Why the Counting Formulas Work

01 · Meet the counting club

Imagine 8 students and 3 positions. The numbers are the same, but a code, a medal podium, and a team are different kinds of outcomes.

AAva

BBen

CCoco

DDiego

EEmi

FFinn

GGia

HHugo

Pick a scene. Watch the chairs—and the number of choices—change.

The swap test: If two selected people swap places, does the outcome change? Yes means roles/order matter. No means it is an unlabeled group.

02 · The formulas are one connected story

Reuse allowed

Every slot keeps all n choices.

nʳ

No reuse

Each filled slot removes one choice.

P(n,r)

No roles

Divide away the r! internal orders.

C(n,r)

Permutation formula

For r labeled positions without repetition, use exactly r descending factors:

P(n,r) = n(n−1)···(n−r+1) = n!/(n−r)!

The denominator (n−r)! cancels the unused tail of n!. It does not correct overcounting.

Combination formula

C(n,r) = P(n,r)/r! = n!/[r!(n−r)!]

The denominator r! has a different job: it removes the repeated orders of the same chosen group.

03 · Six tickets, but only one team

Suppose the chosen team is {Ava, Ben, Coco}. A permutation machine prints six tickets:

ABCABCACBACBBACBACBCABCACABCABCBACBA

ABC

Team ABC
same members, no roles

There are 3! = 6 arrangements of the same three members. Every team is overcounted by the same factor, so divide by 3!.

C(8,3) = [8·7·6] / [3·2·1] = 56

04 · Same people, different answer

Question	Structure	Count
Three-character code; reuse allowed	3 labeled slots, all 8 choices each time	8³ = 512
Gold, silver, bronze	3 labeled roles, no reuse	8·7·6 = 336
Three-person team	No roles; 3! orders describe each team	(8·7·6)/3! = 56

Important: The word “choose” does not guarantee a combination. “Choose a president and a secretary” has two different roles, so order matters.

05 · Follow the formula trail

Pretend you are a formula detective. Start at the magnifying glass, then follow the answer that matches your problem.

START HERECan the same choice be used again?

✅ YES

Choices refill!
Every slot still has n choices.

REUSE ALLOWEDnʳlike a 3-character code

🚫 NO

Try the swap test.
Would swapping two people make a new result?

✅ YES🏅 Roles matterP(n,r)gold ≠ silver

❌ NO🤝 Same teamC(n,r){A, B} = {B, A}

Ella's shortcut: Refill? Use nʳ. No refill? Do the swap test: a new result means P; the same result means C.

What does one outcome look like?
Draw its positions, roles, or group.

Can an object be reused?
If yes and there are r labeled slots, think nʳ.

Does swapping selected objects change the outcome?
If yes, order/roles matter: use P(n,r) or direct multiplication.

If swapping changes nothing, how many orders name the same group?
Divide P(n,r) by r! to get C(n,r).

06 · Useful consequences

Choose what is in—or what is left out

C(n,r) = C(n,n−r)

Choosing 3 students from 8 automatically determines the 5 students left out. This is why C(8,3)=C(8,5).

Choose first, assign roles second

C(n,r)·r! = P(n,r)

Choosing a pair and then assigning speaker/writer gives the same result as assigning those two roles directly.

Why is 0! equal to 1?

One empty arrangement
There is exactly one way to arrange nothing: do nothing.

0! = 1

Repeated objects use the same divide-out idea

BANANA has 6 letters, but swapping its three identical A's or two identical N's creates no new word:

6!/(3!2!) = 60

07 · Your turn

1 · Easy: How many three-digit codes use digits 1–5 if repetition is allowed?

125. Three labeled slots, 5 choices each: 5³.

2 · AMC 8: Eight students run a race. How many gold–silver–bronze results?

336. The roles are distinct: 8·7·6=P(8,3).

3 · AMC 8: How many four-student teams can be chosen from 9 students?

126. No roles: C(9,4)=9·8·7·6/(4·3·2·1).

4 · AMC 10: Choose a four-person committee from 10 students, then choose its chair.

840. C(10,4)·4=210·4. Or choose the chair first: 10·C(9,3).

5 · AMC 10: Choose a seven-person committee from 6 girls and 8 boys with exactly 3 girls.

1400. Choose both required groups: C(6,3)C(8,4)=20·70.