Permutations & Combinations

Strategy/Review

The Decision Rule: Does Rearranging Create a Different Outcome?

This single question is your gateway to choosing the right formula. Ask yourself: "If I pick items A, B, and C, does the order (A,B,C) vs. (C,B,A) count as the same result or different results?"

• If rearranging creates a DIFFERENT outcome → use PERMUTATIONS (nPr). Order matters. Think: passwords, rankings, sequences, arrangements, lineups.
• If rearranging creates the SAME outcome → use COMBINATIONS (nCr). Order doesn't matter. Think: committees, teams, choosing items from a set, hands of cards.

Permutations: Order Matters

Formula: nPr = n! / (n − r)! Where n = total number of items, r = number chosen.

Intuition: You have n choices for the first position, (n−1) for the second, (n−2) for the third, and so on, for r positions. That product is exactly n!/(n−r)!.

Example 1: How many ways can a president, vice president, and secretary be chosen from 10 people? (Order matters — different offices are different roles.) 10P3 = 10! / 7! = 10 × 9 × 8 = 720

Example 2: How many 4-letter "words" (including nonsense) can be formed from the letters A, B, C, D, E, F without repeating letters? 6P4 = 6! / 2! = 6 × 5 × 4 × 3 = 360

Example 3: How many ways can 5 books be arranged on a shelf? 5P5 = 5! / 0! = 5! = 120 (remember 0! = 1)

Combinations: Order Doesn't Matter

Formula: nCr = n! / [r! × (n − r)!] Where n = total number of items, r = number chosen.

Intuition: Start with the permutation count (n!/(n−r)!), then divide by r! because each group of r items can be arranged in r! different orders, and we're counting all those arrangements as ONE combination.

Example 1: How many ways can a 3-person committee be chosen from 10 people? (Order doesn't matter — a committee of {Alice, Bob, Carol} is the same as {Carol, Bob, Alice}.) 10C3 = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720/6 = 120

Example 2: How many different 5-card poker hands from a 52-card deck? 52C5 = 52! / (5! × 47!) = (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1) = 311,875,200 / 120 = 2,598,960

Example 3: From 8 pizza toppings, how many ways to choose 3? 8C3 = 8! / (3! × 5!) = (8 × 7 × 6) / 6 = 56

The Shortcut: Cancel Symmetrically

When computing nCr, use the fact that nCr = nC(n−r). 10C7 = 10C3 = 120. Always choose the smaller r — it makes the arithmetic MUCH easier.

Also: when writing out the product, write the r terms descending from n in the numerator and r! in the denominator, then cancel. 10C3 = (10 × 9 × 8) / (3 × 2 × 1) = (10 × 9 × 8) / 6 = 720/6 = 120.

Recognizing P&C in Disguise

DAT P&C questions are often embedded in probability problems:

Probability = (number of favorable combinations) / (total combinations)

Example: A committee of 4 is randomly chosen from 6 men and 4 women. What's the probability the committee has exactly 2 women?

Favorable: Choose 2 women from 4 AND 2 men from 6. = 4C2 × 6C2 = 6 × 15 = 90

Total: Choose any 4 from 10. = 10C4 = 210

Probability = 90/210 = 3/7.

The Multiplication Principle (And = Multiply, Or = Add):

• "Choose X AND Y" → multiply combinations: 4C2 × 6C2
• "Choose X OR Y" (mutually exclusive scenarios) → add combinations

Factorial Arithmetic Speed Tips

• 0! = 1 (by definition)
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120
• 6! = 720
• 7! = 5,040
• 8! = 40,320

The DAT calculator can handle these, but knowing 5! and 6! by heart speeds you up.

Common Traps

Trap 1 — Using Permutations When You Need Combinations. "How many ways to select 3 students from 15?" If you use 15P3 = 2730, you're counting every group 6 times (3! = 6 ways to order 3 people). The correct answer is 15C3 = 455. The test-makers always include the permutation result as a wrong answer choice.

Trap 2 — Using Combinations When You Need Permutations. "How many 3-digit codes can be formed from digits 1-9 without repetition?" Code 123 ≠ 321. Use 9P3 = 504, not 9C3 = 84. Again, the wrong formula's result will be an answer choice.

Trap 3 — Forgetting to Multiply for Multi-Step Selections. "Choose 2 appetizers from 5 AND 3 entrées from 8." Total = 5C2 × 8C3 = 10 × 56 = 560. Not 5C2 + 8C3 = 66.

Trap 4 — Not Accounting for "At Least" or "At Most." "At least 1 woman on a 3-person committee from 5 men and 4 women." The complement method: P(at least 1 woman) = 1 − P(all men) = 1 − (5C3 / 9C3) = 1 − (10/84) = 74/84 = 37/42.

Trap 5 — Treating "Without Replacement" as "With Replacement." The formulas nPr and nCr assume selection WITHOUT replacement (you can't pick the same person twice). If a problem allows repetition, it's a different formula: n^r (for ordered selections with replacement). The DAT almost always uses without-replacement scenarios, but verify.