₁,…,d₇):d_i∈{1,…,6}, Σ d_i=10}

2STEP 2

Give every die its 1 first

Give each die its minimum 1 first; that uses 7, so writing d_i=1+e_i leaves the extras with e₁+…+e₇=3.

d_i=1+e_i, e_i ≥ 0 → e₁+…+e₇=10-7=3

3STEP 3

Check the top cap never bites

Each die can take at most 5 extra dots, but only 3 exist, so the cap never bites and the dots share freely.

max e_i ≤ 3 ＜ 5 → cap e_i ≤ 5 is automatic

4STEP 4

List the cases by how the 3 dots split

Split by shape: 3 on one die is 7; a 2-and-1 is 7·6=42; three lone 1s is (7·6·5)/3!=35.

3=3: 7; 3=2+1: 7· 6=42; 3=1+1+1: (7· 6· 5)/3!=35

5STEP 5

Add the cases

The three cases don't overlap and cover every split, so add them: 7+42+35=84 — that count of rolls summing to 10 is choice (E).

n=7+42+35=84 → (E)

Answer

The answer 84 matches choice (E) and is a whole number, as n must be. A second count confirms it: spreading 3 identical extra dots among 7 dice with no ceiling can be done in C(3+7-1, 7-1)=C(9, 6)=C(9, 3)=(9· 8· 7)/6=84 ways, exactly the case total 7+42+35. The size is sensible too: 84 out of 6⁷=279936 gives a small probability near 0.0003, which fits a fairly low target sum like 10.

💡Key takeaway

Give every die its smallest value first, then just count how to share the few leftover dots: spreading 3 extra dots among 7 dice gives 7+42+35=84, choice (E).

See what n counts
Give every die its 1 first
Check the top cap never bites
List the cases by how the 3 dots split
Add the cases

See what n counts

Give every die its 1 first

Check the top cap never bites

List the cases by how the 3 dots split

Add the cases

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