Unit 4: Combinatorics & Probability — Class 11 Applied Maths 2026-27 | Boundless Maths
Unit 4 of 7 10 Marks CBSE 2026–27 Combinatorics + Probability

Unit 4: Combinatorics
& Probability

CBSE Class 11 Applied Mathematics · Unit 4 · Free MCQs, Solved Examples & Case Studies

Unit 4 carries 10 marks in the CBSE Class 11 annual exam. Complete free resources: 20 MCQs + AR questions, 13 solved examples and 4 case studies. Covers Combinatorics (Factorial, Fundamental Principle of Counting, Permutation & Combination) and Probability (Random Experiments, Events, Conditional Probability, Independent Events) — fully aligned to CBSE 2026-27.

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Case Studies

Class 11 Applied Maths Unit 4: Combinatorics & Probability — Complete Free Resources (CBSE 2026-27)

This page covers all topics in Unit 4 of CBSE Class 11 Applied Mathematics — carrying 10 marks in the CBSE Class 11 annual exam. You'll find 15 MCQs and 5 Assertion-Reason questions with step-by-step answers, 13 solved examples, and 4 case studies based on real-world contexts. Unit 4 covers two powerful areas: Combinatorics — the mathematics of counting, including Factorial, the Fundamental Principle of Counting, Permutations (nPr) and Combinations (nCr) — and Probability — the mathematics of chance, including Random Experiments, Sample Space, Events, Conditional Probability and Independent Events. Free CBSE 2026-27 aligned practice on permutation and combination problems, probability questions with solutions, and the difference between permutation and combination explained step by step.

Factorial & FPC Permutation nPr Combination nCr Conditional Probability 10 Marks
Unit 4 · 10 Marks

Topics & Key Formulas

Two sections: Combinatorics (4.1) and Probability (4.2). Six topic areas total.

Section A — Combinatorics (4.1)

1. Factorial

Foundation of all counting — the product of all positive integers up to n.

  • n! = n × (n−1) × (n−2) × … × 2 × 1
  • 0! = 1 (by definition)
  • n! = n × (n−1)!
  • Used in all permutation and combination formulas
n! = n × (n−1)!  ·  0! = 1

2. Fundamental Principle of Counting

If task 1 can be done in m ways and task 2 in n ways, both together can be done in m×n ways.

  • Multiplication Principle: sequential tasks → multiply
  • Addition Principle: alternative tasks → add
  • Applies to sequences of independent choices
  • Basis of all combinatorics problems
Sequential tasks: m × n ways

3. Permutation

Arrangement of r objects from n distinct objects where order matters.

  • nPr = n!/(n−r)!
  • n things taken all at a time: nPn = n!
  • Repetition allowed: nʳ arrangements
  • Key: use P when order/arrangement matters
nPr = n!/(n−r)!

4. Combination

Selection of r objects from n distinct objects where order does not matter.

  • nCr = n!/[r!(n−r)!]
  • nCr = nC(n−r) (complementary property)
  • nC0 = nCn = 1  ·  nC1 = n
  • nCr = nPr / r! (removes ordering)
nCr = n!/[r!(n−r)!] = nPr/r!

Section B — Probability (4.2)

5. Random Experiment & Events

An experiment whose outcome cannot be predicted with certainty.

  • Sample Space S: set of all possible outcomes
  • Event: any subset of S
  • Impossible event ∅: P = 0  ·  Sure event S: P = 1
  • Mutually exclusive: A∩B = ∅ → P(A∩B) = 0
  • Exhaustive: A∪B = S
0 ≤ P(E) ≤ 1  ·  P(S) = 1  ·  P(∅) = 0

6. Conditional Probability & Independence

Probability of A given that B has already occurred.

  • P(A|B) = P(A∩B)/P(B), P(B) ≠ 0
  • Multiplication theorem: P(A∩B) = P(A)·P(B|A)
  • Independent events: P(A∩B) = P(A)·P(B)
  • P(A') = 1−P(A)  ·  P(A∪B) = P(A)+P(B)−P(A∩B)
  • Independent ≠ Mutually Exclusive
P(A|B) = P(A∩B)/P(B)  ·  Independent: P(A∩B)=P(A)·P(B)
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Interactive Practice

Practice MCQs — Unit 4: Combinatorics & Probability

15 MCQs + 5 Assertion-Reason questions. Click Show Answer to see the full explanation.

Combinatorics — Factorial & FPC
Q1 Factorial
The value of 5! − 4! is:
(a) 1
(b) 24
(c) 96
(d) 120
Answer: (c) 96
5! = 120, 4! = 24. 5! − 4! = 120 − 24 = 96.
Q2 Factorial
The value of 8!/6! is:
(a) 8
(b) 48
(c) 56
(d) 64
Answer: (c) 56
8!/6! = (8 × 7 × 6!)/6! = 8 × 7 = 56.
Q3 FPC
A student is getting dressed for a school event and can choose from 4 shirts and 3 trousers. How many different outfits can the student put together?
(a) 7
(b) 9
(c) 12
(d) 24
Answer: (c) 12
By the Multiplication Principle: 4 × 3 = 12 outfits.
Permutation
Q4 Permutation
The value of ⁵P₃ is:
(a) 10
(b) 20
(c) 60
(d) 120
Answer: (c) 60
⁵P₃ = 5!/(5−3)! = 5!/2! = (5×4×3×2!)/2! = 5×4×3 = 60.
Q5 Permutation
Five students are lining up for a class photograph. In how many different orders can they stand in a row?
(a) 24
(b) 60
(c) 120
(d) 720
Answer: (c) 120
Arranging all 5: ⁵P₅ = 5! = 5×4×3×2×1 = 120.
Q6 Permutation
How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
(a) 15
(b) 60
(c) 125
(d) 243
Answer: (b) 60
⁵P₃ = 5!/(5−3)! = 5×4×3 = 60.
Combination
Q7 Combination
The value of ⁸C₃ is:
(a) 40
(b) 56
(c) 112
(d) 336
Answer: (b) 56
⁸C₃ = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56.
Q8 Combination
A school wants to form a 3-member organising committee from 6 available teachers. In how many ways can this committee be chosen?
(a) 20
(b) 40
(c) 120
(d) 216
Answer: (a) 20
Order doesn't matter → ⁶C₃ = 6!/(3!×3!) = 720/(6×6) = 20.
Q9 Combination
If ⁿC₂ = 15, the value of n is:
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (b) 6
nC₂ = n(n−1)/2 = 15 → n(n−1) = 30 → n = 6 (since 6×5 = 30).
Q10 P vs C
A cricket coach must pick a playing squad of 4 from a pool of 7 trained players. Since only the final squad composition matters and not any batting order, in how many ways can the coach make this selection?
(a) 28
(b) 35
(c) 140
(d) 840
Answer: (b) 35
Selection only (order doesn't matter) → ⁷C₄ = 7!/(4!×3!) = (7×6×5)/(3×2×1) = 35.
Probability
Q11 Sample Space
A coin is tossed three times in a row. How many elements are there in the sample space of this experiment?
(a) 4
(b) 6
(c) 8
(d) 9
Answer: (c) 8
Each toss: 2 outcomes. Three tosses: 2³ = 8 elements: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Q12 Events
If A and B are mutually exclusive events, then P(A∩B) is:
(a) P(A)·P(B)
(b) P(A)+P(B)
(c) 1
(d) 0
Answer: (d) 0
Mutually exclusive events cannot occur simultaneously → A∩B = ∅ → P(A∩B) = 0.
Q13 Conditional Probability
If P(A) = 1/3, P(B) = 1/4 and P(A∩B) = 1/12, then P(A|B) is:
(a) 1/4
(b) 1/3
(c) 1/2
(d) 3/4
Answer: (b) 1/3
P(A|B) = P(A∩B)/P(B) = (1/12)/(1/4) = (1/12)×4 = 1/3.
Q14 Independent Events
If A and B are independent events with P(A) = 1/2 and P(B) = 1/3, then P(A∩B) is:
(a) 1/2
(b) 1/3
(c) 1/6
(d) 5/6
Answer: (c) 1/6
For independent events: P(A∩B) = P(A)×P(B) = (1/2)×(1/3) = 1/6.
Q15 Probability
A fair die is thrown once. What is the probability of rolling a prime number?
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3
Answer: (c) 1/2
Prime numbers on a die: {2, 3, 5} → 3 favourable outcomes out of 6. P = 3/6 = 1/2.
Assertion-Reason Questions (AR 1–5)
AR 1 Permutation vs Combination
Assertion (A): The number of ways to select 3 students from 5 is 10.

Reason (R): When order does not matter, we use nCr = n!/[r!(n−r)!], so ⁵C₃ = 10.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
A true: ⁵C₃ = 10 ✓. R true and correctly explains A: selection (not arrangement) → use nCr, and the formula correctly gives 10.
AR 2 Permutation
Assertion (A): ⁿPr = nCr × r!

Reason (R): Each combination of r items can be arranged in r! ways, so permutations = combinations × arrangements.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
A true: nPr = n!/(n−r)! = nCr × r! ✓. R true and correctly explains: every selection of r objects gives r! distinct arrangements.
AR 3 Events
Assertion (A): Mutually exclusive events can also be independent events.

Reason (R): For mutually exclusive events P(A∩B) = 0, but for independent events P(A∩B) = P(A)·P(B). If P(A)·P(B) = 0, at least one of them is an impossible event.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (d)
A is false: non-trivial mutually exclusive events (P(A)>0, P(B)>0) cannot be independent since P(A∩B)=0 ≠ P(A)·P(B)>0. R is true and provides the correct explanation for why A is false.
AR 4 Conditional Probability
Assertion (A): If P(A|B) = P(A), then events A and B are independent.

Reason (R): P(A|B) = P(A) means the occurrence of B does not affect the probability of A.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
A true: P(A|B) = P(A) → P(A∩B)/P(B) = P(A) → P(A∩B) = P(A)·P(B) → independent ✓. R true and correctly explains A.
AR 5 Combination Property
Assertion (A): ⁿCr = ⁿC(n−r) for all valid r.

Reason (R): Choosing r objects to include is equivalent to choosing (n−r) objects to exclude.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
A true: nCr = n!/[r!(n−r)!] = nC(n−r) ✓. R true and correctly explains: every selection of r items leaves (n−r) items unselected — a one-to-one correspondence.

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Step-by-Step Solutions

Solved Examples

Click Show Solution to reveal complete working.

Combinatorics
Q1 Factorial
Evaluate: (i) 7!/5!   (ii) (n+1)!/n!   (iii) 10!/(7! × 3!)
(i) 7!/5! = (7×6×5!)/5! = 7×6 = 42
(ii) (n+1)!/n! = (n+1)×n!/n! = n+1
(iii) 10!/(7!×3!) = (10×9×8×7!)/(7!×6) = 720/6 = 120
✓ (i) 42  |  (ii) n+1  |  (iii) 120
Q2 FPC
A restaurant offers 3 starters, 5 main courses and 2 desserts. In how many ways can a person choose a 3-course meal (one from each)?
Choices are sequential and independent → apply Multiplication Principle
Total = 3 × 5 × 2 = 30 ways
✓ 30 different 3-course meal combinations
Q3 Permutation
In how many ways can 4 different books be arranged on a shelf? If one particular book must always be at the first position, how many arrangements are possible?
(i) All 4 books in a row: ⁴P₄ = 4! = 24 ways
(ii) 1st position fixed → arrange remaining 3 books: 3! = 6 ways
✓ (i) 24 arrangements  |  (ii) 6 arrangements with fixed book
Q4 Permutation
How many 3-letter words (with or without meaning) can be formed from the letters of MATHS, if no letter is repeated?
MATHS has 5 distinct letters: M, A, T, H, S
Arranging 3 out of 5 (order matters): ⁵P₃ = 5!/(5−3)! = 5!/2! = 5×4×3 = 60
✓ 60 three-letter words possible
Q5 Combination
A group of 4 students is to be chosen from 7 boys and 4 girls such that: (i) there is no restriction, (ii) exactly 2 girls are included.
(i) No restriction: choose 4 from 11 → ¹¹C₄ = 11!/(4!×7!) = (11×10×9×8)/(4×3×2×1) = 330
(ii) Exactly 2 girls: choose 2 from 4 girls × choose 2 from 7 boys
= ⁴C₂ × ⁷C₂ = 6 × 21 = 126
✓ (i) 330 ways  |  (ii) 126 ways
Q6 P vs C
Explain with an example the difference between Permutation and Combination. Find ⁶P₂ and ⁶C₂ and verify ⁶P₂ = ⁶C₂ × 2!
Permutation (order matters): AB ≠ BA. Example: arranging 2 letters from {A,B,C,D,E,F}
Combination (order doesn't matter): {A,B} = {B,A}. Example: selecting 2 letters from same set
⁶P₂ = 6!/(6−2)! = 6×5 = 30
⁶C₂ = 6!/(2!×4!) = (6×5)/(2×1) = 15
Verify: ⁶C₂ × 2! = 15 × 2 = 30 = ⁶P₂ ✓
✓ ⁶P₂ = 30, ⁶C₂ = 15, and ⁶P₂ = ⁶C₂ × 2! verified
Probability
Q7 Sample Space
A coin is tossed twice. Write the sample space. Find the probability of: (i) exactly one head, (ii) at least one tail.
S = {HH, HT, TH, TT}, n(S) = 4
(i) Exactly one head: {HT, TH} → P = 2/4 = 1/2
(ii) At least one tail: {HT, TH, TT} → P = 3/4 or 1 − P(HH) = 1 − 1/4 = 3/4
✓ (i) 1/2  |  (ii) 3/4
Q8 Conditional Probability
If P(A) = 0.6, P(B) = 0.5 and P(A∩B) = 0.35, find P(A|B) and P(B|A).
P(A|B) = P(A∩B)/P(B) = 0.35/0.5 = 0.7
P(B|A) = P(A∩B)/P(A) = 0.35/0.6 = 7/12 ≈ 0.583
✓ P(A|B) = 0.7  |  P(B|A) = 7/12
Q9 Events
A die is thrown. Event A = {1,2,3}, Event B = {3,4,5}. Check if A and B are independent events.
P(A) = 3/6 = 1/2, P(B) = 3/6 = 1/2
A∩B = {3} → P(A∩B) = 1/6
P(A)·P(B) = (1/2)×(1/2) = 1/4
Since 1/6 ≠ 1/4, A and B are NOT independent (dependent events).
✓ A and B are dependent events
Q10 Multiplication Theorem
A bag contains 4 red and 6 black balls. Two balls are drawn one after the other without replacement. Find the probability that both are red.
P(1st red) = 4/10 = 2/5
P(2nd red | 1st red) = 3/9 = 1/3 (3 red left, 9 total)
P(both red) = P(1st red) × P(2nd red | 1st red) = (2/5) × (1/3) = 2/15
✓ P(both red) = 2/15
Q11 Independent Events
A and B are independent events. P(A) = 1/4, P(B) = 2/5. Find: (i) P(A∩B), (ii) P(A∪B), (iii) P(A'∩B').
(i) P(A∩B) = P(A)·P(B) = (1/4)×(2/5) = 1/10
(ii) P(A∪B) = P(A)+P(B)−P(A∩B) = 1/4+2/5−1/10 = 5/20+8/20−2/20 = 11/20
(iii) A', B' also independent: P(A'∩B') = P(A')·P(B') = (3/4)×(3/5) = 9/20
✓ (i) 1/10  |  (ii) 11/20  |  (iii) 9/20
Q12 Combinatorics + Probability
A committee of 2 men and 3 women is to be formed from 5 men and 4 women. In how many ways can this be done? If a committee is chosen at random, what is the probability that it contains a particular woman?
Total ways = ⁵C₂ × ⁴C₃ = 10 × 4 = 40
Ways with particular woman W fixed: ⁵C₂ × ³C₂ = 10 × 3 = 30 (choose 2 more women from remaining 3)
P(W in committee) = 30/40 = 3/4
✓ 40 total ways  |  P(particular woman included) = 3/4
Q13 At Least / At Most
Three students A, B, C independently try to solve a problem. Their probabilities of success are 1/3, 1/4, 1/5. Find: (i) P(all solve), (ii) P(none solves), (iii) P(at least one solves).
P(A)=1/3, P(B)=1/4, P(C)=1/5 · P(A')=2/3, P(B')=3/4, P(C')=4/5
(i) P(all) = (1/3)×(1/4)×(1/5) = 1/60
(ii) P(none) = (2/3)×(3/4)×(4/5) = 24/60 = 2/5
(iii) P(at least one) = 1 − P(none) = 1 − 2/5 = 3/5
✓ (i) 1/60  |  (ii) 2/5  |  (iii) 3/5

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4-Mark Questions

Case Studies

Board-pattern case questions covering Combinatorics and Probability. Click Show Answers for each case.

🎫

Case Study 1: School Fair Card Game (Probability)

At a school fair, a student picks one card at random from a well-shuffled deck of 52 playing cards. Event A = "the card is a red card". Event B = "the card is a face card" (Jack, Queen or King).
(i)

Find P(A) and P(B).

(ii)

Given the card is a face card, find the probability it is red.

(iii)

Are events A and B independent? Justify with calculation.

(iv)

Find P(A∪B) — probability of drawing a red card or a face card.

(i)Red cards = 26 → P(A) = 26/52 = 1/2. Face cards = 12 → P(B) = 12/52 = 3/13.
(ii)Red face cards = 6. P(A|B) = P(A∩B)/P(B) = (6/52)/(12/52) = 1/2.
(iii)P(A∩B) = 6/52 = 3/26. P(A)·P(B) = (1/2)(3/13) = 3/26. Since P(A∩B) = P(A)·P(B), A and B are independent.
(iv)P(A∪B) = P(A)+P(B)−P(A∩B) = 26/52+12/52−6/52 = 32/52 = 8/13.
🏫

Case Study 2: Student Council Selection (Combinatorics)

A school wants to form a 5-member student council from 8 boys and 6 girls.
(i)

In how many ways can the council be formed with no restriction?

(ii)

In how many ways can the council have exactly 3 boys and 2 girls?

(iii)

In how many ways can the council include at least 4 girls?

(iv)

If the council is selected at random, find the probability that it has exactly 3 boys and 2 girls.

(i)Choose 5 from 14: ¹⁴C₅ = 14!/(5!×9!) = 2,002 ways.
(ii)⁸C₃ × ⁶C₂ = 56 × 15 = 840 ways.
(iii)At least 4 girls = (4 girls,1 boy) + (5 girls,0 boys)
= ⁶C₄×⁸C₁ + ⁶C₅×⁸C₀ = 15×8 + 6×1 = 120+6 = 126 ways.
(iv)P = 840/2002 = 60/143 ≈ 0.42.
🎒

Case Study 3: Coloured Balls — Sequential Draws (Probability)

A bag contains 5 red, 4 white and 3 blue balls. Two balls are drawn one after the other without replacement.
(i)

Find the probability that the first ball is red and the second is white.

(ii)

Find the probability that both balls drawn are of the same colour.

(iii)

Given the second ball is blue, find the probability the first was also blue.

(iv)

Find the probability that the two balls are of different colours.

(i)P(R then W) = (5/12)×(4/11) = 20/132 = 5/33.
(ii)P(both red)+P(both white)+P(both blue) = 20/132+12/132+6/132 = 38/132 = 19/66.
(iii)P(2nd blue) = 33/132 = 1/4. P(1st blue|2nd blue) = (6/132)/(1/4) = 2/11.
(iv)1 − P(same colour) = 1 − 19/66 = 47/66.
🔑

Case Study 4: Password Creation (Combinatorics)

A company requires employees to create a 4-character password using the digits 0–9, with no digit repeated.
(i)

How many such passwords are possible?

(ii)

How many passwords start with the digit 5?

(iii)

How many passwords have all even digits (0,2,4,6,8)?

(iv)

If a password is chosen at random from all valid passwords, find the probability that it starts with 5.

(i)Order matters, no repetition: ¹⁰P₄ = 10!/6! = 10×9×8×7 = 5,040.
(ii)1st digit fixed (5), arrange remaining 3 from 9 digits: ⁹P₃ = 9×8×7 = 504.
(iii)5 even digits available, arrange 4: ⁵P₄ = 5!/1! = 120.
(iv)P = 504/5040 = 1/10.

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Score Full Marks

Exam Tips — Unit 4: Combinatorics & Probability

What separates 9-mark answers from 6-mark answers in this unit.

✅ Tip 1 — Permutation vs Combination

Always ask "does order matter?" first. Arranging, ranking, forming numbers, seating → Permutation (P). Selecting a team, committee, group → Combination (C). This single question prevents the most common error in the entire unit.

✅ Tip 2 — FPC Problems

Break the task into independent stages and multiply. Write each stage clearly: "Stage 1: choose shirt (4 ways), Stage 2: choose trouser (3 ways)" before multiplying. This earns method marks even if the final answer has an arithmetic slip.

✅ Tip 3 — Conditional Probability

Always write the formula P(A|B) = P(A∩B)/P(B) before substituting values. Many students skip straight to numbers and lose the formula mark. Double-check which event is the "given" condition — it goes in the denominator.

✅ Tip 4 — Independent vs Mutually Exclusive

These are NOT the same thing. To check independence: verify P(A∩B) = P(A)·P(B). To check mutually exclusive: verify A∩B = ∅. Non-trivial mutually exclusive events can never be independent — this is a favourite Assertion-Reason trap.

❌ Common Mistakes to Avoid in Unit 4

  • Using nPr formula when the question asks for selection (should use nCr)
  • Forgetting that 0! = 1, not 0
  • Confusing P(A|B) with P(B|A) — check which event is "given"
  • Assuming mutually exclusive events are automatically independent (they are NOT, except in trivial cases)
  • Forgetting "without replacement" changes probabilities for subsequent draws
  • Not verifying P(A)·P(B) = P(A∩B) before claiming independence
  • Mixing up nCr = nC(n−r) — useful shortcut often missed
Common Questions

Frequently Asked Questions

Questions students ask most about Class 11 Applied Maths Unit 4.

Unit 4 covers two sections: Combinatorics (Factorial, Fundamental Principle of Counting, Permutation nPr, Combination nCr, the difference between Permutation and Combination) and Probability (Random Experiment, Sample Space, Types of Events, Conditional Probability P(A|B), Independent Events). The unit carries 10 marks.
Permutation is an arrangement where order matters: nPr = n!/(n−r)!. Combination is a selection where order does not matter: nCr = n!/[r!(n−r)!]. Quick test: if rearranging gives a different outcome (like a password or a race finish), use P. If only the group composition matters (like a committee), use C.
Conditional probability P(A|B) is the probability of event A occurring given that event B has already occurred. Formula: P(A|B) = P(A∩B)/P(B), valid only when P(B) ≠ 0. It updates our probability estimate for A based on new information about B.
Unit 4 (Combinatorics & Probability) carries 10 marks in the CBSE Class 11 Applied Maths annual exam. Questions appear as 1-mark MCQs and Assertion-Reason, 2–3 mark short answers, and 4-mark case studies.
The Fundamental Principle of Counting (FPC) states: if one task can be done in m ways and a second independent task in n ways, both together can be done in m × n ways. This is also called the Multiplication Principle and is the foundation of all permutation and combination problems.
Mutually exclusive events cannot occur together: P(A∩B) = 0 (e.g. rolling a 2 and a 5 on one die). Independent events do not affect each other's probability: P(A∩B) = P(A)·P(B). A common exam trap: non-trivial mutually exclusive events (where both have probability > 0) can never be independent, since 0 ≠ P(A)·P(B) when both probabilities are positive.
0! = 1 by convention, so that formulas like nPr = n!/(n−r)! and nCr = n!/[r!(n−r)!] work correctly when r = n. For example, nCn should equal 1 (there's only one way to choose all n items), and this only works if 0! = 1 in the denominator.
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