MathsCBSE Class 11 · Permutations and Combinations

Combinations: Selections, Conditions, Geometry Counts and Grouping (Class 11)

Updated 17 October 2026 · Fact-checked

The number of ways of choosing r objects out of n distinct objects is nCr = n!/(r!(n − r)!), with nCr = nC(n−r). For "at least" conditions, add the cases or subtract the complement from the total. For geometry, choose points and remove the invalid sets (for example collinear triples). For arrangement after selection, multiply by r!. To divide into equal unlabelled groups, divide by the factorial of the number of equal groups.

For an 'exactly' selection, count each group's choices separately with nCr, then multiply: 10 × 6 = 60.

Illustration drawn for QuestionBuddy.

Understand Combinations: Selections, Committees and Grouping Problems

Selections. When order does not matter, the number of ways of choosing $r$ out of $n$ distinct objects is
$ $^{n} C_{r} = \frac{n !}{r ! ( n - r )!} = \frac{^{n} P _{r}}{r !},^{n} C_{r} =^{n} C_{n - r},^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r} .$ $
If $^{n} C_{a} =^{n} C_{b}$ then $a = b$ or $a + b = n$ .

Conditions on a selection.

*Particular objects always included:* remove them and choose the rest.
*Particular objects excluded:* remove them from the pool.
*At least / at most:* add the cases (for example 2 women, 3 women, 4 women) or take total minus complement ("at least one" = total $-$ none).
*Exactly:* multiply the independent selections from each group, e.g. $^{5} C_{2} \times^{4} C_{2}$ .

Geometry.

Lines from $n$ points, no three collinear: $^{n} C_{2}$ . With $m$ of the points collinear, the lines number $^{n} C_{2} -^{m} C_{2} + 1$ .
Triangles from $n$ points with $m$ collinear: $^{n} C_{3} -^{m} C_{3}$ .
Diagonals of an $n$ -sided polygon: $^{n} C_{2} - n = \frac{n ( n - 3 )}{2}$ .

Selection then arrangement. Choose first (combinations), then arrange the selected objects (multiply by the factorial). Example: 4-letter words with 3 consonants from 7 and 1 vowel from 4: $^{7} C_{3} \times^{4} C_{1} \times 4! = 3360$ .

Dividing into groups. To split $n$ objects into groups of given sizes, choose the groups one after another (multinomial $\frac{n !}{a ! b ! c !}$ ). If some groups have the same size and are unlabelled, divide by the factorial of the number of such groups. Eight players into two unlabelled teams of 4: $\frac{^{8} C _{4}}{2} = 35$ ; into two labelled teams (team A, team B): $70$ . Into groups of sizes 3, 3 and 2: $\frac{8 !}{3 ! 3 ! 2 ! 2 !} = 280$ .

Permutation or combination? Ask "does the order of the chosen objects matter?" Teams, committees, hands of cards, handshakes: combination. Seats, codes, ranks: permutation.

Combinations: Selections, Committees and Grouping Problems: formulas to remember

Combinations: $n C r = n! / (r! (n - r)!)$; Order does not matter.
Symmetry: $n C r = n C (n - r)$; Choosing r is the same as leaving n − r.
Pascal identity: $n C r + n C (r - 1) = (n + 1) C r$; Used in proofs and sums.
At least one: $T o t a l - n o n e$; Complement method.
Triangles from points: $n C 3 - m C 3$; m collinear points out of n.
Diagonals: $n C 2 - n = n (n - 3) /2$; Polygon with n sides.

How to solve Combinations: Selections, Committees and Grouping Problems questions

Use these steps for combination problems with conditions.

1Decide whether order matters; if not, use combinations.
2Split the objects into groups (men and women, red and blue) and note the conditions on each group.
3For "at least" or "at most", list the cases or use total minus complement, whichever is shorter.
4For "exactly", multiply the independent selections from each group.
5If objects must also be arranged, multiply the combination count by the factorial of the number selected.
6For unlabelled equal groups, divide by the factorial of the number of equal groups.

Combinations: Selections, Committees and Grouping 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.

Handshakes

Suppose 12 people at a meeting each shake hands with every other person exactly once.

How many handshakes take place?

Try it first, then show the solutionSolution

A handshake is an unordered pair of people.
$^{12} C_{2} = \frac{12 \times 11}{2} = 66$ .

Answer: 66 handshakes.

Memory hook: A handshake has no order, so use a combination.

Choosing toppings

Suppose a pizza shop offers 8 toppings and a "special" uses exactly 3 different toppings.

How many different specials are possible, and how many include cheese (one of the 8)?

Try it first, then show the solutionSolution

All specials: $^{8} C_{3} = \frac{8 \times 7 \times 6}{6} = 56$ .
Specials with cheese: choose the other 2 toppings from 7: $^{7} C_{2} = 21$ .
Check: specials without cheese $^{7} C_{3} = 35$ , and $35 + 21 = 56$ .

Answer: 56 specials; 21 include cheese.

Memory hook: Fix the compulsory item, then choose the rest.

Tricks to remember Combinations: Selections, Committees and Grouping Problems

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

Order test
Does swapping two chosen objects give a different outcome? Yes: permutation. No: combination.
Use it when: Choosing the right formula.
At least means complement
At least one = total − none; at most one = none + exactly one.
Use it when: Condition questions.
Pick the smaller r
Use nCr = nC(n−r) to compute 20C18 as 20C2.
Use it when: Arithmetic.
Equal groups divide
Unlabelled equal groups: divide by the factorial of the number of such groups.
Use it when: Grouping.

Combinations: Selections, Committees and Grouping Problems: solved examples

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

Example 1: Equal combinations · 2 marks · Very short answerBoard level

If $^{n} C_{8} =^{n} C_{12}$ , find $n$ and then $^{22} C_{n}$ .

Try it first, then show the solutionSolution

$^{n} C_{a} =^{n} C_{b}$ with $a \neq = b$ gives $a + b = n$ , so $n = 8 + 12 = 20$ . (1 mark)
$^{22} C_{20} =^{22} C_{2} = \frac{22 \times 21}{2} = 231$ . (1 mark)

Answer: $n = 20$ ; $^{22} C_{20} = 231$

Example 2: At least one woman · 3 marks · Short answerBoard level

A committee of 4 is to be chosen from 6 men and 5 women so that it contains at least one woman. In how many ways can this be done?

Try it first, then show the solutionSolution

Total committees from 11 people: $^{11} C_{4} = 330$ . (1 mark)
Committees with no woman (all men): $^{6} C_{4} = 15$ . (1 mark)
At least one woman: $330 - 15 = 315$ . (1 mark)

Answer: 315

Example 3: Triangles and lines · 3 marks · Short answerHarder board level

There are 12 points in a plane, of which 5 are collinear and no other three are collinear. Find the number of (i) triangles, (ii) distinct straight lines they determine.

Try it first, then show the solutionSolution

(i) Any 3 of the 12 points give a triangle except 3 chosen from the 5 collinear points: $^{12} C_{3} -^{5} C_{3} = 220 - 10 = 210$ . (1½ marks)
(ii) Pairs: $^{12} C_{2} = 66$ , but the $^{5} C_{2} = 10$ pairs from the collinear points all lie on one line: $66 - 10 + 1 = 57$ . (1½ marks)

Answer: (i) 210 (ii) 57

Example 4: Selection followed by arrangement · 4 marks · Short answerHarder board level

How many 4-letter words (meaningful or not, no repetition) can be formed from 7 consonants and 4 vowels if each word has exactly 3 consonants and 1 vowel?

Try it first, then show the solutionSolution

Choose 3 consonants: $^{7} C_{3} = 35$ . (1 mark)
Choose 1 vowel: $^{4} C_{1} = 4$ . (1 mark)
Arrange the 4 chosen letters: $4! = 24$ . (1 mark)
Total $= 35 \times 4 \times 24 = 3360$ . (1 mark)

Answer: 3360

Example 5: Dividing into groups · 5 marks · Long answerHOTS / competency

(a) In how many ways can 8 players be divided into two teams of 4 (i) if the teams are named A and B, (ii) if the teams are not named? (b) In how many ways can the 8 players be divided into three groups of sizes 3, 3 and 2?

Try it first, then show the solutionSolution

(a)(i) Choose the 4 for team A: $^{8} C_{4} = 70$ . (1 mark)
(a)(ii) Unnamed teams: each division is counted twice (swapping the two teams), so $\frac{70}{2} = 35$ . (1 mark)
(b) Multinomial: $\frac{8 !}{3 ! 3 ! 2 !} = \frac{40320}{72} = 560$ for labelled groups. (1 mark)
The two groups of size 3 are interchangeable, so divide by $2!$ : $\frac{560}{2} = 280$ . (1 mark)
Check: the group of 2 can be chosen in $^{8} C_{2} = 28$ ways; the other 6 split into two unlabelled threes in $\frac{^{6} C _{3}}{2} = 10$ ways; $28 \times 10 = 280$ . (1 mark)

Answer: (a)(i) 70 (ii) 35 (b) 280

Where students lose marks in Combinations: Selections, Committees and Grouping Problems

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

Mistake: Using permutations for a selection
Order of a chosen group does not matter.
Wrong working
Committees of 3 from 10: $^{10} P_{3} = 720$ .
Right working
$^{10} C_{3} = 120$ .
Do this: Ask "does swapping two chosen people change the outcome?"
Usually costs: The answer.
Mistake: Counting at least as exactly
Only one case is included.
Wrong working
"At least one woman" taken as exactly one woman.
Right working
Add the cases for 1, 2, 3, 4 women, or use total minus none.
Do this: Use the complement method.
Usually costs: The whole answer.
Mistake: Forgetting to remove collinear triples
Three collinear points do not form a triangle.
Wrong working
12 points, triangles $^{12} C_{3} = 220$ .
Right working
$220 -^{5} C_{3} = 210$ .
Do this: Subtract the invalid selections.
Usually costs: 1 to 2 marks.
Mistake: Not dividing for equal unlabelled groups
Swapping equal groups gives the same division.
Wrong working
Two teams of 4 from 8 (unnamed): 70.
Right working
$\frac{70}{2} = 35$ .
Do this: Check whether the groups are named.
Usually costs: The final mark.

Exam tips

Write which groups you are choosing from (men, women) before using any formula.
For "at least one" questions, total minus none is nearly always shortest.
Show the combination in factorial form once, then evaluate; numbers alone earn fewer method marks.
In geometry counts state what you subtract and why (collinear points do not give triangles).
For grouping, say whether the groups are labelled; it decides whether you divide.

Practice questions on Combinations: Selections, Committees and Grouping 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) 240
Hint
$^{10} C_{3} = \frac{10 \times 9 \times 8}{3 !}$ .
Answer
(b). $\frac{720}{6} = 120$ .
Q2 · 1 mark · MCQHarder board level
The number of diagonals of a regular decagon is:
(a) 45
(b) 35
(c) 70
(d) 90
Hint
Diagonals $=^{n} C_{2} - n$ .
Answer
(b). $^{10} C_{2} - 10 = 45 - 10 = 35$ .
Q3 · 3 marks · Short answerHarder board level
A box contains 5 red and 4 blue balls. In how many ways can 4 balls be chosen so that at least 2 are red?
Hint
Add the cases 2R2B, 3R1B, 4R, or use total minus complement.
Answer
Cases: $^{5} C_{2}^{4} C_{2} = 60$ , $^{5} C_{3}^{4} C_{1} = 40$ , $^{5} C_{4} = 5$ ; total $105$ . Check: $^{9} C_{4} - (^{4} C_{4} +^{5} C_{1}^{4} C_{3}) = 126 - 21 = 105$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The number of ways of choosing 3 persons from 10 equals the number of ways of choosing 7 persons from 10.
Reason (R): Every choice of 3 persons to include corresponds to exactly one choice of 7 persons to leave out.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Recall $^{n} C_{r} =^{n} C_{n - r}$ .
Answer
(a). $^{10} C_{3} =^{10} C_{7} = 120$ , and R gives the counting reason.
Q5 · 4 marks · Short answerHOTS / competency
A cricket team of 11 is chosen from 15 players: 3 wicket-keepers, 2 star batters $A$ and $B$ , and 10 others. The team must have exactly one wicket-keeper and must include both $A$ and $B$ . In how many ways can the team be chosen?
Hint
Fix the compulsory members first.
Answer
Choose the keeper: $^{3} C_{1} = 3$ . Include $A$ and $B$ . The remaining $11 - 1 - 2 = 8$ places come from the 10 others (not keepers, not $A$ , $B$ ): $^{10} C_{8} = 45$ . Total $3 \times 45 = 135$ .

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.What is nCr?Show answer
n!/(r!(n − r)!).
2.What is nCr in terms of nC(n−r)?Show answer
They are equal.
3.How many diagonals does an n-gon have?Show answer
n(n − 3)/2.
4.At least one means what, in terms of the total?Show answer
Total minus none.
5.How many ways to choose 2 from 6?Show answer
15.

Combinations: Selections, Committees and Grouping Problems: frequently asked questions

How do I decide between nPr and nCr?

Use nCr when the order of the selected objects does not matter and nPr when it does. Selection first and then arrangement is nCr × r!.

What does "at least" mean in committee problems?

It means that number or more, so add all cases from the minimum upward, or subtract the cases that break the condition from the total.

Why divide by 2 when dividing 8 players into two teams?

Because choosing team A first and team B second counts every division twice when the teams are not named.

Is dividing into groups in the syllabus?

It is a natural extension of the selection principles in the chapter and a frequent school and JEE-foundation question.