PERMUTATIONS & COMBINATIONS

Overview

Counting problems ask "in how many ways can something happen?" The key split is whether order matters. When order matters (arranging, ordering, lining up) we use permutations; when order does not matter (choosing, picking a subset, forming a committee) we use combinations.

AMC10 loves these topics because the same situation can be counted two different ways — and picking the wrong one is a classic error. Circular arrangements and problems with repeated elements are recurring AMC favorites, as are "stars and bars" distribution problems that disguise themselves as word problems.

A fluent AMC solver has all six techniques in this guide automatic, and can quickly identify which one a problem is testing based on keywords like "arrange," "choose," "distribute," or "committee."

Key facts

• Permutation (order matters): arrange $r$ items from $n$ distinct items. $P(n,r) = \frac{n!}{(n-r)!}$ Use when the problem asks for sequences, codes, or ordered arrangements.

• Combination (order doesn't matter): choose $r$ items from $n$ distinct items. $\binom{n}{r} = \frac{n!}{r!\,(n-r)!}$ Use when the problem asks for subsets, teams, or unordered groups.

• Relationship: $\dbinom{n}{r} = \dfrac{P(n,r)}{r!}$ — divide out the $r!$ orderings you don't care about.

• Circular arrangements of $n$ distinct objects (rotations equivalent): $(n-1)!$ Use when objects sit around a table, ring, or necklace and no seat is labeled. Fix one object to break rotational symmetry, then arrange the rest linearly.

• Arrangements with repeats (multinomial): arrange $n$ objects where group 1 has $k_1$ identical copies, group 2 has $k_2$, etc. $\frac{n!}{k_1!\,k_2!\cdots}$ Use when some objects are identical (repeated letters, identical tiles, same color balls).

• Stars and bars (distributing identical objects): number of ways to distribute $n$ identical objects into $k$ distinct bins with each bin receiving $\ge 0$ objects: $\binom{n+k-1}{k-1}$ Use when objects are indistinguishable and you just care how many go to each recipient.

• Committee with roles: if some positions are distinct (e.g., president, treasurer), multiply combinations by the ways to assign roles. E.g., choose 1 chair from $n$ people then a committee of $r$ from the remaining $n-1$: $n \cdot \binom{n-1}{r}$.

Worked example 1

How many distinct arrangements are there of the letters in the word LEVEL?

The letters are L, E, V, E, L — 5 total with L repeated twice and E repeated twice.

If all five were distinct we'd have $5! = 120$ arrangements. But swapping the two L's produces the same word, as does swapping the two E's. So we divide:

$\frac{5!}{2!\cdot 2!} = \frac{120}{4} = 30.$

There are 30 distinct arrangements.

Sanity check (small case): For the word "AAB" the formula gives $3!/2! = 3$: AAB, ABA, BAA — three arrangements. ✓

Worked example 2

A club has 7 members. They want to elect a president and then form a 3-person advisory board (from the remaining 6 members, unordered). How many ways can this be done?

Step 1 — Choose president: 7 options.

Step 2 — Choose the board: After the president is chosen, 6 members remain. We pick an unordered group of 3: $\binom{6}{3} = 20.$

Total: $7 \times 20 = 140.$

Why not just $\binom{7}{4}$? Because the president is a labeled role — swapping the president with a board member gives a different outcome. Labeled roles force you to treat that selection as ordered (or multiply it in separately).

Common traps

• Forgetting to divide by symmetry. Arranging 5 people in a circle is not $5!$ but $(5-1)! = 24$ — rotations are identical.

• Treating identical objects as distinct. The letters in MISSISSIPPI are not all different. Always check for repeated elements before applying $n!$.

• Stars and bars miscount. The formula $\binom{n+k-1}{k-1}$ counts distributions where bins can be empty. If each bin needs at least 1, give each bin 1 first (reducing to $n-k$ objects) and then apply stars and bars: $\binom{(n-k)+k-1}{k-1} = \binom{n-1}{k-1}$.

• Roles vs. membership. A "committee of 5" is a combination. A "committee of 5 with a president and secretary" is a combination for 3 regular members times a permutation (or separate choices) for the 2 roles.

• Double-counting complementary events. When you use complementary counting ("total minus bad cases"), make sure your "bad" cases are correctly enumerated and do not overlap.