1.9 Permutations, Combinations, Probability & Statistics

Permutations, Combinations, Probability & Statistics

Combinatorics, probability, and data interpretation represent a critical component of the ACET. Rapid distinction between ordered arrangements and unordered subsets is required.

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1. Fundamental Counting Principle & Permutations

Fundamental Counting Principle (FCP)

If Event 1 can occur in $m$ ways and Event 2 can occur independently in $n$ ways, the total number of combined outcomes is $m \times n$.

Permutations (Order Matters!)

A permutation is an ordered arrangement of elements.

$$\textbf{Permutations of } n \textbf{ items taken } k \textbf{ at a time:} \quad P(n, k) = \frac{n!}{(n - k)!}$$ $$\textbf{Permutations with Repetition (Distinguishable):} \quad P = \frac{n!}{n_1! \cdot n_2! \cdots n_k!}$$ $$\textbf{Circular Permutations of } n \textbf{ distinct items:} \quad P_{\text{circular}} = (n - 1)!$$ $$\textbf{Circular Permutations (Keyrings/Necklaces):} \quad P_{\text{keyring}} = \frac{(n - 1)!}{2}$$

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2. Combinations (Order Does NOT Matter!)

A combination is a selection of items without regard to order.

$$\textbf{Combinations of } n \textbf{ items taken } k \textbf{ at a time:} \quad C(n, k) = \binom{n}{k} = \frac{n!}{k!(n - k)!}$$

Key Combinatorial Properties

• $\binom{n}{k} = \binom{n}{n - k}$
• $\binom{n}{0} = \binom{n}{n} = 1$
• Pascal's Identity: $\binom{n}{k} + \binom{n}{k + 1} = \binom{n + 1}{k + 1}$

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3. Probability Axioms & Compound Rules

Theoretical probability of event $A$ in sample space $S$:

$$P(A) = \frac{n(A)}{n(S)}, \quad 0 \le P(A) \le 1$$

Probability Rules

1. Complement Rule: $P(A') = 1 - P(A)$
2. Addition Rule (Any two events): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
3. Addition Rule (Mutually Exclusive): $P(A \cup B) = P(A) + P(B)$
4. Multiplication Rule (Independent Events): $P(A \cap B) = P(A) \cdot P(B)$
5. Conditional Probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0)$

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4. Descriptive Statistics & Normal Distribution

Measures of Central Tendency

• Mean ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{N}$
• Weighted Mean: $\bar{x}_w = \frac{\sum (w_i x_i)}{\sum w_i}$
• Median: Middle value when dataset is sorted.
• Mode: Most frequently occurring value.

Measures of Dispersion & Z-Scores

• Range $= X_{\text{max}} - X_{\text{min}}$
• Standard Deviation ($\sigma$): $\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}$
• Standardized Z-Score: $z = \frac{x - \mu}{\sigma}$

Empirical Rule (68-95-99.7 Rule for Normal Distributions)

• $\approx 68\%$ of data lies within $\mu \pm 1\sigma$
• $\approx 95\%$ of data lies within $\mu \pm 2\sigma$
• $\approx 99.7\%$ of data lies within $\mu \pm 3\sigma$

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Worked Step-by-Step ACET Exam Problems

Problem 1: Committee Selection with Group Constraints

A committee of 5 people is to be chosen from a group of 6 men and 4 women. In how many ways can the committee be formed if it must contain at least 3 women?

Step 1: Break down into mutually exclusive cases. Since the committee has 5 members, "at least 3 women" means either 3 women OR 4 women.

Step 2: Case 1 — Exactly 3 women and 2 men.

• Ways to select 3 women from 4: $\binom{4}{3} = 4$.
• Ways to select 2 men from 6: $\binom{6}{2} = \frac{6 \times 5}{2} = 15$.

$$\text{Ways for Case 1} = 4 \times 15 = 60$$

Step 3: Case 2 — Exactly 4 women and 1 man.

• Ways to select 4 women from 4: $\binom{4}{4} = 1$.
• Ways to select 1 man from 6: $\binom{6}{1} = 6$.

$$\text{Ways for Case 2} = 1 \times 6 = 6$$

Step 4: Sum all mutually exclusive cases.

$$\text{Total Ways} = 60 + 6 = 66$$

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Problem 2: Conditional Probability Without Replacement

A box contains 5 red marbles and 3 blue marbles. Two marbles are drawn sequentially without replacement. What is the probability that both marbles drawn are red?

Step 1: Determine probability of 1st draw.

$$P(R_1) = \frac{5}{5 + 3} = \frac{5}{8}$$

Step 2: Determine conditional probability of 2nd draw given 1st draw was red. Remaining red marbles $= 4$; total remaining marbles $= 7$.

$$P(R_2 \mid R_1) = \frac{4}{7}$$

Step 3: Apply multiplication rule for dependent events.

$$P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

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ACET Speed Tactics & Traps Summary