Module 12 of 46Lesson 1 of 7 · HL · AA 1.10 — Counting principles; permutations and combinations; extended binomial theorem

Course map · Topic 1: Number and Algebra

1.10 Counting principles; permutations and combinations; extended binomial theorem

HL only12 worked questions

Counting tells you how many ways something can happen without listing them all — the key question is always “does order matter?”. The extended binomial theorem lets you expand brackets with negative or fractional powers as an infinite series.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[3 marks]

Find the number of different arrangements of the letters of the word GABORONE.

Show the full step-by-step solution

Step 1🧠 ThinkCount the letters and spot repeats

G, A, B, O, R, O, N, E: 8 letters, and the letter O appears twice. Swapping the two O's gives the same word, so every arrangement has been counted twice — divide by

2!

Step 2✍️ By hand

\dfrac{8!}{2!}

= \dfrac{40\,320}{2} = 20\,160

Final answer

20\,160

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

A club in Maun has 8 members.

(a) Find the number of ways a group of 3 members can be chosen to attend a workshop.[2]
(b) Find the number of ways a chairperson, a secretary and a treasurer can be chosen (one person per post).[2]

Show the full step-by-step solution

Step 1🧠 ThinkDoes order matter?

In (a) the 3 people just go together — order does not matter: combination. In (b) the posts are different, so Neo as chair and Mpho as secretary is different from the swap: permutation.

Part (a)

Step 1✍️ By hand

\binom{8}{3} = \dfrac{8\times7\times6}{3\times2\times1}

= 56

Part (b)

Step 1✍️ By hand

8 choices for chair, then 7 for secretary, then 6 for treasurer.

{}^{8}P_{3} = 8\times7\times6

= 336

💡 Remember

336 = 56\times3!

: each group of 3 can be given the 3 posts in

3! = 6

ways.

Final answer (a)

56

(b)

336

Question 3●○○ Warm-up🖩 GDC · Paper 2[3 marks]

A netball squad has 12 players. A team of 7 is chosen for a match.

(a) Find the number of different teams that can be chosen.[1]
(b) Each of the 7 players is then given a different position (GS, GA, WA, C, WD, GD, GK). Find the number of ways of choosing and positioning the 7 players.[2]

Show the full step-by-step solution

Step 1🧠 ThinkTeam = combination, positions = permutation

(a) only asks who plays:

\binom{12}{7}

. (b) also asks where they play, so order matters:

{}^{12}P_{7}

Step 2🖩 On the GDCnCr( and nPr( on a Calculator page

menu💻 click menu on keypad›5: Probability›3: Combinations›12,7)›enter💻 Enter›menu💻 click menu on keypad›5: Probability›2: Permutations›12,7)›enter💻 Enter

792 teams; 3 991 680 line-ups

Final answer (a)

792

(b)

3\,991\,680

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[6 marks]

Consider

\sqrt{1+4x}

(a) Find the first three terms of its expansion in ascending powers of $x$ .[3]
(b) State the values of $x$ for which the expansion is valid.[1]
(c) Use your answer to (a) with $x = 0.01$ to find an approximation for $\sqrt{1.04}$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkSquare root = power one half

\sqrt{1+4x} = (1+4x)^{\frac{1}{2}}

. Use the extended binomial theorem with

n = \dfrac{1}{2}

and “

x

” replaced by

4x

Step 2✍️ By hand

1 + \dfrac{1}{2}(4x) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}(4x)^{2} + \dots

= 1 + 2x + \left(-\dfrac{1}{8}\right)(16x^{2}) + \dots

= 1 + 2x - 2x^{2} + \dots

Part (b)

Step 1✍️ By hand

|4x| \lt 1 \Rightarrow |x| \lt \dfrac{1}{4}

Part (c)

Step 1✍️ By hand

\sqrt{1.04} \approx 1 + 2(0.01) - 2(0.01)^{2}

= 1 + 0.02 - 0.0002 = 1.0198

✅ Check it At home you can check on the GDC:

\sqrt{1.04} = 1.019804\ldots

Final answer (a)

1 + 2x - 2x^{2}

(b)

|x| \lt \dfrac{1}{4}

(c)

1.0198

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

Find the first three terms in the expansion of

(2+x)^{-3}

in ascending powers of

x

, and state the interval of validity.

Show the full step-by-step solution

Step 1🧠 ThinkMake the first number 1

The theorem needs

(1 + \text{something})^{n}

. Take out the 2:

(2+x)^{-3} = 2^{-3}\left(1+\dfrac{x}{2}\right)^{-3} = \dfrac{1}{8}\left(1+\dfrac{x}{2}\right)^{-3}

Step 2✍️ By hand

\dfrac{1}{8}\left[1 + (-3)\left(\dfrac{x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(\dfrac{x}{2}\right)^{2} + \dots\right]

M1M1

= \dfrac{1}{8}\left[1 - \dfrac{3x}{2} + \dfrac{3x^{2}}{2} + \dots\right]

= \dfrac{1}{8} - \dfrac{3x}{16} + \dfrac{3x^{2}}{16} + \dots

Valid for

\left|\dfrac{x}{2}\right| \lt 1

, i.e.

|x| \lt 2

Final answer

\dfrac{1}{8} - \dfrac{3}{16}x + \dfrac{3}{16}x^{2}

, valid for

|x| \lt 2

⚠️ Common mistake Writing

2(1+\dfrac{x}{2})^{-3}

: the 2 must be raised to the power

-3

too, giving

\dfrac{1}{8}

Question 6●●○ Standard✍️ + 🖩 · Paper 2[5 marks]

Seven friends, including Neo and Kabo, sit in a row of 7 seats at the cinema in Gaborone.

(a) Find the number of ways they can sit.[1]
(b) Find the number of ways they can sit if Neo and Kabo must sit next to each other.[2]
(c) Find the number of ways they can sit if Neo and Kabo must not sit next to each other.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

7! = 5040

Part (b)

Step 1🧠 ThinkGlue Neo and Kabo together

Treat [Neo Kabo] as one block. Now there are 6 “things” to arrange:

6!

ways. Inside the block they can be Neo–Kabo or Kabo–Neo:

2!

ways.

Step 2✍️ By hand

6!\times2! = 720\times2 = 1440

M1A1

Part (c)

Step 1✍️ By hand

Not together = all arrangements − together.

5040 - 1440 = 3600

M1A1

Step 2🖩 On the GDCCheck with the factorial

7›menu💻 click menu on keypad›5: Probability›1: Factorial (!)›enter💻 Enter

5040, 1440, 3600

Final answer (a)

5040

(b)

1440

(c)

3600

Question 7●●○ Standard✍️ + 🖩 · Paper 2[5 marks]

A committee of 4 is chosen from 6 men and 5 women. Find the number of committees that contain at least two women.

Show the full step-by-step solution

Step 1🧠 ThinkSplit into exact cases

“At least two women” means exactly 2, exactly 3, or exactly 4 women. For each case choose the women and the men separately, multiply (product principle), then add the cases (addition principle).

Step 2✍️ By hand

2W 2M:

\binom{5}{2}\binom{6}{2} = 10\times15 = 150

3W 1M:

\binom{5}{3}\binom{6}{1} = 10\times6 = 60

4W:

\binom{5}{4} = 5

Total

= 150 + 60 + 5 = 215

M1A1

Step 3🖩 On the GDCCheck another way: total − (0 or 1 woman)

330 − 115 = 215 ✔

Final answer

215

⚠️ Common mistake Choosing 2 women then any 2 of the other 9 people:

\binom{5}{2}\binom{9}{2} = 360

counts some committees more than once.

Question 8●●● Exam-level✍️ + 🖩 · Paper 2[6 marks]

Consider the letters of the word STATISTICS.

(a) Find the number of different arrangements of all ten letters.[2]
(b) Find the number of arrangements in which the three S's are together.[2]
(c) Find the number of arrangements that begin with A.[2]

Show the full step-by-step solution

Step 1🧠 ThinkCount each letter

S × 3, T × 3, I × 2, A × 1, C × 1 — that is 10 letters.

Part (a)

Step 1✍️ By hand

\dfrac{10!}{3!\,3!\,2!}

= 50\,400

Part (b)

Step 1✍️ By hand

Glue SSS into one block. Now arrange 8 items: [SSS], T, T, T, I, I, A, C. The S's inside the block are identical, so no extra factor.

\dfrac{8!}{3!\,2!} = 3360

M1A1

Part (c)

Step 1✍️ By hand

Fix A first. Arrange the other 9 letters: S×3, T×3, I×2, C.

\dfrac{9!}{3!\,3!\,2!} = 5040

M1A1

Step 2🖩 On the GDCFactorials on the GDC

(a), (b), (c)

Final answer (a)

50\,400

(b)

3360

(c)

5040

Question 9●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

The expansion of

(1+ax)^{n}

, where

n\in\mathbb{Z}

, begins

1 - 6x + 27x^{2} + \dots

(a) Find the value of $n$ and the value of $a$ .[4]
(b) Find the coefficient of $x^{3}$ .[1]
(c) State the values of $x$ for which the expansion is valid.[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkMatch the coefficients

(1+ax)^{n} = 1 + nax + \dfrac{n(n-1)}{2}a^{2}x^{2} + \dots

. So

na = -6

and

\dfrac{n(n-1)}{2}a^{2} = 27

Step 2✍️ By hand

na = -6 \Rightarrow a = -\dfrac{6}{n}

\dfrac{n(n-1)}{2}\times\dfrac{36}{n^{2}} = 27

\dfrac{18(n-1)}{n} = 27 \Rightarrow 18n - 18 = 27n

n = -2

a = -\dfrac{6}{-2} = 3

💡 Remember A negative

n

means the expansion never stops — that is why the question says “begins”.

Part (b)

Step 1✍️ By hand

\dfrac{(-2)(-3)(-4)}{3!}(3x)^{3} = -4\times27x^{3} = -108x^{3}

Part (c)

Step 1✍️ By hand

|3x| \lt 1 \Rightarrow |x| \lt \dfrac{1}{3}

Final answer (a)

n = -2

a = 3

(b)

-108

(c)

|x| \lt \dfrac{1}{3}

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[6 marks]

A group of 10 volunteers in Kasane includes a married couple, Thato and Lesedi. A team of 5 is chosen to help at a conservation project.

(a) Find the number of different teams.[1]
(b) Find the number of teams that include both Thato and Lesedi.[2]
(c) The couple refuse to be in the team together. Find the number of possible teams.[2]
(d) Find the number of teams that include neither of them.[1]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

\binom{10}{5} = 252

Part (b)

Step 1🧠 ThinkPut the couple in first

Both are in, so 3 more places are filled from the other 8 people.

Step 2✍️ By hand

\binom{8}{3} = 56

M1A1

Part (c)

Step 1✍️ By hand

252 - 56 = 196

M1A1

Part (d)

Step 1✍️ By hand

Choose all 5 from the other 8:

\binom{8}{5} = 56

Step 2🖩 On the GDCCheck with nCr(

252, 56, 196, 56

Final answer (a)

252

(b)

56

(c)

196

(d)

56

Question 11●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

Four-digit numbers are formed from the digits 1, 2, 3, 4, 5, 6, 7. No digit may be used more than once.

(a) Find how many of these numbers are even.[2]
(b) Find how many of these numbers are even and greater than 5000.[4]

Show the full step-by-step solution

Step 1🧠 ThinkFill the restricted places first

Draw four boxes: _ _ _ _. Always fill the box with a rule first (the last digit must be even), then the others.

Part (a)

Step 1✍️ By hand

Last digit: 2, 4 or 6 → 3 ways. Then

6\times5\times4

for the others.M1

3\times6\times5\times4 = 360

Part (b)

Step 1🧠 ThinkTwo rules — and they share the digit 6

The first digit must be 5, 6 or 7. But 6 is also even, so split into cases: first digit 6, or first digit 5 or 7.

Step 2✍️ By hand

First digit 6: last digit 2 or 4 (2 ways); middle

5\times4

1\times5\times4\times2 = 40

M1A1

First digit 5 or 7: last digit 2, 4, 6 (3 ways):

2\times5\times4\times3 = 120

Total

= 40 + 120 = 160

Final answer (a)

360

(b)

160

Question 12●●● Exam-level✍️ + 🖩 · Paper 3[10 marks]

Investigating square roots with the extended binomial theorem.

(a) Show that the first four terms of the expansion of $(1+x)^{\frac{1}{2}}$ are $1 + \dfrac{1}{2}x - \dfrac{1}{8}x^{2} + \dfrac{1}{16}x^{3}$ .[3]
(b) Kagiso puts $x = 1$ in these four terms to estimate $\sqrt{2}$ . Find his estimate and explain why it is poor.[2]
(c) Show that $\sqrt{2} = \dfrac{7}{5}\left(1+\dfrac{1}{49}\right)^{\frac{1}{2}}$ .[2]
(d) Hence use the four terms to estimate $\sqrt{2}$ , and use your GDC to find the error in this estimate.[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

1 + \dfrac{1}{2}x + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}x^{2} + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}x^{3}

= 1 + \dfrac{1}{2}x + \dfrac{-\frac{1}{4}}{2}x^{2} + \dfrac{\frac{3}{8}}{6}x^{3}

= 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^{2} + \dfrac{1}{16}x^{3}

Part (b)

Step 1✍️ By hand

1 + \dfrac{1}{2} - \dfrac{1}{8} + \dfrac{1}{16} = 1.4375

The series is only valid for

|x| \lt 1

;

x = 1

is on the edge, so the terms shrink very slowly and four terms are not enough (

\sqrt{2} = 1.41421\ldots

).R1

Part (c)

Step 1✍️ By hand

\left(1+\dfrac{1}{49}\right)^{\frac{1}{2}} = \sqrt{\dfrac{50}{49}} = \dfrac{\sqrt{25\times2}}{7} = \dfrac{5\sqrt{2}}{7}

M1A1

\dfrac{7}{5}\times\dfrac{5\sqrt{2}}{7} = \sqrt{2}

Part (d)

Step 1🧠 ThinkA tiny x makes the series converge fast

Now

x = \dfrac{1}{49}

, which is very small, so each new term is about 50 times smaller than the one before.

Step 2🖩 On the GDCEvaluate the estimate and the error

7/5·(1+1/2·(1/49)−1/8·(1/49)^2+1/16·(1/49)^3)›ctrl+enter💻 Ctrl+Enter›ans−√(2)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ square root: ctrl then x² on the keypad, or type sqrt(

Press ctrl enter for a decimal (in Auto mode the TI first gives a fraction).

error ≈ 9.35 × 10⁻⁹

Step 3✍️ By hand

Estimate

\approx 1.41421357

Error

\approx 9.35\times10^{-9}

— accurate to 8 decimal placesA1

Final answer (b)

1.4375

;

x = 1

is at the edge of validity, convergence is slow (d)

\sqrt{2} \approx 1.41421357

, error

\approx 9.35\times10^{-9}

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 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.

🎯 What you must know

Product principle: if one choice can be made in $m$ ways and then another in $n$ ways, together there are $m\times n$ ways. Addition principle: for separate cases (“or”), add the counts.
Arrangements of $n$ different objects in a row: $n!$ . If some are identical, divide: $\dfrac{n!}{p!\,q!\dots}$ for $p$ of one kind, $q$ of another…
Permutations (order matters): ${}^{n}P_{r} = \dfrac{n!}{(n-r)!}$ . Combinations (order does not matter): ${}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}$ .
“Must be together”: glue them into one block, arrange the blocks, then arrange inside the block. “Not together” = total − together.
“At least one / at least two”: add the separate cases, or use total − (the cases you don't want).
Extended binomial theorem ( $n\in\mathbb{Q}$ ): $(1+x)^{n} = 1 + nx + \dfrac{n(n-1)}{2!}x^{2} + \dfrac{n(n-1)(n-2)}{3!}x^{3} + \dots$ , an infinite series valid for $|x| \lt 1$ .
For $(a+bx)^{n}$ , first take out $a^{n}$ : $(a+bx)^{n} = a^{n}\left(1+\dfrac{b}{a}x\right)^{n}$ , valid for $\left|\dfrac{bx}{a}\right| \lt 1$ .

Permutations

{}^{n}P_{r} = \dfrac{n!}{(n-r)!}

📘 In the formula booklet

Combinations

{}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}

📘 In the formula booklet

Extended binomial theorem

(a+b)^{n} = a^{n}\left(1 + n\left(\dfrac{b}{a}\right) + \dfrac{n(n-1)}{2!}\left(\dfrac{b}{a}\right)^{2} + \dots\right)

n\in\mathbb{Q}

📘 In the formula booklet

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Factorial n!

menu💻 click menu on keypad›5: Probability›1: Factorial (!)

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 5: Probability.
Choose 1: Factorial (!).

The menu path

5040, 1440, 3600

Worked example: Question 6 — Check with the factorial

Permutations nPr(n,r)

menu💻 click menu on keypad›5: Probability›2: Permutations›12,7)›enter💻 Enter

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 5: Probability.
Choose 2: Permutations.
Type 12,7).
Press enter.

The menu path

792 teams; 3 991 680 line-ups

Worked example: Question 3 — nCr( and nPr( on a Calculator page

Combinations nCr(n,r)

menu💻 click menu on keypad›5: Probability›3: Combinations›12,7)›enter💻 Enter

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 5: Probability.
Choose 3: Combinations.
Type 12,7).
Press enter.

The menu path

792 teams; 3 991 680 line-ups

Worked example: Question 3 — nCr( and nPr( on a Calculator page

⚠️ Where students lose marks

Using ${}^{n}P_{r}$ when order does not matter (a committee, a hand of cards) — use ${}^{n}C_{r}$ .
Forgetting to divide by $p!$ for repeated letters, or forgetting to arrange the people inside a block.
Counting “at least 2” as $\binom{5}{2}\times\binom{9}{2}$ — this double-counts. Split into exact cases.
Extended binomial: forgetting to take $a^{n}$ outside, or forgetting the validity condition $|x| \lt \dots$ .
Sign slips with negative $n$ : $\dfrac{(-2)(-3)}{2!} = +3$ , $\dfrac{(-2)(-3)(-4)}{3!} = -4$ .

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · AA 1.10 — Counting principles; permutations and combinations; extended binomial theorem

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

1.10 Counting principles; permutations and combinations; extended binomial theorem

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks