5.3 Permutations, Combinations & Counting Principles

5.3 Permutations, Combinations & Counting Principles

Combinatorics and counting principles measure the total number of possible outcomes for complex multi-stage choices, arrangements, and selections. Mastering the distinction between permutations (where order matters) and combinations (where order does not matter), alongside overlapping set formulas, is vital for high scores on GRE Quantitative Reasoning.

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The Fundamental Counting Principle (FCP)

If an experiment consists of $k$ sequential stages, where Stage 1 can occur in $n_1$ ways, Stage 2 in $n_2$ ways, ..., and Stage $k$ in $n_k$ ways, then the total number of distinct outcomes is:

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

Example: A student creates a 4-digit PIN where the first digit cannot be 0 or 1 (8 options: 2-9), and the remaining three digits can be any digit 0-9 (10 options each). Total PINs = $8 \times 10 \times 10 \times 10 = 8,000$.

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Factorial Notation & Permutations (Order Matters)

Factorial Notation ($n!$)

For any positive integer $n$, $n!$ (n-factorial) is the product of all positive integers from $n$ down to $1$:

$n! = n \times (n-1) \times (n-2) \times \dots \times 1$

Special Definition: $0! = 1$.

Permutations ($nPr$)

A permutation is an arrangement of $r$ objects selected from a pool of $n$ distinct objects in a specific order.

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

When arranging all $n$ objects, $nPn = n!$.

Worked Example (Permutation): A board of directors with 10 members must elect a President, Vice President, and Secretary. How many different slates of officers are possible?

• Order matters because roles are distinct (President $ eq$ Secretary).
• $n = 10, r = 3$.
• $10P3 = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720$.

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Combinations (Order Does NOT Matter)

A combination is a selection of $r$ objects from a set of $n$ distinct objects where the order of selection is irrelevant.

$nCr = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Key Combination Properties & Pascal's Identity

1. Symmetry: $\binom{n}{r} = \binom{n}{n-r}$. Choosing $r$ items to include is mathematically identical to choosing $n-r$ items to leave out (e.g., $\binom{10}{8} = \binom{10}{2} = 45$).
2. Boundary Values: $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$.

Worked Example (Combination): A manager must select a committee of 3 employees from a team of 8. How many different committees can be chosen?

• Order does not matter (a committee of Alice, Bob, Carol is the same as Carol, Alice, Bob).
• $n = 8, r = 3$.
• $8C3 = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

Multi-Group Selection Problems

When making independent selections across multiple groups, calculate combinations for each group and multiply using the Fundamental Counting Principle.

Worked Example: Select 2 engineers from a group of 5, AND 3 technicians from a group of 6.

• Engineer selections = $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$.
• Technician selections = $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
• Total Teams = $10 \times 20 = 200$.

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Indistinguishable Permutations (Counting with Repetition)

When arranging $n$ items where some items are identical, simple factorial $n!$ overcounts because swapping identical items produces no new visual sequence. The formula for distinct arrangements is:

$\text{Distinct Permutations} = \frac{n!}{k_1! \cdot k_2! \cdot \dots \cdot k_m!}$

where $k_1, k_2, \dots, k_m$ are the frequencies of each duplicate item.

Worked Example: How many unique 7-letter arrangements can be made using the letters in "ARRANGE"?

• Total letters $n = 7$.
• Duplicate frequencies: A appears 2 times, R appears 2 times, N, G, E appear 1 time each.
• $\text{Distinct Arrangements} = \frac{7!}{2! \times 2!} = \frac{5040}{2 \times 2} = 1260$.

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Decision Matrix: Permutations vs. Combinations

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Overlapping Sets & Venn Diagrams

2-Set Overlapping Formula

For two sets $A$ and $B$ within a universal set:

$N(A \cup B) = N(A) + N(B) - N(A \cap B)$ $\text{Total Population} = N(A) + N(B) - N(A \cap B) + N(\text{Neither})$

Worked Example: Out of 100 students, 65 study French, 45 study Spanish, and 20 study both. How many study neither?

• $N(\text{French} \cup \text{Spanish}) = 65 + 45 - 20 = 90$.
• $N(\text{Neither}) = 100 - 90 = 10$.

3-Set Venn Diagram Formula

For three overlapping sets $A, B, C$:

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

• Subtract pairwise intersections because they are counted twice when adding individual sets.
• Add back the triple intersection $N(A \cap B \cap C)$ because subtracting all pairwise intersections removed the triple overlap entirely.