Introduction to Combinations

Learn how to count selections where order doesn't matter using the combinations formula.

Advanced25 minLesson
Printable worksheet

Definition

A combination is a selection of items where the order does not matter.
If you're choosing rr items from a group of nn items, the number of combinations is:
C(n,r)=(nr)=n!r!(n−r)!C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}
Key difference from permutations:
  • Permutation: Order matters (arranging)
  • Combination: Order doesn't matter (selecting)
Example: Choosing 2 people from Alice, Bob, and Carol:
  • {Alice, Bob} is the same as {Bob, Alice}
  • There are only 3 combinations: {A,B}, {A,C}, {B,C}

Try it now

Which formula represents the number of combinations when choosing rr items from nn items?

Worked Examples

A club has 5 members. How many ways can you choose a 2-person committee?
1

Identify n and r

n=5n = 5 (total members), r=2r = 2 (to choose) → n=5n = 5, r=2r = 2
2

Apply the combination formula

C(5,2)=5!2!(5−2)!=5!2!×3!C(5,2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \times 3!} → Formula set up
3

Calculate factorials

5×4×3!2×1×3!=5×42\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} → 202\frac{20}{2}
4

Simplify

202=10\frac{20}{2} = 10 → 1010 ways

Common Mistakes

Confusing combinations with permutations

Why it's wrong: Permutations count arrangements (order matters), combinations count selections (order doesn't matter). P(5,2)=20P(5,2) = 20 but C(5,2)=10C(5,2) = 10.

Correct: Ask yourself: 'Does rearranging change the outcome?' If choosing a committee, {A,B} = {B,A}, so use combinations.

Forgetting to divide by r!r! in the formula

Why it's wrong: The division by r!r! removes the duplicate orderings of the same selection.

Correct: Always use the full formula: C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!}

Calculating factorials incorrectly

Why it's wrong: Large factorials can be overwhelming. Remember that most terms cancel out.

Correct: Simplify before multiplying: 7!5!×2!=7×62=21\frac{7!}{5! \times 2!} = \frac{7 \times 6}{2} = 21

Interactive Visual

Choosing a group

Choose 2 of 5: A, B, C, D, E

10 groups

P(5, 2) = 5 × 4 = 20

C(5, 2) = P(5, 2) / 2! = 20 / 2 = 10

Every group of 2 can be ordered in 2! = 2 ways, so the number of ordered lists is divided by 2!. Tap a group to see its orders.

Pascal's triangle

n=011n=1112n=21214n=313318n=41464116n=51510105132n=6161520156164n=7172135352171128n=818285670562881256

C(8, 3) = 56 = 21 + 35

Row 8: 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256 = 2⁸

Tap a number: it is the sum of the two numbers above it, and it equals C(n, k).

Interactive Sandbox

Expression Calculator

Try these:

History

No calculations yet

Practice Problems

16 problems
Problem 1 of 16
Easy
Which formula represents the number of combinations when choosing rr items from nn items?

Why It Matters

Combinations are essential for understanding probability and making real-world decisions:
  • Lottery: How many ways can 6 numbers be drawn from 49?
  • Team Selection: How many ways can you pick 5 players from 12 for a starting lineup?
  • Menu Planning: How many 3-course meals from 10 dishes?
  • Science: How many ways to select samples for experiments?
Understanding combinations helps you calculate probabilities and make informed choices!

Real World Applications

Sports Team Selection

Coaches use combinations to understand how many different team lineups are possible.

Example:

A basketball coach choosing 5 starters from 12 players: C(12,5)=792C(12,5) = 792 possible lineups.

1Try It Yourself

A soccer coach has 18 players and needs to choose 11 for the starting lineup.

How many different starting lineups are possible?

Step 1: Write the mathematical expression

Use C(18,11)C(18, 11):

Card Games

Combinations calculate the number of possible hands in card games.

Example:

A 5-card poker hand from 52 cards: C(52,5)=2,598,960C(52,5) = 2,598,960 possible hands.

2Try It Yourself

How many different 7-card hands are possible in a game?

Calculate C(52,7)C(52,7).

Step 1: Write the mathematical expression

Set up the formula:

Menu Planning

Restaurants use combinations to calculate meal options and combo deals.

Example:

Choosing 3 appetizers from 10 options: C(10,3)=120C(10,3) = 120 different selections.

3Try It Yourself

A catering company offers 15 different dishes. A customer wants to select 4 for their event.

How many different 4-dish selections are possible?

Step 1: Write the mathematical expression

Calculate C(15,4)C(15, 4):

Key Takeaways

  • 1Combinations count selections where order doesn't matter
  • 2Formula: C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!}, read as 'n choose r'
  • 3Key difference: Permutations = arrangements (order matters), Combinations = selections (order doesn't)
  • 4C(n,r)=C(n,n−r)C(n,r) = C(n, n-r): Choosing what to include equals choosing what to exclude

Frequently Asked Questions

Use combinations when order doesn't matter (selecting a committee, choosing toppings). Use permutations when order matters (assigning positions, arranging a lineup in order).
Use combinations when order doesn't matter (selecting a committee, choosing toppings). Use permutations when order matters (assigning positions, arranging a lineup in order).
Choosing rr items to include is the same as choosing n−rn-r items to exclude. For example, C(5,2)=C(5,3)=10C(5,2) = C(5,3) = 10.
The exclamation mark means factorial. 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120. It counts all possible arrangements of a set.

Glossary

Combination
A selection of items where order does not matter
Factorial
The product of all positive integers up to n, written as n!n!
Binomial coefficient
Another name for C(n,r)C(n,r), written as (nr)\binom{n}{r}
n choose r
Common way to read C(n,r)C(n,r), meaning 'the number of ways to choose r from n'

Formula Card

Combinations Formula

C(n,r)=(nr)=n!r!(n−r)!C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}

n = total items, r = items chosen. Symmetry: $C(n,r) = C(n, n-r)$. Special cases: $C(n,0) = C(n,n) = 1$

Today's challengeOne problem, three tries

More in This Topic

Introduction to Permutations

Advanced25 min