4.4 Permutations, Combinations & Probability Principles

4.4 Permutations, Combinations & Probability Principles

Combinatorics and discrete probability represent advanced Quantitative Reasoning topics on the HEC Higher Education Aptitude Test (HAT). These concepts evaluate a candidate's structural analytical capacity to count arrangements, evaluate multi-stage selection committees, and quantify likelihoods under probabilistic constraints. Mastered systematically, permutation, combination, and probability items can be solved rapidly using factorial properties and complementary shortcuts without manually listing outcome sample spaces.

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The Fundamental Counting Principle

The Fundamental Counting Principle provides the foundation for all combinatorics rules.

1. Multiplication Principle (Sequential Independent Choices)

If an operation consists of $k$ sequential stages, where Stage $1$ can be performed in $n_1$ ways, Stage $2$ in $n_2$ ways, $\dots$, and Stage $k$ in $n_k$ ways, the total number of composite ways to perform the operation is:

$$\text{Total Outcomes} = n_1 \times n_2 \times n_3 \times \dots \times n_k$$

Worked Example: A security passcode requires $2$ uppercase letters (from $26$) followed by $3$ non-zero digits (from $1$ to $9$). How many distinct passcodes can be created if repetition is allowed?

• Total Passcodes $= 26 \times 26 \times 9 \times 9 \times 9 = 676 \times 729 = 492,804$.

2. Addition Principle (Mutually Exclusive Alternatives)

If a choice can be made either from Set A (containing $m$ options) or from Set B (containing $n$ options), and Set A and Set B share no common elements (disjoint sets), the total number of available choices is:

$$\text{Total Choices} = m + n$$

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Permutations: Linear, Indistinguishable & Circular Arrangements

A permutation is an ordered arrangement of items. Order is paramount ($AB \neq BA$).

Linear Permutations

The number of ways to arrange $r$ distinct items chosen from a set of $n$ distinct items is:

$$P(n,r) = \frac{n!}{(n-r)!}$$

When arranging all $n$ items ($r = n$), $P(n,n) = n!$.

Permutations with Indistinguishable (Repeated) Items

If a set of $n$ items contains $p$ identical items of type 1, $q$ identical items of type 2, and $r$ identical items of type 3, the number of unique linear arrangements is:

$$\text{Distinct Permutations} = \frac{n!}{p! \cdot q! \cdot r!}$$

Worked Example: How many distinct $8$-letter arrangements can be formed using all the letters of the word PAKISTAN?

• Total letters $n = 8$.
• Letter frequencies: P:1, A:2, K:1, I:1, S:1, T:1, N:1 (letter 'A' repeats $p=2$ times).
• Distinct Arrangements $= \frac{8!}{2!} = \frac{40,320}{2} = 20,160$.

Circular Permutations

When arranging $n$ distinct items in a closed circle (where rotating the entire circle does not create a new relative ordering):

1. Fixed Circular Orientation (e.g., seating people around a round table): $$\text{Circular Permutations} = (n - 1)!$$
2. Flippable Circular Orientation (e.g., threading distinct beads on a necklace or keys on a ring, where flipping the ring over yields identical arrangements): $$\text{Necklace Permutations} = \frac{(n - 1)!}{2}$$

Worked Example: In how many ways can $6$ committee members be seated around a circular conference table?

• Ways $= (6 - 1)! = 5! = 120$.

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Combinations & Mathematical Properties

A combination is an unordered selection of items. Order does NOT matter ($AB = BA$).

$$\text{Combinations } C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

Essential Algebraic Identities of Combinations

1. Symmetry Property: $C(n,r) = C(n, n-r)$. (e.g., $C(10,8) = C(10,2) = \frac{10 \times 9}{2 \times 1} = 45$).
2. Pascal's Identity: $C(n,r) + C(n, r-1) = C(n+1, r)$.
3. Sum of Combinations: $\sum_{k=0}^n C(n,k) = C(n,0) + C(n,1) + \dots + C(n,n) = 2^n$.
4. Boundary Values: $C(n,0) = 1, \quad C(n,1) = n, \quad C(n,n) = 1$.

Committee Selection Problems with Restrictions

Worked Example: A committee of $5$ members is to be selected from $6$ men and $5$ women. In how many ways can the committee be formed if it must contain exactly $3$ men and $2$ women?

1. Select $3$ men from $6$: $C(6,3) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
2. Select $2$ women from $5$: $C(5,2) = \frac{5 \times 4}{2 \times 1} = 10$.
3. Total Committee Selection $= 20 \times 10 = 200 \text{ ways}$.

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Basic Probability Axioms & Theoretical Rules

Classical probability quantifies the likelihood of an event $A$ occurring within a sample space $S$ containing equally likely outcomes:

$$P(A) = \frac{N(A)}{N(S)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$$

Kolmogorov Axioms of Probability

1. Probability Bounds: For any event $A$, $0 \le P(A) \le 1$.
2. Certain Event: $P(S) = 1$; Impossible Event: $P(\emptyset) = 0$.

Multiplication Rule for Independent Events

Two events $A$ and $B$ are independent if the occurrence of $A$ does not alter the probability of $B$:

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

Addition Rule for Mutually Exclusive Events

Two events $A$ and $B$ are mutually exclusive if they cannot occur simultaneously ($A \cap B = \emptyset$):

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

General Addition Rule (Non-Mutually Exclusive Events)

If events $A$ and $B$ can overlap:

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

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Complementary Probability & "At Least One" Problems

The Complementary Probability Law states:

$$P(A^c) = P(\text{Not } A) = 1 - P(A)$$

Strategic Shortcut for "At Least One" Problems

Calculating the probability of "at least one success" across $n$ trials directly can be complex, as it requires summing $P(1) + P(2) + \dots + P(n)$. Using the complement rule makes it trivial:

$$P(\text{At least one success}) = 1 - P(\text{Zero successes})$$

Worked Example: A fair coin is tossed $5$ times. What is the probability of getting at least one Head?

• Probability of getting Zero Heads (all Tails) $= \left(\frac{1}{2}\right)^5 = \frac{1}{32}$.
• $P(\text{At least one Head}) = 1 - P(\text{Zero Heads}) = 1 - \frac{1}{32} = \frac{31}{32}$.