AMC 10 · 2025 · #11

Grade 7 probability

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

derangementcombinations-basicprobability-basic principle-of-inclusion-exclusion ↑ Prerequisites: permutations-basic

📏 Long solution 💡 3 insights

Problem

Six students are each randomly matched to one of six tutors on Monday, one student per tutor. On Tuesday the same six students are randomly matched to the same six tutors again, one per tutor. Find the probability that exactly 2 of the students end up with their Monday tutor a second time.

Pick an answer.

(A)

$\frac{1}{16}$

(B)

$\frac{3}{16}$

(C)

$\frac{1}{4}$

(D)

$\frac{3}{8}$

(E)

$\frac{1}{2}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

Two separate random days sound complicated, so Tool #15 (Organize Information in More Ways) rewrites them as one object: fix Monday as the reference, and Tuesday becomes a single random reshuffle of the students. Then "met the same tutor both days" simply means a student stays put — a fixed point of the reshuffle. Tool #2 (Make a Systematic List) counts the equally likely outcomes: all 6! reshuffles, and the C(6, 2) choices of which two stay. Tool #16 (Count the Complement) handles the hard part — the other four must all move, an "everybody switches" count called a derangement. Tool #7 (Identify Subproblems) keeps these three counts separate so each stays easy.

1STEP 1

Turn two days into one shuffle

Freeze Monday, and Tuesday is one random reshuffle of the six students: same tutor twice means a student who stays in place.

same tutor both days ⇔ student stays fixed in Tuesday's reshuffle

2STEP 2

Count all equally likely outcomes

Fill the tutors one at a time with 6, 5, 4, 3, 2, 1 choices: 720 equally likely Tuesday matchings, the denominator.

6! = 6·5·4·3·2·1 = 720

3STEP 3

Pick which 2 students stay

First choose which two students keep their Monday tutor: order does not matter, so C(6, 2) = 15 ways.

C(6, 2) = (6·5)/(2·1) = 15

4STEP 4

Force the other 4 to all move

The other 4 must all switch tutors — a derangement. Inclusion–exclusion on 24 gives 24 - 24 + 12 - 4 + 1 = 9 ways.

24 - 24 + 12 - 4 + 1 = 9

5STEP 5

Combine into favorable outcomes

Pick the stayers, then derange the rest — the two stages multiply: 15 · 9 = 135 favorable Tuesday matchings.

C(6, 2) · D₄ = 15 · 9 = 135

6STEP 6

Divide and simplify

Favorable over total is 135/720, and cancelling the common factor 45 leaves 3/16 — choice (B).

P = 135/720 = (45 · 3)/(45 · 16) = 3/16 = (B)

Answer

3/16

The number of favorable outcomes, 135, is less than the total 720, so the probability is under 1 — sanity passes. It should also be small, since needing exactly two returns (no more, no fewer) is a fairly special event, and 3/16≈ 0.19 fits that. A cross-check: the fraction of all six-item reshuffles with a given number of fixed points, using C(6, k)D₆-k, gives counts 265,264,135,40,15,0,1 for k=0,1,2,3,4,5,6, which sum to 720 exactly — and the k=2 entry is our 135. This confirms both the count and that no outcomes were double-counted, so 3/16 is right.

💡Key takeaway

Freeze the first day and treat the second as a reshuffle: pick which 2 students stay (15 ways), make the other 4 all move (9 ways), and divide 15 · 9=135 by 720 to get 3/16.

Turn two days into one shuffle
Count all equally likely outcomes
Pick which 2 students stay
Force the other 4 to all move
Combine into favorable outcomes
Divide and simplify