HL · Counting & partial fractions 1 · Counting principles: permutations & combinations

Module 8 of 46Lesson 1 of 7 · HL · Counting & partial fractions 1 · Counting principles: permutations & combinations

Counting principles, extended binomial theorem & partial fractions · Essential understanding 1

Counting principles: permutations & combinations

IB AA AHL 1.10with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Counting questions all come down to two moves: multiply when you make choices one after the other (this AND that), and add when you pick one route or the other (this OR that). Permutations count selections where order matters; combinations count selections where it does not.

By the end of this lesson you can:

use the multiplication (product) rule and the addition rule
use $n!$ , $^nP_r$ and $^nC_r$ and know when to use each
decide whether order matters in a question
count “at least” and “at most” selections, using the complement when it is quicker
use $^nC_r = {}^nC_{n-r}$ and solve simple equations that contain $^nC_r$

🧑🏾 Imagine this…

Kagiso is packing for a trip to Kasane. She has 3 shirts and 2 pairs of trousers. Pick a shirt (3 ways) and then pick trousers (2 ways): 3 × 2 = 6 different outfits. Easy. Now she can travel either by bus (4 different departures) or by combi (5 different departures): that is 4 + 5 = 9 ways to leave. Everything in this lesson is built from those two ideas.

Multiplication rule

if a task has stages, and stage 1 can be done in

m

ways and stage 2 in

n

ways (whatever you did in stage 1), then the whole task can be done in

m \times n

ways.

Addition rule

if a task can be done in one way or another, and the two lists share nothing, add the numbers of ways.

Factorial

n! = n \times (n-1) \times \dots \times 2 \times 1

, with

0! = 1

. It counts the ways to put

n

different things in a row.

Permutation

an arrangement where order matters. Choosing and arranging

r

things from

n

different things.

Combination

a selection where order does not matter: just a group of

r

things from

n

★ The formulas

Permutations: $^nP_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$ (r factors).
Combinations: $^nC_r = \dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!} = \dfrac{^nP_r}{r!}$ . Each group of $r$ can be arranged in $r!$ ways, so dividing by $r!$ removes the order.
Symmetry: $^nC_r = {}^nC_{n-r}$ (choosing the $r$ people to include is the same as choosing the $n-r$ to leave out). Also $^nC_0 = {}^nC_n = 1$ and $^nC_1 = n$ .
Complement: “at least one” = (all possibilities) − (none). Use it whenever it saves work.
The combination formula $^nC_r$ is in the IB formula booklet (as the binomial coefficient); learn $^nP_r = \dfrac{n!}{(n-r)!}$ as well, because it is quick to use.

⚠️ Mistakes that lose marks

Order test. Ask: “if I swap two of the chosen things, is it a different answer?” Yes (president, secretary, PIN code) → permutation. No (a committee, a hand of cards) → combination.
Add or multiply? “and then” → multiply. “either … or” → add. Never add stages that happen one after the other.
Do not forget repeats allowed. A 4-digit PIN where digits can repeat has $10^4$ possibilities, not $^{10}P_4$ .
In “at least” questions, check the cases carefully: “at least 2 girls” leaves out both 0 girls and 1 girl.

Prof. Kitso says

Paper 1 has no calculator, so write

^nC_r

as a product and cancel:

^{10}C_3 = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

. Never work out

10!

and divide.

🧪 Try it yourselfAdd, multiply, permute or combine?

Worked examples

Example 1.1Multiplication and addition rules

A Gaborone restaurant offers 4 starters, 6 main courses and 3 desserts. A “full meal” is one starter, one main and one dessert. A “light meal” is one main course or one dessert.

(a) How many different full meals are there?
(b) How many different light meals are there?
(c) How many full meals are there if Neo refuses to have the dessert called “malva pudding”?

🧠 Think it through

A full meal is a starter AND a main AND a dessert, so multiply.
A light meal is a main OR a dessert, so add.
If one dessert is ruled out, only 2 desserts remain.

✍️ By hand

(a)

4 \times 6 \times 3 = 72

(M1)(A1)

(b)

6 + 3 = 9

(A1)

(c)

4 \times 6 \times 2 = 48

(A1)

✅ Answer (a) 72 (b) 9 (c) 48

Example 1.2Permutations: order matters

A school entrepreneurship fair gives prizes for 1st, 2nd and 3rd place to different stalls. There are 10 stalls.

(a) In how many ways can the three prizes be awarded?
(b) In how many ways can the three prizes be awarded if stall A must win one of the three prizes?

🧠 Think it through

Each prize is different, so order matters: $^{10}P_3$ .
For (b), first choose WHICH prize A gets (3 ways), then fill the other two prizes from the remaining 9 stalls.

✍️ By hand

(a)

^{10}P_3 = 10 \times 9 \times 8 = 720

(M1)(A1)

(b) A gets one of 3 prizes: 3 ways. The other two prizes:

9 \times 8 = 72

ways. Total

3 \times 72 = 216

(M1)(M1)(A1)

Check: stall A wins nothing in

^{9}P_3 = 504

ways, and

720 - 504 = 216

✓

🔢 On your TI-Nspire CX II

To check (a): menu💻 click menu on keypad▸5: Probability▸2: Permutations then type nPr(10,3) enter ↵💻 Enter. For (b) check 3*nPr(9,2).

✅ Answer (a) 720 (b) 216

Example 1.3Combinations and “at least”

A debate team of 5 is chosen from 6 boys and 5 girls.

(a) How many teams are possible?
(b) How many teams contain at least 2 girls?

🧠 Think it through

A team is a group, so order does not matter: combinations.
“At least 2 girls” = all teams − (0 girls) − (exactly 1 girl).
0 girls: all 5 from the 6 boys. 1 girl: 1 of 5 girls AND 4 of 6 boys.

✍️ By hand

(a)

^{11}C_5 = \dfrac{11 \times 10 \times 9 \times 8 \times 7}{5!} = 462

(A1)

(b) 0 girls:

^6C_5 = 6

. 1 girl:

^5C_1 \times {}^6C_4 = 5 \times 15 = 75

. (M1)(A1)

At least 2 girls:

462 - 6 - 75 = 381

(M1)(A1)

Direct check:

^5C_2{}^6C_3 + {}^5C_3{}^6C_2 + {}^5C_4{}^6C_1 + {}^5C_5{}^6C_0 = 200 + 150 + 30 + 1 = 381

✓

🔢 On your TI-Nspire CX II

menu💻 click menu on keypad▸5: Probability▸3: Combinations gives nCr(n,r). Type the check:

✅ Answer (a) 462 (b) 381

Example 1.4Counting digits and numbers

How many 4-digit numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7 if no digit may be used more than once, and the number is

(a) any such number?
(b) even?
(c) greater than 5000?

🧠 Think it through

Fill the four places one by one and count the choices, starting with the place that has a restriction.
Even: the last digit must be 2, 4 or 6, so fill the units place first.
Greater than 5000: the first digit must be 5, 6 or 7.

✍️ By hand

(a)

^7P_4 = 7 \times 6 \times 5 \times 4 = 840

(b) units digit: 3 choices; then the other three places from the remaining 6 digits:

6 \times 5 \times 4

. Total

3 \times 120 = 360

(M1)(A1)

6 \times 5 \times 4 = 120

. Total

3 \times 120 = 360

(M1)(A1)

✅ Answer (a) 840 (b) 360 (c) 360

Always fill the restricted place first. If you fill the free places first, you will not know how many digits are left for the restricted one.

Example 1.5Combination equations and symmetry

(a) Show that

^nC_2 = \dfrac{n(n-1)}{2}

. (b) Find

n

^nC_2 = 45

. (c) Given

^{14}C_5 = 2002

, write down

^{14}C_9

and

^{14}C_4 + {}^{14}C_5

(hence find

^{15}C_5

🧠 Think it through

Write $^nC_2 = \dfrac{n!}{2!\,(n-2)!}$ and cancel $(n-2)!$ .
Set $\dfrac{n(n-1)}{2} = 45$ , get a quadratic, reject the negative root.
Symmetry: $^{14}C_9 = {}^{14}C_5$ . Pascal: $^{14}C_4 + {}^{14}C_5 = {}^{15}C_5$ .

✍️ By hand

(a)

^nC_2 = \dfrac{n!}{2!(n-2)!} = \dfrac{n(n-1)(n-2)!}{2\,(n-2)!} = \dfrac{n(n-1)}{2}

(M1)(AG)

(b)

n(n-1) = 90 \Rightarrow n^2 - n - 90 = 0 \Rightarrow (n-10)(n+9) = 0

, so

n = 10

(as

n>0

) (M1)(A1)

(c)

^{14}C_9 = {}^{14}C_5 = 2002

because

9 = 14 - 5

^{14}C_4 = \dfrac{14 \times 13 \times 12 \times 11}{24} = 1001

, so

^{15}C_5 = 1001 + 2002 = 3003

. (Choose 5 of 15: either include the last person,

^{14}C_4

ways, or leave them out,

^{14}C_5

ways.)

🔢 On your TI-Nspire CX II

Check with the calculator:

^{15}C_5

= 3003

Challenge questions

⭐ Challenge 1.6Even numbers from six digitsPaper 1 style

Three-digit numbers are formed from the digits 1, 2, 3, 4, 5, 6 with no digit repeated. How many of them are even? How many are even and greater than 400?

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Fill the units place first: 3 choices (2, 4 or 6).
For “greater than 400”, the hundreds digit is 4, 5 or 6, but if the units digit is 4 or 6 you must watch what is left.
Split into cases by the hundreds digit, or by the units digit.

✍️ By hand

Even: units 3 ways, then

5 \times 4

3 \times 20 = 60

Hundreds 4: units 2 or 6 (2 ways), tens from the remaining 4 digits:

2 \times 4 = 8

Hundreds 5: units 3 ways, tens 4 ways: 12. Hundreds 6: units 2 or 4 (2 ways), tens 4 ways: 8.

Total

8 + 12 + 8 = 28

✅ Answer 60 even numbers; 28 of them are greater than 400.

⭐ Challenge 1.7Choosing a delegationPaper 2 style

A school sends a delegation of 6 to a maths conference. There are 8 Form 5 students and 7 Form 4 students to choose from. The delegation must have at least 2 students from each form.

(a) How many delegations are possible with no restrictions?
(b) How many delegations satisfy the “at least 2 from each form” rule?
(c) Two of the Form 5 students, Lesedi and Thato, are best friends and will only go together or not at all. How many of the delegations in (b) respect this?

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) $^{15}C_6$ .
(b) Cases (Form 5, Form 4): (2,4), (3,3), (4,2). Add them. Or subtract the bad cases (6,0), (5,1), (1,5), (0,6).
(c) Count delegations containing both friends plus those containing neither, case by case.

✍️ By hand
(a) 
(b) 
(c) Both friends in: choose the other Form 5 students from the remaining 6. (F5,F4) = (2,4): ; (3,3): ; (4,2): . Sum 560.
Neither in: the other 6 Form 5 students. (2,4): ; (3,3): ; (4,2): . Sum 1540.
Total .

🔢 On your TI-Nspire CX II

Use nCr( for each product and add on the home screen.

✅ Answer (a) 5005 (b) 4410 (c) 2100

Next: Lesson 2 · Arrangements with restrictions

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Counting & partial fractions 1 · Counting principles: permutations & combinations

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Counting principles: permutations & combinations

Worked examples

Challenge questions