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⚡ Self-Study Edition

Sets, Probability
& Combinations

20 carefully crafted questions with instant feedback. Focus on the tricky spots that most students get wrong.

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SECTION 01

Set Theory & Probability

🧠

⚡ Quick Memory Points — Sets

Core words you MUST know to explain sets clearly:
union → OR → ∪ → "everything" intersection → AND → ∩ → "overlap" complement → NOT → Aᶜ → "outside A" subset → A⊆B → "A fits inside B" ∅ = empty set = no elements n(A) = number of elements in A n(A∪B) = n(A)+n(B)−n(A∩B) Venn diagram = visual overlap tool
Inclusion-Exclusion
n(A∪B) = n(A)+n(B)−n(A∩B)

Always subtract the overlap or you double-count it!

Complement
n(Aᶜ) = n(U) − n(A)

U = Universal set. Everything NOT in A.

Probability
P(A) = n(A) / n(U)

Favorable outcomes ÷ Total outcomes.

Mutually Exclusive
P(A∩B) = 0 → P(A∪B) = P(A)+P(B)

Events that CANNOT happen at the same time.

Set Theory · Q01 Easy
In a class of 30 students, 18 play football and 15 play basketball. 8 students play both sports. How many students play football or basketball (at least one)?
Use: n(F∪B) = n(F) + n(B) − n(F∩B). Don't forget to subtract the overlap!
📖 Explanation
Using the Inclusion-Exclusion Principle:
n(F∪B) = n(F) + n(B) − n(F∩B) = 18 + 15 − 8 = 25
Key insight: Without subtracting the 8 who play both, we'd count them twice. Always subtract the intersection!
Correct answer: A) 25
Set Theory · Q02 Easy
Universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Set A = {2, 4, 6, 8, 10}. What is A' (the complement of A)?
Complement = everything in U that is NOT in A.
📖 Explanation
A' = U − A = all elements in U that are not in A.
U = {1,2,3,4,5,6,7,8,9,10}, A = even numbers = {2,4,6,8,10}
A' = odd numbers = {1, 3, 5, 7, 9}
Correct answer: B)
Set Theory · Q03 ⚠ Tricky
In a survey: n(U) = 50, n(A) = 30, n(B) = 25, n(A∩B) = 10.
How many students are in neither A nor B?
Step 1: Find n(A∪B). Step 2: Neither = n(U) − n(A∪B).
📖 Explanation
Step 1: n(A∪B) = 30 + 25 − 10 = 45
Step 2: Neither = n(U) − n(A∪B) = 50 − 45 = 5
Common mistake: Students forget the "neither" region exists outside both circles in the Venn diagram.
Correct answer: B) 5
Probability · Q04 Easy
A bag contains 4 red balls, 3 blue balls, and 5 green balls. A ball is picked at random. What is the probability of picking a red or blue ball?
Red and blue are mutually exclusive (can't be both at once). P(A or B) = P(A) + P(B).
📖 Explanation
Total balls = 4 + 3 + 5 = 12
P(red) = 4/12, P(blue) = 3/12
Since mutually exclusive: P(red or blue) = 4/12 + 3/12 = 7/12
Correct answer: A) 7/12
Probability · Q05 ⚠ Tricky
P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2.
Are events A and B independent?
Independence test: P(A∩B) = P(A) × P(B). Check if this equation holds!
📖 Explanation
Independence test: Is P(A∩B) = P(A) × P(B)?
P(A) × P(B) = 0.4 × 0.5 = 0.20
P(A∩B) = 0.20 ✓ They are equal!
→ A and B ARE independent.
Key point: Independent ≠ Mutually Exclusive. Events can overlap AND be independent.
Correct answer: A)
Probability · Q06 ⚠ Tricky
A card is drawn from a standard 52-card deck. Event A = "the card is a King", Event B = "the card is a Heart".
What is P(A∪B)?
There is ONE card that is both a King AND a Heart (King of Hearts). Don't forget to subtract it!
📖 Explanation
P(A) = 4/52 (4 Kings), P(B) = 13/52 (13 Hearts)
P(A∩B) = 1/52 (King of Hearts)
P(A∪B) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
Trap: Students often forget the King of Hearts exists in both categories.
Correct answer: B) 16/52
Probability · Q07 Medium
A fair die is rolled twice. What is the probability of getting a 6 on the first roll and an even number on the second?
These are independent events. Multiply the probabilities: P(both) = P(first) × P(second).
📖 Explanation
P(6 on first) = 1/6
P(even on second) = 3/6 = 1/2 (even numbers: 2,4,6)
Since independent: P(both) = 1/6 × 1/2 = 1/12
Correct answer: A) 1/12
Probability · Q08 ⚠ Tricky
A box has 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability that both are red?
"Without replacement" means after removing the first ball, the total count decreases. Use conditional probability: P(R₁ and R₂) = P(R₁) × P(R₂ | R₁).
📖 Explanation
P(1st red) = 5/8
P(2nd red | 1st was red) = 4/7 (now 4 red remain out of 7 total)
P(both red) = 5/8 × 4/7 = 20/56 = 5/14
Common mistake: Using 5/8 × 5/8 (treating it as "with replacement"). Always check!
Note: 10/56 = 5/28, which is not the same as 5/14. Answer B is correct.
Correct answer: B) 5/14
Set Theory · Q09 Medium
Which of the following is always true for any two sets A and B within universal set U?
Think about De Morgan's Laws: (A∪B)' = A'∩B' and (A∩B)' = A'∪B'.
📖 Explanation
This is De Morgan's First Law: (A∪B)' = A' ∩ B'
In plain English: "NOT (A or B)" = "NOT A AND NOT B"
Memory trick: When you flip the complement over a union/intersection, the symbol FLIPS (∪ ↔ ∩).
Correct answer: C)
Probability · Q10 ⚠ Tricky
P(A) = 0.6, P(B|A) = 0.3. What is P(A∩B)?
Conditional probability formula: P(B|A) = P(A∩B) / P(A). Rearrange to find P(A∩B).
📖 Explanation
From the conditional probability formula:
P(B|A) = P(A∩B) / P(A)
→ P(A∩B) = P(B|A) × P(A) = 0.3 × 0.6 = 0.18
Key word: P(B|A) means "probability of B GIVEN A has happened." Always multiply to get the joint probability.
Correct answer: C) 0.18
SECTION 02

Combinations & Permutations

🔢

⚡ Quick Memory Points — Counting

The ONE question to ask yourself every time:
ORDER matters? → Permutation nPr ORDER doesn't matter? → Combination nCr nPr = n! / (n−r)! nCr = n! / (r! × (n−r)!) nCr = nPr / r! 0! = 1 (memorize this!) nC0 = 1, nCn = 1 Password/code → Permutation Committee/team → Combination
Permutation (order matters)
nPr = n! / (n−r)!

Arranging r items from n, where position matters (e.g., race results).

Combination (order doesn't matter)
nCr = n! / (r!(n−r)!)

Choosing r items from n, where position doesn't matter (e.g., team selection).

Factorial
n! = n × (n−1) × ... × 1

5! = 120, 4! = 24, 3! = 6, 2! = 2, 1! = 1, 0! = 1

Fundamental Counting
Total = m × n × p × ...

Multiply choices at each independent step.

Combination · Q11 Easy
How many ways can you choose 3 students from a group of 7 to form a committee? (The committee has no specific roles — just members.)
Committee = no order needed → Combination! Use ⁷C₃ = 7! / (3! × 4!).
📖 Explanation
⁷C₃ = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 210/6 = 35
Tip: Cancel common factors before multiplying: (7×6×5)/(3×2×1) = 7×1×5/1 = 35
Correct answer: B) 35
Permutation · Q12 Easy
In a race with 8 runners, how many different ways can 1st, 2nd, and 3rd place be awarded?
1st ≠ 2nd ≠ 3rd — positions matter! Use ⁸P₃ = 8 × 7 × 6.
📖 Explanation
Positions matter → Permutation!
⁸P₃ = 8!/(8−3)! = 8!/5! = 8 × 7 × 6 = 336
Think: 8 choices for 1st, then 7 for 2nd, then 6 for 3rd.
Correct answer: C) 336
Combination · Q13 ⚠ Tricky
A student must choose 2 subjects from Math, Physics, Chemistry, and Biology. How many different subject combinations are possible?
Choosing {Math, Physics} is the same as {Physics, Math} — order doesn't matter. Use ⁴C₂.
📖 Explanation
⁴C₂ = 4! / (2! × 2!) = (4 × 3) / (2 × 1) = 12/2 = 6
Listing them: {M,P}, {M,C}, {M,B}, {P,C}, {P,B}, {C,B} → yes, 6 pairs.
Correct answer: D) 6
Permutation · Q14 Medium
How many 4-digit codes can be made using digits 1–9 if no digit can repeat?
A code like 1234 is different from 4321 → order matters → Permutation! ⁹P₄ = 9 × 8 × 7 × 6.
📖 Explanation
⁹P₄ = 9 × 8 × 7 × 6 = 3024
9 choices for 1st digit, 8 for 2nd (no repeat), 7 for 3rd, 6 for 4th.
Trap: 9⁴ = 6561 is wrong — that allows repetition.
Correct answer: C) 3024
Combination · Q15 ⚠ Tricky
A team of 5 is chosen from 6 boys and 4 girls. How many ways can the team be formed if it must include exactly 3 boys and 2 girls?
Solve in two parts: choose 3 from 6 boys AND choose 2 from 4 girls. Then multiply the results.
📖 Explanation
Boys: ⁶C₃ = 6!/(3!3!) = 20
Girls: ⁴C₂ = 4!/(2!2!) = 6
Total = 20 × 6 = 120
Key principle: Independent choices multiply! (AND = multiply)
Correct answer: C) 120
Counting · Q16 Medium
A menu has 3 starters, 4 main courses, and 2 desserts. How many different 3-course meals (one from each category) are possible?
Fundamental Counting Principle: multiply the number of options at each step.
📖 Explanation
Starters × Mains × Desserts = 3 × 4 × 2 = 24
The Fundamental Counting Principle: if event 1 has m ways and event 2 has n ways, together there are m×n ways.
Correct answer: B) 24
Combination · Q17 ⚠ Tricky
What is the value of 10C8?
Hint: Look for a shortcut before calculating!
Key identity: nCr = nC(n−r). So ¹⁰C₈ = ¹⁰C₂. This makes the calculation much easier!
📖 Explanation
Use the symmetry property: ¹⁰C₈ = ¹⁰C₂
¹⁰C₂ = (10 × 9) / (2 × 1) = 90/2 = 45
This works because choosing 8 to INCLUDE is the same as choosing 2 to EXCLUDE.
Correct answer: A) 45
Mixed · Q18 Medium
A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. What is the probability that both are blue?
Method 1: Multiply probabilities step by step. Method 2: Use combinations — ⁶C₂ / ¹⁰C₂.
📖 Explanation
Method 1: P(B₁) × P(B₂|B₁) = 6/10 × 5/9 = 30/90 = 1/3
Method 2: ⁶C₂ / ¹⁰C₂ = 15/45 = 1/3
Both methods give 1/3. Always double-check with the second method if time permits!
Correct answer: A) 1/3
Permutation · Q19 ⚠ Tricky
How many ways can the letters of the word "MATH" be arranged?
Now, how many arrangements start with the letter M?
Total arrangements: 4! (all 4 letters). If M is fixed at the start, arrange the remaining 3 letters: 3!.
📖 Explanation
Total arrangements of MATH (4 different letters): 4! = 4×3×2×1 = 24
Starting with M: Fix M, arrange A,T,H in the remaining 3 spots: 3! = 6
Shortcut: Fraction starting with M = 1/4 of total (since M is 1 of 4 letters) → 24/4 = 6 ✓
Correct answer: B) Total: 24, Starting with M: 6
Mixed · Q20 ⚠ Tricky
There are 5 boys and 4 girls in a club. A committee of 3 is chosen at random. What is the probability that the committee contains at least one girl?
"At least one" is tricky to count directly. Use the complement: P(at least 1 girl) = 1 − P(no girls at all).
📖 Explanation
Total ways to pick 3 from 9: ⁹C₃ = 84
P(no girls) = all 3 from 5 boys = ⁵C₃ = 10
P(no girls) = 10/84 = 5/42
P(at least 1 girl) = 1 − 5/42 = 37/42
Key strategy: "At least one" → always try COMPLEMENT first! Much faster.
Correct answer: A) 37/42