MathsISC Class 11 · Permutations and Combinations

Combinations: ⁿCᵣ, Selections with Conditions, Geometry and Mixed Problems

Updated 17 October 2026 · Fact-checked

The number of ways of selecting r objects from n distinct objects is ⁿCᵣ = n!/(r!(n − r)!) = ⁿPᵣ/r!. Also ⁿCᵣ = ⁿCₙ₋ᵣ and ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ. For problems with conditions, split into disjoint cases or subtract the unwanted selections from the total.

Every pair of points makes one line, so n points give nC2 lines; remove the n sides to get the diagonals.

Illustration drawn for QuestionBuddy.

Understand Combinations, Selections and Mixed Problems

Combinations. A selection in which order does not matter: $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ . Properties: $^{n} C_{0} =^{n} C_{n} = 1$ ; $^{n} C_{r} =^{n} C_{n - r}$ ; $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ ; if $^{n} C_{x} =^{n} C_{y}$ then $x = y$ or $x + y = n$ . Also $^{n} P_{r} =^{n} C_{r} \cdot r!$ (select, then arrange).

Conditional selections. "At least k of a kind": add the cases k, k + 1, ..., or use the complement. "A particular object is included": choose the rest from n − 1. "A particular object is excluded": choose all from n − 1.

Geometry. The number of lines from n points (no three collinear) is $^{n} C_{2}$ ; of triangles $^{n} C_{3}$ ; of diagonals of an n-gon $^{n} C_{2} - n$ . If m of the points are collinear, subtract the lines or triangles that cannot be formed: triangles $^{n} C_{3} -^{m} C_{3}$ .

Selection of any number. From n distinct objects, selecting at least one gives $2^{n} - 1$ ways. From identical objects of several kinds with $p$ , $q$ , $r$ items, selecting at least one gives $(p + 1) (q + 1) (r + 1) - 1$ .

Mixed problems. First select, then arrange: for example choose 5 letters in $^{n} C_{5}$ ways and arrange them in $5!$ ways. Organise the case work carefully, and avoid counting the same selection twice.

Combinations, Selections and Mixed Problems: formulas to remember

Combination: $^nC_r = \frac{n!}{r!(n - r)!}$ | $^nC_r = ^nC_{n - r}$; Order does not matter.
Pascal's rule: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$; Used to simplify sums.
Geometry: $Lines: $^nC_2$ | Triangles: $^nC_3$ | Diagonals: $^nC_2 - n$$; No three collinear.
At least one selection: $Distinct: $2^n - 1$ | Identical kinds: $(p+1)(q+1)(r+1) - 1$$; Includes all subsets except the empty one.

Combinations, Selections and Mixed Problems in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Choosing a committee

Suppose a club has 6 men and 4 women and wants a committee of 4 with at least 2 women.

Find the number of ways.

Try it first, then show the solutionSolution

Case 1: 2 women and 2 men: $^{4} C_{2} \cdot^{6} C_{2} = 6 \times 15 = 90$ .
Case 2: 3 women and 1 man: $^{4} C_{3} \cdot^{6} C_{1} = 4 \times 6 = 24$ .
Case 3: 4 women: $^{4} C_{4} = 1$ . Total $= 90 + 24 + 1 = 115$ .

Answer: 115 committees.

Memory hook: Add the disjoint cases.

Tricks to remember Combinations, Selections and Mixed Problems

Short memory aids. Say each one aloud once, then use it on a question the same day.

Pick, then arrange
ⁿPᵣ = ⁿCᵣ × r!.
Use it when: Mixed problems.
Complement
When "at least" gives many cases, subtract the unwanted selections from the total.
Use it when: Counting efficiently.

Combinations, Selections and Mixed Problems: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A property · 1 mark · MCQBoard level

If $^{n} C_{8} =^{n} C_{2}$ then n equals:
(a) 6
(b) 8
(c) 10
(d) 16

Try it first, then show the solutionSolution

$8 + 2 = n$ , so $n = 10$ .

Answer: (c) 10

Example 2: Committee with a condition · 3 marks · Short answerBoard level

A committee of 4 is to be chosen from 6 men and 4 women with at least 2 women. In how many ways can this be done?

Try it first, then show the solutionSolution

2 women, 2 men: $^{4} C_{2} \cdot^{6} C_{2} = 6 \times 15 = 90$ . (1 mark)
3 women, 1 man: $^{4} C_{3} \cdot^{6} C_{1} = 4 \times 6 = 24$ ; 4 women: 1. (1 mark)
Total $= 90 + 24 + 1 = 115$ . (1 mark)

Answer: 115

Example 3: Diagonals and triangles · 4 marks · Short answerHarder board level

Find the number of diagonals of a 12-sided polygon. From 10 points in a plane, of which 4 are collinear, how many triangles can be formed?

Try it first, then show the solutionSolution

Lines joining 12 vertices: $^{12} C_{2} = 66$ ; subtract the 12 sides: 54 diagonals. (2 marks)
Triangles from 10 points: $^{10} C_{3} = 120$ . (1 mark)
Subtract the degenerate ones from the 4 collinear points: $^{4} C_{3} = 4$ ; triangles $= 116$ . (1 mark)

Answer: 54 diagonals; 116 triangles.

Example 4: Selection of any number of items · 4 marks · Short answerHarder board level

Find the number of ways of selecting at least one fruit from 3 identical apples, 2 identical oranges and 1 mango. Find also the number of subsets of 5 distinct objects with at least one object.

Try it first, then show the solutionSolution

For identical kinds, the number of choices of apples is 0 to 3: 4 ways; oranges 0 to 2: 3 ways; mango: 2 ways. (2 marks)
Total selections $= 4 \times 3 \times 2 = 24$ ; excluding the empty selection gives 23. (1 mark)
For 5 distinct objects: $2^{5} - 1 = 31$ . (1 mark)

Answer: 23; 31.

Example 5: Select and arrange · 5 marks · Long answerHOTS / competency

In how many ways can 3 vowels and 2 consonants be chosen from the letters of the word EQUATION, and how many 5-letter arrangements of the chosen letters are possible?

Try it first, then show the solutionSolution

Vowels: E, U, A, I, O (5); consonants: Q, T, N (3). (1 mark)
Selection: $^{5} C_{3} \cdot^{3} C_{2} = 10 \times 3 = 30$ . (2 marks)
Each selection of 5 letters can be arranged in $5! = 120$ ways. (1 mark)
Total words $= 30 \times 120 = 3600$ . (1 mark)

Answer: 30 selections; 3600 arrangements.

Where students lose marks in Combinations, Selections and Mixed Problems

The slips examiners see most often, and what to write instead.

Mistake: Using permutations for selections
Counting committees as ordered.
Do this: If order does not matter, use ⁿCᵣ.
Usually costs: 2 marks.
Mistake: Double-counting cases
Counting "at least 2 women" by choosing 2 first and then the rest freely.
Do this: Use disjoint cases or the complement.
Usually costs: 2 marks.
Mistake: Forgetting to subtract the empty selection
Giving 2ⁿ for "at least one".
Do this: Subtract 1.
Usually costs: 1 mark.

Exam tips

Decide whether order matters.
Split into disjoint cases and add.
Use symmetry ⁿCᵣ = ⁿCₙ₋ᵣ to simplify.
Check totals using small examples.

Practice questions on Combinations, Selections and Mixed Problems

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $^{10} C_{3}$ is:
(a) 720
(b) 120
(c) 30
(d) 90
Hint
$\frac{10 \times 9 \times 8}{6}$ .
Answer
(b) 120
Q2 · 2 marks · Short answerBoard level
Find n if $^{n} C_{2} = 45$ .
Hint
$\frac{n ( n - 1 )}{2} = 45$ .
Answer
$n (n - 1) = 90$ , so $n = 10$ .
Q3 · 3 marks · Short answerHarder board level
In how many ways can a team of 11 be chosen from 15 players if one particular player must be in the team and another must be left out?
Hint
Remove both from the pool.
Answer
The remaining 10 places come from $15 - 2 = 13$ players: $^{13} C_{10} =^{13} C_{3} = 286$ .
Q4 · 3 marks · Short answerHarder board level
How many handshakes occur among 12 people if everyone shakes hands once with everyone else?
Hint
Choose pairs.
Answer
$^{12} C_{2} = 66$ .
Q5 · 3 marks · Short answerHarder board level
A bag has 5 red and 4 blue balls. In how many ways can 4 balls be drawn so that at least one is blue?
Hint
Total minus all red.
Answer
Total $^{9} C_{4} = 126$ ; all red $^{5} C_{4} = 5$ ; at least one blue $= 121$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a quiz team of 5 is formed from 7 girls and 5 boys, with at least 3 girls.
(i) Find the number of teams with exactly 3 girls. (1 mark)
(ii) Find the number with exactly 4 girls. (1 mark)
(iii) Find the number with 5 girls. (1 mark)
(iv) Find the total number of valid teams. (1 mark)
Hint
Sum of three cases.
Answer
(i) $^{7} C_{3} \cdot^{5} C_{2} = 35 \times 10 = 350$ . (ii) $^{7} C_{4} \cdot^{5} C_{1} = 35 \times 5 = 175$ . (iii) $^{7} C_{5} = 21$ . (iv) $350 + 175 + 21 = 546$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.ⁿC₂?Show answer
n(n − 1)/2.
2.ⁿCᵣ in terms of ⁿPᵣ?Show answer
ⁿPᵣ/r!.
3.Subsets of an n-set?Show answer
2ⁿ.
4.Diagonals of an n-gon?Show answer
ⁿC₂ − n.

Combinations, Selections and Mixed Problems: frequently asked questions

What is the difference between permutations and combinations?

In permutations order matters (an arrangement); in combinations it does not (a selection).

Why is ⁿCᵣ = ⁿCₙ₋ᵣ?

Choosing r objects to include is the same as choosing the n − r objects to leave out.

Is 0! = 1 needed?

Yes, so that ⁿC₀ = ⁿCₙ = 1.