Counting and Probability

What This Module Covers

This module covers the three main counting frameworks the GRE tests — the Fundamental Counting Principle, permutations, and combinations — plus probability rules including the complement, addition, and multiplication rules. You'll also see how overlapping sets connect to probability problems. Every concept here shows up on both the multiple-choice and Numeric Entry question types.

Why It Matters on the GRE

Counting and probability questions trip up students more than almost any other topic, not because the math is hard, but because one wrong assumption — "does order matter here?" — sends you down the wrong path entirely. The GRE loves to mix counting with probability in a single question, so you need both toolkits working together. The good news: once you have a reliable decision process, these questions become among the most straightforward on the test.

Core Concepts

1. The Fundamental Counting Principle

If you have one decision to make with m options and a second independent decision with n options, the total number of ways to make both decisions is m × n. The principle extends naturally — just keep multiplying across each independent decision. This is always your starting point before reaching for any formula.

Must Know: Total outcomes = (choices for step 1) × (choices for step 2) × (choices for step 3) × … Multiply across every independent slot.

Example: You're packing 4 shirts and 3 pairs of pants. How many outfits can you make?

1. Shirts: 4 options
2. Pants: 3 options
3. Total outfits: 4 × 3 = 12

No formula needed — just multiply the slots.

2. Permutations — When Order Matters

A permutation is an ordered arrangement. Think: passwords, race finishes, assigned seats — being first is different from being second.

Must Know: P(n, r) = n! / (n − r)!, where n = total items and r = positions you're filling. Equivalently, just count down: n × (n−1) × (n−2) × … for r terms.

Example: You have 5 books and 3 shelf slots. How many arrangements are possible?

1. Slot 1: 5 choices
2. Slot 2: 4 remaining choices (one book is already placed)
3. Slot 3: 3 remaining choices
4. Total: 5 × 4 × 3 = 60

Using the formula: P(5, 3) = 5! / (5−3)! = 120 / 2 = 60. Same answer.

Ask yourself first: "If I swapped two of the chosen items, would I get a different outcome?" If yes, it's a permutation.

3. Combinations — When Order Doesn't Matter

A combination is a selection where the arrangement is irrelevant. A committee of Alice, Bob, and Carol is the same committee no matter what order you list their names.

Must Know: C(n, r) = n! / (r! × (n−r)!). The r! in the denominator divides out all the orderings you'd get if you treated the selection as a permutation — because those orderings are all the same outcome here.

Example: Choose 3 people from a group of 10 to form a committee.

1. Start with permutations: 10 × 9 × 8 = 720 ordered arrangements
2. We overcounted — every group of 3 was counted 3! = 6 times (once for each ordering)
3. Divide out the overcounting: 720 / 6 = 120

Using the formula: C(10, 3) = 10! / (3! × 7!) = 120.

The key question every time: "Does order matter?" If yes → permutation. If no → combination.

4. Basic Probability and the Complement Rule

Probability = (favorable outcomes) / (total equally likely outcomes). The result is always between 0 and 1, inclusive. When all outcomes are equally likely, counting is the tool that gets you favorable outcomes and total outcomes.

Must Know: P(not A) = 1 − P(A). This is the complement rule, and it's one of the most powerful shortcuts on the GRE. Use it any time the direct calculation requires adding up many cases.

"At least one" problems: Computing P(at least one) directly means adding up P(exactly 1) + P(exactly 2) + … — that's tedious. The complement is almost always one calculation: P(at least one) = 1 − P(none).

Example: P(rolling at least one 6 on two dice) = 1 − P(no 6 on either die) = 1 − (5/6 × 5/6) = 1 − 25/36 = 11/36.

5. Combined Probability — Addition and Multiplication Rules

Addition rule (P(A or B)):

• Mutually exclusive events (can't both happen): P(A or B) = P(A) + P(B). Rolling a 2 or a 5 on one die — these can't both occur on one roll.
• Non-exclusive events: P(A or B) = P(A) + P(B) − P(A and B). You must subtract the overlap to avoid counting it twice.

Must Know: For non-exclusive events, always subtract P(A and B). Forgetting this step is the single most common error in GRE probability.

Multiplication rule (P(A and B)):

• Independent events (outcome of one doesn't affect the other): P(A and B) = P(A) × P(B).
• Dependent events (drawing without replacement is the classic case): The second probability changes based on what happened first. Multiply P(first event) × P(second event | first event happened).

Example: Drawing 2 red cards in a row from a standard deck without replacement: P = (26/52) × (25/51) = 25/102.

6. Overlapping Sets — Inclusion-Exclusion

When two groups share members, the formula is: |A or B| = |A| + |B| − |A and B|. In word problems this becomes: Total = Group A + Group B − Both + Neither.

Must Know: If a problem gives you two overlapping groups and a total, set up the inclusion-exclusion equation and solve for the unknown. Draw a two-circle Venn diagram if it helps you track the regions.

Common Traps

• Permutation when order doesn't matter: Using P(n, r) for a committee or selection problem overcounts by a factor of r!. Always ask "does order matter?" before choosing your formula.
• Forgetting to subtract P(A and B): In P(A or B) for non-exclusive events, skipping the subtraction double-counts the overlap — a very common source of wrong answers.
• Assuming independence when it isn't: "Without replacement" problems are dependent — the pool shrinks after each draw. Never apply the simple multiplication rule (P(A) × P(B)) unless the draws are truly independent.
• Computing "at least one" directly: This requires summing multiple cases. Use the complement: 1 − P(none) is almost always a single, fast calculation.
• Factorial arithmetic errors: On GRE problems, most factorials cancel cleanly. Write out the fraction before computing, and cancel terms before multiplying.

GRE Strategy

• Before writing anything, decide: counting question or probability question? Then decide: permutation or combination? These two decisions determine everything.
• On any "at least one" probability question, immediately write P = 1 − P(none) and work from there.
• When a problem involves drawing from a group without replacement, write out the chain of fractions (first draw, second draw, …) and multiply — don't look for a shortcut that assumes independence.
• Check that your final probability answer is between 0 and 1. If it isn't, you made an arithmetic or setup error.
• For counting problems, use the Fundamental Counting Principle slot-by-slot before reaching for the permutation or combination formula — it's faster and less error-prone on straightforward problems.

Worked Example

Question: A bag contains 4 red marbles and 6 blue marbles. If 2 marbles are drawn randomly without replacement, what is the probability that both are red?

(A) 2/15 (B) 4/25 (C) 1/5 (D) 2/5 (E) 4/15

Solution:

Step 1 — Count the total ways to draw 2 marbles from 10.

Order doesn't matter here (we just care whether both are red, not which one came out first). Use combinations:

C(10, 2) = 10! / (2! × 8!) = (10 × 9) / (2 × 1) = 45

Step 2 — Count the favorable ways to draw 2 red marbles from the 4 red ones.

C(4, 2) = 4! / (2! × 2!) = (4 × 3) / (2 × 1) = 6

Step 3 — Calculate the probability.

P(both red) = favorable / total = 6 / 45 = 2/15

The answer is (A).

Verification: Check that 6/45 simplifies to 2/15. GCD(6, 45) = 3 → 6/3 = 2 and 45/3 = 15. Confirmed.

Why the other choices are wrong:

• (B) 4/25 comes from incorrectly treating the draws as independent: (4/10) × (4/10) = 16/100 = 4/25. This ignores that after drawing one red marble, only 9 marbles remain — the draws are dependent.
• (C) 1/5 likely results from computing (4/10) × (3/9) = 12/90 = 2/15 incorrectly simplified, or from a partial setup error in the combination count.
• (D) 2/5 is the probability that a single marble drawn is red (4 out of 10), not both marbles — a scope error.
• (E) 4/15 could come from using P(4,2) / P(10,2) = 12/90 = 2/15 but making an arithmetic mistake along the way, or from a denominator error in the combination setup.

Key takeaway from this problem: Recognize "without replacement" immediately — it means dependent events and combinations for the count. Never multiply (4/10) × (4/10) when the pool is shrinking.