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Counting: The Counting Principle, Permutations and Combinations

Count outcomes fast by multiplying choices, and know when order matters (permutations) or not (combinations).

Practice this skill 25 min Counting, combinations and permutations Leer en español
Key idea

What you'll learn

In this lesson you will learn how to count the number of possible outcomes without listing every single one. You will use three tools:

  1. The fundamental counting principle — multiply the number of choices at each step.
  2. Permutations — count arrangements when order matters (1st, 2nd, 3rd place; a PIN code; seats in a row).
  3. Combinations — count groups when order does not matter (a committee, pizza toppings, a hand of cards).

Why it matters on the GED: counting questions appear in the Quantitative Problem Solving part of the test, usually as short word problems: "How many different outfits…?", "How many ways can a manager choose…?", "How many license plates…?". They also feed into probability (the next lesson), because a probability is often (number of ways to win) ÷ (number of possible outcomes).

Real life: counting tells you how many password or PIN combinations exist (that is why longer passwords are safer), how many meal deals a restaurant can advertise, or how many different work schedules a manager can build.

Words to know

  • Outcome — one possible result. Example: "Thin crust, Large" is one outcome when ordering a pizza.
  • Stage (or step) — one decision in a multi-step choice. Example: choosing a crust is one stage; choosing a size is another.
  • Fundamental counting principle — if one stage has m choices and the next has n choices, together there are m \times n outcomes. Example: 2 crusts and 3 sizes give 2 \times 3 = 6 pizzas.
  • Repetition allowed — the same item may be used again. Example: a PIN like 7-7-2-7.
  • Without repetition — once an item is used, it is gone. Example: a person can't be both president and treasurer.
  • Factorial (n!) — multiply n by every whole number below it down to 1. Example: 4! = 4 \times 3 \times 2 \times 1 = 24. By definition, 0! = 1.
  • Permutation — an arrangement where order matters. Example: Ana-Ben-Cal and Cal-Ben-Ana are different finishing orders.
  • Combination — a selection where order does not matter. Example: the committee {Ana, Ben, Cal} is the same as {Cal, Ben, Ana}.
  • {}_nP_r — the number of permutations of r items chosen from n. Example: {}_5P_2 = 20.
  • {}_nC_r — the number of combinations of r items chosen from n. Example: {}_5C_2 = 10.

All Math words in the glossary

1. The fundamental counting principle

When a choice happens in stages, and every choice in one stage can be paired with every choice in the next, multiply:

\text{total outcomes} = (\text{choices at stage 1}) \times (\text{choices at stage 2}) \times (\text{choices at stage 3}) \times \cdots

Example: A pizza comes in 2 crusts (thin, thick) and 3 sizes (S, M, L). The tree diagram below shows every outcome. There are 2 \times 3 = 6.

Tree diagram: from Start, two branches for crust (Thin, Thick). Each crust splits into three branches for size (Small, Medium, Large), giving 6 outcomes: Thin-Small, Thin-Medium, Thin-Large, Thick-Small, Thick-Medium, Thick-Large. 2 times 3 = 6.StartThinS1. Thin, SmallM2. Thin, MediumL3. Thin, LargeThickS4. Thick, SmallM5. Thick, MediumL6. Thick, LargeCrust (2 choices)Size (3 choices)

You do not need to draw the tree on the GED — it just shows why we multiply: each of the 2 crust branches splits into 3 size branches.

Example: A sandwich: 3 breads, 4 meats, 2 cheeses, one of each → 3 \times 4 \times 2 = 24 sandwiches.

"And" versus "or" inside a stage. Multiply when you choose one thing and another. Add when a single stage lets you pick from one list or another list.

Example: A lunch special is a starter (4 soups or 3 salads) and a main (5 choices). Starters: 4 + 3 = 7. Total: 7 \times 5 = 35.

Two stages for a lunch special. Stage 1, the starter, is 4 soups or 3 salads, so 4 + 3 = 7 starters (add for "or" inside a stage). Stage 2 is 5 mains. Multiply between stages: 7 times 5 = 35 lunch specials.Stage 1: starter4soups3saladsor4 + 3 = 7 starters (add)×Stage 2: main5mains5 choices"or" inside a stage → add; "and" between stages → multiply7 × 5 = 35 lunch specials

Repetition allowed. If items can be reused, every stage has the full number of choices.

Example: A 4-digit PIN using digits 0–9: 10 \times 10 \times 10 \times 10 = 10^4 = 10{,}000.

Example: A license plate with 3 letters then 3 digits: 26^3 \times 10^3 = 17{,}576 \times 1{,}000 = 17{,}576{,}000.

Restrictions first. If one stage has a special rule (must be odd, can't start with 0, a certain person must be first), fill that stage first, then the others.

Example: 3-digit codes from digits 1–9, no repeats, and the code must be odd. Last digit: 5 choices (1, 3, 5, 7, 9). First digit: 8 left. Middle: 7 left. Total: 5 \times 8 \times 7 = 280.

Three slots for a 3-digit code using digits 1 to 9 with no repeats that must be odd. The restricted last slot is filled first with 5 choices (1, 3, 5, 7, 9). Then the first digit has 8 choices and the middle digit 7. 8 times 7 times 5 = 280 codes.Restriction first: 3-digit odd code, digits 1–9, no repeats81st digitfill 2nd×72nd digitfill 3rd×53rd digitfill 1st: oddLast digit must be odd: 1, 3, 5, 7 or 9 → 5 choices.Digits 1–9, no repeats: 8 are left for the 1st digit, then 7.8 × 7 × 5 = 280 codes

2. Permutations: when order matters

If you arrange all n different items in a row, the first spot has n choices, the next n-1, and so on down to 1. That product is n! ("n factorial").

Example: 5 people line up for a photo: 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 ways.

n1234567
n!126241207205,040

Factorials grow very fast — 10! is already 3,628,800.

If you fill only r spots from n items (no repeats), stop multiplying after r factors:

{}_nP_r = \underbrace{n \times (n-1) \times \cdots}_{r \text{ factors}} = \frac{n!}{(n-r)!}

Example: A club of 10 people elects a president, a vice-president and a treasurer. The titles are different, so order matters: {}_{10}P_3 = 10 \times 9 \times 8 = 720.

Three slots labelled President, Vice-president and Treasurer for a club of 10 people. The slots hold 10, 9 and 8 choices because a person cannot hold two offices. 10 times 9 times 8 = 720 permutations.Club of 10 people, 3 different offices10President×9Vice-president×8TreasurerDifferent titles → order matters. Each person used once,so the choices go down by 1 at each spot.₁₀P₃ = 10 × 9 × 8 = 720

Example: Gold, silver and bronze among 6 runners: 6 \times 5 \times 4 = 120.

Clue words for "order matters": first/second/third, president/vice-president, line up, arrange, seat, code, password, schedule in order, rank.

3. Combinations: when order does not matter

In a combination you only care which items are chosen, not the order. Every group of r items can be arranged in r! orders, and a permutation counts each of those separately. So divide them out:

{}_nC_r = \frac{{}_nP_r}{r!} = \frac{n!}{r!\,(n-r)!}

Example: Choose 3 toppings from 8 for a pizza. Order doesn't matter (pepperoni-onion-olive is the same pizza as olive-onion-pepperoni).

{}_8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
The 6 orders of the same three toppings P, N, O (pepperoni, onion, olive): PNO, PON, NPO, NOP, OPN and ONP, 3! = 6 in all, all point to one group, the same pizza. A permutation counts each group 6 times, so divide: 8 times 7 times 6 = 336, divided by 6 = 56 combinations.3! = 6 orders (permutations)PNOPONNPONOPOPNONPsame pizzaPNO1 combinationP = pepperoni, N = onion,O = olive₈C₃ = (8 × 7 × 6) ÷ 3! = 336 ÷ 6 = 56 pizzas

Example: Choose 2 people from 5 to work Saturday: \dfrac{5 \times 4}{2 \times 1} = 10.

Quick facts:

  • {}_nC_1 = n and {}_nC_n = 1.
  • {}_nC_r = {}_nC_{n-r}. Choosing 3 people to go is the same as choosing 7 to stay: {}_{10}C_3 = {}_{10}C_7 = 120.
  • A combination is always smaller than or equal to the matching permutation.

Combining groups. If you choose from two separate groups, find each combination and multiply (counting principle).

Example: A committee needs 2 managers from 5 and 3 workers from 7: {}_5C_2 \times {}_7C_3 = 10 \times 35 = 350.

How to decide: Ask, "If I swap two of the chosen items, do I get a different result?" If yes → permutation. If no → combination.

SituationOrder matters?Tool
3 officers with different titlesYes{}_nP_r
3-person committee, all equalNo{}_nC_r
Arrange all 6 books on a shelfYes6!
4-digit PIN (repeats OK)Yes10^4
Pick 2 side dishes from 6No{}_6C_2 = 15
Formula

Formulas (memorize these)

\text{Counting principle: } m \times n \times p \times \cdots n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 {}_nP_r = \frac{n!}{(n-r)!} \qquad\qquad {}_nC_r = \frac{n!}{r!\,(n-r)!}
  • n = how many items you can choose from.
  • r = how many you actually choose (or how many spots you fill).
  • ! = factorial.

These formulas are not on the GED formula sheet, so you must remember them — or, better, remember the idea: multiply the choices for each spot; if order doesn't matter, divide by r!. The TI-30XS calculator also has {}_nP_r, {}_nC_r and ! built in (see the calculator tip).

Worked example

Worked example 1 (easy): outfits

Dana has 4 shirts, 3 pairs of pants and 2 pairs of shoes. How many different outfits (one of each) can she make?

  1. There are 3 stages: shirt, pants, shoes.
  2. Multiply the choices: 4 \times 3 \times 2 = 24.
Tree diagram for Dana's outfits. Start splits into 4 shirts; each shirt splits into 3 pants (A, B, C); each pants choice splits into 2 shoes (sneakers, boots). The 24 end branches are numbered 1 to 24, so 4 times 3 times 2 = 24 outfits.Shirt (4)Pants (3)Shoes (2)StartShirt 1Pants ASneakers1Boots2Pants BSneakers3Boots4Pants CSneakers5Boots6Shirt 2Pants ASneakers7Boots8Pants BSneakers9Boots10Pants CSneakers11Boots12Shirt 3Pants ASneakers13Boots14Pants BSneakers15Boots16Pants CSneakers17Boots18Shirt 4Pants ASneakers19Boots20Pants BSneakers21Boots22Pants CSneakers23Boots244 × 3 × 2 = 24 outfits

Answer: 24 outfits.

Check: For each of the 4 shirts there are 3 \times 2 = 6 pants-and-shoes pairs, and 4 \times 6 = 24 ✓

Worked example

Worked example 2: officers (order matters)

A bowling league of 12 members chooses a captain and a co-captain. How many ways can this be done?

  1. The two jobs are different, so order matters → permutation.
  2. Captain: 12 choices. Co-captain: 11 left (one person can't hold both jobs).
  3. {}_{12}P_2 = 12 \times 11 = 132.
Two slots, Captain with 12 choices and Co-captain with 11 choices, because the same person cannot hold both jobs. 12 times 11 = 132 ways.Bowling league of 12 members12Captain×11Co-captainTwo different jobs → order matters.The captain can't also be co-captain, so 11 are left.₁₂P₂ = 12 × 11 = 132 ways

Answer: 132 ways.

Check: If order didn't matter we'd get 132 \div 2 = 66 pairs; each pair can be assigned to the two jobs in 2 ways, 66 \times 2 = 132 ✓

Worked example

Worked example 3: a committee (order does not matter)

A school board must choose 4 parents from 9 volunteers to serve on an equal-member committee. How many different committees are possible?

  1. All committee members are equal → order doesn't matter → combination.
  2. {}_9C_4 = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}.
  3. Numerator: 9 \times 8 \times 7 \times 6 = 3{,}024. Denominator: 4! = 24.
  4. 3{,}024 \div 24 = 126.
Nine volunteers numbered 1 to 9; four of them (2, 4, 5, 8) are highlighted as one committee, which is the same committee in any order. 9C4 = (9 times 8 times 7 times 6) divided by (4 times 3 times 2 times 1) = 3,024 divided by 24 = 126 committees.9 volunteers — choose 4 equal members123456789Committee {2, 4, 5, 8}= {8, 5, 4, 2}: order doesn't matter₉C₄ = (9 × 8 × 7 × 6) ÷ (4 × 3 × 2 × 1)= 3,024 ÷ 24 = 126 committees

Answer: 126 committees.

Check: {}_9C_4 should equal {}_9C_5: \dfrac{9 \times 8 \times 7 \times 6 \times 5}{120} = \dfrac{15{,}120}{120} = 126 ✓

Worked example

Worked example 4 (real world): passwords

A website requires a password of 2 capital letters followed by 4 digits (0–9). Letters and digits may repeat. How many passwords are possible?

  1. Six stages: 2 letter spots and 4 digit spots.
  2. Each letter spot: 26 choices. Each digit spot: 10 choices.
  3. 26 \times 26 \times 10 \times 10 \times 10 \times 10 = 676 \times 10{,}000 = 6{,}760{,}000.
Six slots for a password: two letter slots with 26 choices each and four digit slots with 10 choices each. 26 squared = 676 and 10 to the 4th = 10,000; 676 times 10,000 = 6,760,000 passwords.Password: 2 capital letters, then 4 digits (repeats allowed)26letter×26letter×10digit×10digit×10digit×10digit26² = 67610⁴ = 10,000676 × 10,000 = 6,760,000 passwordsEvery spot keeps all its choices because repeats are allowed.

Answer: 6,760,000 passwords.

Check: 26^2 = 676 and 10^4 = 10{,}000; 676 \times 10{,}000 = 6{,}760{,}000 ✓

Worked example

Worked example 5 (GED-hard): work backwards

A manager says there are exactly 28 different ways to choose 2 employees from her team to attend a training. How many employees are on the team?

  1. Choosing 2 people with no order is {}_nC_2 = \dfrac{n(n-1)}{2}.
  2. Set it equal to 28: \dfrac{n(n-1)}{2} = 28, so n(n-1) = 56.
  3. Find two consecutive whole numbers whose product is 56: 8 \times 7 = 56.
  4. So n = 8.
Eight points numbered 1 to 8 for 8 employees, with a line joining every pair of points. There are 8 times 7 divided by 2 = 28 lines, one for each way to choose 2 employees. One line, between 1 and 4, is highlighted in red.123456788 employeesEach line = 1 pair(red line: pair 1 & 4)Lines: 8 × 7 ÷ 2= 28 pairsn(n − 1) ÷ 2 = 28→ n = 8

Answer: 8 employees.

Check: {}_8C_2 = \dfrac{8 \times 7}{2} = 28 ✓ (Tip: on multiple choice, just test each answer choice in \frac{n(n-1)}{2}.)

Calculator tip

Calculator tip: TI-30XS MultiView

The TI-30XS has counting tools under the prb key:

  • Factorial: type the number, press prb, choose !, then enter. Example: 6 prb (choose !) enter → 720.
  • Permutations: type n, press prb, choose nPr, type r, press enter. Example: 10 nPr 3 → 720.
  • Combinations: type n, press prb, choose nCr, type r, press enter. Example: 8 nCr 3 → 56.
  • Repetition allowed: use the ^ key. Example: 10 ^ 4 enter → 10000.

When NOT to use it: for small problems like 3 \times 4 \times 2, mental math is faster. And the calculator can't decide which tool to use — you must decide "order matters or not?" and "repeats allowed or not?" first. Many counting questions are short enough to do by hand.

Common mistake

Common mistakes

  • Adding instead of multiplying. 3 breads and 4 meats is not 3 + 4 = 7 sandwiches. Each bread goes with each meat: 3 \times 4 = 12. (Add only for an "or" choice inside one stage.)
  • Using a combination when order matters. Electing president, VP and treasurer from 10 is 10 \times 9 \times 8 = 720, not {}_{10}C_3 = 120 — the titles are different.
  • Using a permutation when order doesn't matter. 3 toppings from 8 is {}_8C_3 = 56, not 8 \times 7 \times 6 = 336. You must divide by 3! = 6.
  • Allowing repeats when they are not allowed (or the reverse). A person can't be two officers, so the choices go down: 10, 9, 8. A PIN can repeat digits, so it stays 10, 10, 10, 10.
  • Multiplying n \times r. "Choose 3 from 8" is not 8 \times 3 = 24. Use the formula.
  • Forgetting a restriction or filling it last. For "the code must be odd," fill the last digit first (5 odd choices), then the others.
Test strategy

GED test strategy

  • Read the question twice and underline the key words: arrange, order, first/second (permutation) vs. group, committee, choose, select (combination). Look for "may repeat" or "different".
  • Estimate. For the same n and r, a combination is smaller than a permutation. If your answer is huge for a simple committee question, you probably forgot to divide by r!.
  • Eliminate. Distractors are usually the "other" calculation: the sum of the choices, the permutation instead of the combination, or n^r with repeats. Compute the right one and look for it.
  • Work backwards when the total is given: test each answer choice in the formula.
  • Draw a tiny tree or list if you're unsure — for small numbers, listing outcomes is a reliable check.
  • Time: most counting questions take under 90 seconds. Use the prb key to save time on factorials.

Check yourself

Counting, combinations and permutationsNo calculator

A restaurant lets you build a burrito bowl with one rice, one protein, and one salsa. There are 2 kinds of rice, 5 proteins, and 4 salsas.

How many different burrito bowls can be made?

Answer choices
Hint

Each rice can be paired with every protein, and each of those pairs with every salsa.

Check yourself

Counting, combinations and permutationsFill in

A storage locker uses a 4-letter code. Each letter can be A, B, C, D, or E, and letters may repeat (for example, B B A E is allowed).

How many different codes are possible?

Type your answer in the box. Fractions like 3/4 are fine.

Hint

How many choices are there for each of the 4 positions? Does using a letter once stop you from using it again?

Summary

  • Counting principle: multiply the number of choices at each stage ("and" → multiply, "or" inside one stage → add).
  • Repeats allowed: every stage keeps all n choices → n^r.
  • Order matters, no repeats: {}_nP_r = n(n-1)(n-2)\cdots (r factors). All n items in a row: n!.
  • Order does not matter: {}_nC_r = \dfrac{{}_nP_r}{r!}.
  • Handle restrictions first. These formulas are not on the formula sheet — the prb key on the TI-30XS has !, nPr and nCr.
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