Advanced Quantitative & Reasoning

🎲 Permutations, Combinations & Probability

Day 3 | ⏱️ Part of 3-Hour Advanced Quant Block | TCS Interview Prep


📌 Subtopics

Permutations

  • Arrangements of objects (order matters)
  • Permutations with repetition
  • Circular permutations

Combinations

  • Selections of objects (order doesn’t matter)
  • Combinations with restrictions

Probability

  • Basic probability (favorable/total outcomes)
  • Addition and multiplication rules
  • Independent & dependent events
  • Conditional probability basics

📖 Comprehensive Notes

1. Permutations (Arrangements)

Permutation is an arrangement of a set of objects in a specific order. Order matters.

  • Without Repetition: Number of ways to arrange rr objects from a set of nn distinct objects: nPr=n!(n−r)!nPr = \frac{n!}{(n-r)!}
    • Arrangement of nn distinct objects in a row: n!n!.
  • With Repetition: Number of arrangements of nn objects where pp are of one kind, qq of another, and rr of another: Arrangements=n!p!×q!×r!\text{Arrangements} = \frac{n!}{p! \times q! \times r!}
  • Circular Permutations: Number of ways to arrange nn distinct objects in a circle: Circular arrangements=(n−1)!\text{Circular arrangements} = (n - 1)!
    • If the direction (clockwise vs. counter-clockwise) does not matter (e.g., beads in a necklace): Arrangements=(n−1)!2\text{Arrangements} = \frac{(n-1)!}{2}

Example: In how many ways can the letters of the word “APPLE” be arranged? Total letters n=5n = 5. ‘P’ is repeated 22 times (p=2p = 2). Ways=5!2!=1202=60\text{Ways} = \frac{5!}{2!} = \frac{120}{2} = 60 ways.

2. Combinations (Selections)

Combination is a selection of items from a group. Order does not matter.

  • Formula: Number of ways to select rr items from nn distinct items: nCr=n!r!(n−r)!=nPrr!nCr = \frac{n!}{r!(n-r)!} = \frac{nPr}{r!}
  • Properties:
    • nCr=nC(n−r)nCr = nC(n-r)
    • nC0=nCn=1nC0 = nCn = 1
    • nC1=nnC1 = n

Example: In how many ways can a committee of 33 members be chosen from a group of 88 people? n=8n = 8, r=3r = 3. 8C3=8×7×63×2×1=568C3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 ways.

3. Probability Basics & Rules

Probability is the measure of the likelihood that an event will occur. Probability of an event P(E)=Favorable OutcomesTotal Outcomes\text{Probability of an event } P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

  • Range: 0≤P(E)≤10 \le P(E) \le 1.
  • Complement: P(E)+P(E′)=1P(E) + P(E') = 1 (where E′E' is the event that EE does not occur).
  • Addition Rule: For any two events AA and BB: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
    • For mutually exclusive events (P(A∩B)=0P(A \cap B) = 0): P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B).
  • Multiplication Rule: For independent events AA and BB: P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)

Example: A card is drawn at random from a pack of 5252 cards. Find the probability that it is a King or a Heart. Total cards =52= 52. P(King)=4/52P(\text{King}) = 4/52, P(Heart)=13/52P(\text{Heart}) = 13/52. P(King ∩ Heart)=1/52P(\text{King } \cap \text{ Heart}) = 1/52 (the King of Hearts). P(King ∪ Heart)=452+1352−152=1652=413P(\text{King } \cup \text{ Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}.

4. Conditional Probability

Conditional probability is the probability of event AA occurring, given that event BB has already occurred. P(A∣B)=P(A∩B)P(B)P(A | B) = \frac{P(A \cap B)}{P(B)}

Example: A die is rolled. If the number shown is odd (event B), what is the probability that it is a prime number (event A)? Out of 66 outcomes:

  • Odd numbers {1,3,5}  ⟹  P(B)=3/6=1/2\{1, 3, 5\} \implies P(B) = 3/6 = 1/2.
  • Prime numbers {2,3,5}\{2, 3, 5\}.
  • Odd and Prime {3,5}  ⟹  P(A∩B)=2/6=1/3\{3, 5\} \implies P(A \cap B) = 2/6 = 1/3.
  • P(A∣B)=1/31/2=23P(A | B) = \frac{1/3}{1/2} = \frac{2}{3}.

🌐 Learning Resources

ResourceLink
IndiaBIX — Permutation & Combinationindiabix.com
IndiaBIX — Probabilityindiabix.com
GFG — Permutations & Combinationsgeeksforgeeks.org
GFG — Probabilitygeeksforgeeks.org
PrepInsta — TCS P&C and Probabilityprepinsta.com
Khan Academy — Probabilitykhanacademy.org
Khan Academy — Counting & Combinatoricskhanacademy.org

PlatformLink
PrepInsta — P&C Questionsprepinsta.com
IndiaBIX — P&C Problemsindiabix.com
IndiaBIX — Probability Problemsindiabix.com

📝 Key Formulas

PERMUTATIONS:
  nPr = n! / (n-r)!
  Circular = (n-1)!
  With repetition: n! / (p! × q! × r!)

COMBINATIONS:
  nCr = n! / (r! × (n-r)!)
  nCr = nC(n-r)
  nC0 = nCn = 1

PROBABILITY:
  P(E) = Favorable outcomes / Total outcomes
  0 ≤ P(E) ≤ 1
  P(E) + P(E') = 1
  P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
  P(A ∩ B) = P(A) × P(B)  [if independent]