9.2 Permutations, Combinations, Classical Probability, Independent Events & Odds

Fundamental Principles of Counting

All combinatorial and probability calculations originate from two fundamental counting axioms:

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|                        FUNDAMENTAL COUNTING PRINCIPLES                      |
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| MULTIPLICATION PRINCIPLE (AND): Sequential stages with stated numbers of choices                |
|   If Step 1 has m ways AND Step 2 has n ways ---> Total = m * n             |
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| ADDITION PRINCIPLE (OR): Mutually exclusive alternatives                    |
|   If Alternative 1 has m ways OR Alternative 2 has n ways ---> Total = m + n |
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• Multiplication Rule: If an operation consists of $k$ sequential stages such that the first stage can be performed in $n_1$ ways, the second in $n_2$ ways, and the $k$-th in $n_k$ ways, and each earlier choice permits the stated number of choices at the next stage, the total number of composite ways is:

$$N = n_1 \times n_2 \times n_3 \times \dots \times n_k$$

• Addition Rule: If a task can be accomplished either by method $A$ (in $m$ mutually exclusive ways) or by method $B$ (in $n$ mutually exclusive ways), where both methods cannot occur simultaneously, the task can be completed in:

$$N = m + n \quad \text{ways}$$

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Permutations (Arrangements Where Order Matters)

A permutation is an ordered arrangement of a specified number of items selected from a given set.

1. Linear Permutations of Distinct Objects

The number of ways to arrange $r$ distinct items selected from $n$ distinct items without replacement is:

$$^n P_r = \frac{n!}{(n - r)!} = n(n - 1)(n - 2)\dots(n - r + 1)$$

• Special values: $^n P_0 = 1$, $^n P_1 = n$, $^n P_n = n!$, $0! = 1$.

2. Circular Permutations

When arranging $n$ distinct objects in a closed loop or circle:

• Distinct Orientations (e.g., seating people around a circular conference table):

$$P_{\text{circular}} = (n - 1)!$$

• Indistinguishable Orientations (e.g., beads on a necklace, keys on a ring where clockwise and counterclockwise arrangements coincide upon flipping):

$$P_{\text{necklace}} = \frac{(n - 1)!}{2}$$

3. Permutations with Identical (Repeated) Elements

The number of distinct permutations of $n$ objects, of which $p$ are of one identical kind, $q$ of a second identical kind, and $r$ of a third identical kind, is:

$$P = \frac{n!}{p! \cdot q! \cdot r!}$$

Example: The number of distinct arrangements of the 10 letters in the word LOCOMOTIVE (Total $n = 10$; letters O appears 3 times): $\frac{10!}{3!} = \frac{3,628,800}{6} = 604,800$.

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Combinations (Selections Where Order Does Not Matter)

A combination is an unordered grouping or selection of items chosen from a set.

Formula & Core Algebraic Identities

The number of combinations of $n$ distinct objects taken $r$ at a time is:

$$^n C_r = \binom{n}{r} = \frac{n!}{r! (n - r)!} = \frac{^n P_r}{r!}$$

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Classical Probability Foundations

If an experiment has a finite, non-empty sample space $S$ consisting of equally likely elementary outcomes, and an event $E \subseteq S$ comprises $n(E)$ favorable outcomes, the classical probability of $E$ is:

$$P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible exhaustive outcomes}}$$

Axiomatic Boundaries & Complementarity

1. Probability Range: $0 \le P(E) \le 1$.
   • $P(\emptyset) = 0$ (Impossible event).
   • $P(S) = 1$ (Sure or certain event).
2. Complement Rule: If $E'$ (or $\bar{E}$) is the non-occurrence of $E$:

$$P(E') = 1 - P(E) \iff P(E) + P(E') = 1$$

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Compound Events, Addition Rule & Conditional Independence

1. The Addition Theorem of Probability

• For Any Two Events $A$ and $B$ (Union / OR):

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

• Mutually Exclusive Events ($A \cap B = \emptyset$): Events that cannot occur simultaneously ($P(A \cap B) = 0$):

$$P(A \cup B) = P(A) + P(B)$$

• For Three Non-Exclusive Events:

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$$

2. Independent Events & The Multiplication Theorem

Two events $A$ and $B$ are statistically independent if the occurrence of one provides zero information about the likelihood of the other:

$$P(A \cap B) = P(A) \times P(B)$$

• At Least One Event Occurs: If $A, B,$ and $C$ are mutually independent events, the probability that at least one occurs is:

$$P(\text{At least one}) = 1 - P(A' \cap B' \cap C') = 1 - [P(A') \times P(B') \times P(C')]$$

3. Conditional Probability

The conditional probability of event $A$ occurring given that event $B$ has already occurred ($P(B) > 0$) is:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$

If $A$ and $B$ are independent, $P(A \mid B) = \frac{P(A) \times P(B)}{P(B)} = P(A)$.

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Standard Discrete Probability Models

1. Coin Tossing

For $n$ unbiased coins tossed simultaneously, total outcomes $n(S) = 2^n$. The probability of obtaining exactly $r$ heads obeys the binomial expansion:

$$P(r \text{ heads}) = \frac{^n C_r}{2^n}$$

2. Dice Rolling

• One standard 6-sided die: $n(S) = 6$.
• Two fair dice thrown together: $n(S) = 6 \times 6 = 36$.
• Sum of Two Dice Distribution: The sum $S$ ranges from 2 to 12 in a symmetric pyramid pattern:

Each paired row below gives the ways and probability for each individual sum, not the union of both sums.

3. Standard 52-Card Pack Structure

Total Deck = 52 Cards
 ├── Red Cards (26):   Hearts (13), Diamonds (13)
 └── Black Cards (26): Spades (13), Clubs (13)

Per Suit (13 cards):   Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King
Face Cards (12 total): 4 Jacks, 4 Queens, 4 Kings (3 per suit)
Aces plus Face Cards: 4 Aces + 12 Face Cards = 16

4. Urn / Ball Drawing: With vs. Without Replacement

• With Replacement: Trials are independent; total outcomes and composition remain constant across draws.
• Without Replacement: Trials are dependent; combinations ($^n C_r$) evaluate joint selections directly in one step.

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Odds in Favor and Odds Against

Probability can be expressed as a ratio of favorable outcomes to unfavorable outcomes.

Let an event $E$ have $a$ favorable outcomes and $b$ unfavorable outcomes, with total exhaustive outcomes $n(S) = a + b$:

• Odds in Favor of Event $E$:

$$\text{Odds in Favor} = a : b = \frac{P(E)}{1 - P(E)} = \frac{P(E)}{P(E')}$$

• Odds Against Event $E$:

$$\text{Odds Against} = b : a = \frac{1 - P(E)}{P(E)} = \frac{P(E')}{P(E)}$$

• Converting Odds to Probability:

$$P(E) = \frac{a}{a + b}, \quad P(E') = \frac{b}{a + b}$$

Example: If the odds against an operational delay on a section are $7 : 3$, then unfavorable outcomes $b = 7$ (on-time) and favorable outcomes $a = 3$ (delay). The probability of delay is $P(E) = \frac{3}{3 + 7} = \frac{3}{10} = 0.30$.

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Worked Examples

Worked Example 1: Operational Safety Committee Combinatorics

Problem: A divisional safety inspection team of 5 officers is to be formed from a pool of 6 Section Controllers and 4 Track Engineers. In how many ways can the committee be selected if: (a) There are no restrictions? (b) The committee must contain exactly 3 Section Controllers and 2 Track Engineers? (c) The committee must contain at least 4 Section Controllers?

• Step 1: Unrestricted Selection (Part a): Total officers $n = 6 + 4 = 10$; committee size $r = 5$.

$$N = {^{10} C_5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \text{ ways}$$

• Step 2: Exactly 3 Section Controllers and 2 Track Engineers (Part b): Apply multiplication rule between independent sub-selections:

$$N = {^6 C_3} \times {^4 C_2}$$ $${^6 C_3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $${^4 C_2} = \frac{4 \times 3}{2 \times 1} = 6$$ $$N = 20 \times 6 = 120 \text{ ways}$$

• Step 3: At Least 4 Section Controllers (Part c): "At least 4" implies either 4 SCs OR 5 SCs. Evaluate disjoint cases and sum via Addition Rule:
  • Case 1 (4 SC and 1 TE):

$${^6 C_4} \times {^4 C_1} = {^6 C_2} \times 4 = \left( \frac{6 \times 5}{2} \right) \times 4 = 15 \times 4 = 60 \text{ ways}$$

• Case 2 (5 SC and 0 TE):

$${^6 C_5} \times {^4 C_0} = 6 \times 1 = 6 \text{ ways}$$ $$\text{Total} = 60 + 6 = 66 \text{ ways}$$

• Conclusion: (a) 252 ways, (b) 120 ways, (c) 66 ways.

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Worked Example 2: Independent Reliability of Railway Signalling Blocks

Problem: Two adjacent automatic block sections operate independent electronic axle-counters. The probability that Block A clears correctly on train approach is $0.80$, while the probability that Block B clears correctly is $0.75$. Calculate the probability that: (a) Both block sections clear correctly. (b) Exactly one block section clears correctly. (c) At least one block section clears correctly.

• Step 1: Define Events and Complement Probabilities: Let $P(A) = 0.80 \implies P(A') = 1 - 0.80 = 0.20$. Let $P(B) = 0.75 \implies P(B') = 1 - 0.75 = 0.25$. Since the mechanisms are independent, $P(A \cap B) = P(A) \times P(B)$.
• Step 2: Both Clear Correctly (Part a):

$$P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60$$

• Step 3: Exactly One Clears Correctly (Part b): This corresponds to $(A \cap B') \cup (A' \cap B)$:

$$P(\text{Exactly One}) = [P(A) \times P(B')] + [P(A') \times P(B)]$$ $$P(\text{Exactly One}) = (0.80 \times 0.25) + (0.20 \times 0.75) = 0.20 + 0.15 = 0.35$$

• Step 4: At Least One Clears Correctly (Part c): Use complement of neither clearing:

$$P(\text{At least one}) = 1 - [P(A') \times P(B')] = 1 - (0.20 \times 0.25) = 1 - 0.05 = 0.95$$

• Conclusion: (a) 0.60, (b) 0.35, (c) 0.95.

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Worked Example 3: Selection Without Replacement and Odds Against

Problem: An operational junction store contains 5 red signal aspect bulbs, 4 green aspect bulbs, and 3 yellow aspect bulbs (total 12 bulbs). A maintenance technician selects 2 bulbs at random without replacement. Find: (a) The probability that both selected bulbs are red. (b) The probability that one bulb is green and one is yellow. (c) The odds against both bulbs being red.

• Step 1: Total Exhaustive Outcomes $n(S)$: Choosing 2 bulbs out of 12:

$$n(S) = {^{12} C_2} = \frac{12 \times 11}{2 \times 1} = 66$$

• Step 2: Probability Both Are Red (Part a): Favorable selections: choosing 2 red bulbs from 5 available:

$$n(E) = {^5 C_2} = \frac{5 \times 4}{2 \times 1} = 10$$ $$P(\text{Both Red}) = \frac{10}{66} = \frac{5}{33} \approx 0.1515$$

• Step 3: Probability One Green and One Yellow (Part b): Choosing 1 green from 4 AND 1 yellow from 3:

$$n(E) = {^4 C_1} \times {^3 C_1} = 4 \times 3 = 12$$ $$P(\text{One Green, One Yellow}) = \frac{12}{66} = \frac{2}{11} \approx 0.1818$$

• Step 4: Odds Against Both Bulbs Being Red (Part c): Probability of event $P = \frac{5}{33}$, so favorable outcomes $a = 5$ and unfavorable outcomes $b = 33 - 5 = 28$.

$$\text{Odds Against} = b : a = 28 : 5$$

• Conclusion: (a) $5/33$, (b) $2/11$, (c) $28 : 5$.

The classical favorable-outcomes ratio requires equally likely elementary outcomes. The coin and dice models assume independent fair trials; the necklace reflection formula assumes at least three distinct objects. Repeated-color sampling without replacement requires accounting for the changed composition.