6.3 Counting Techniques, Permutations, Combinations & Probability

Why This Topic Matters for the Exam

Assessment target Q.8 is the smallest domain on the entire GED Math blueprint — just two indicators, Q.8.a ("Use counting techniques to solve problems and determine combinations and permutations") and Q.8.b ("Determine the probability of simple and compound events") — out of 62 total leaf indicators. Small does not mean skippable: counting and probability items appear reliably within the 45%-weighted Quantitative Problem Solving portion of the test, and because the underlying logic (multiply choices, decide if order matters, divide favorable by total) is compact, these are some of the highest-value points per minute of study on the whole exam.

Core Terms: Counting Techniques

The Fundamental Counting Principle states that if one event can happen in m ways and a second, independent event can happen in n ways, then both events together can happen in m × n ways. This extends to any number of stages: multiply the number of choices at every stage.

A permutation is a uniquely ordered arrangement of objects — order matters. A combination is a selection of objects where order does not matter. This is the single most important distinction on this topic, and the GED's own glossary defines permutation with an explicit warning: "do not substitute combination."

(Recall that n!, read "n factorial," means n × (n − 1) × (n − 2) × ... × 1; for example 4! = 4 × 3 × 2 × 1 = 24.)

Worked Example 1: Fundamental Counting Principle

A diner offers 3 sandwich choices, 4 side choices, and 2 drink choices. How many different complete meals (one sandwich, one side, one drink) are possible?

Multiply the choices at each stage: 3 × 4 × 2 = 24 different meals. Note there is no need to worry about order here — each meal is simply a combination of one item from each category, and the counting principle handles it directly.

Worked Example 2: Permutation — Order Matters

Five runners compete in a race. In how many different ways can 1st, 2nd, and 3rd place be awarded?

Because 1st, 2nd, and 3rd are distinct, ordered positions, this is a permutation of 5 runners taken 3 at a time: 5P3 = 5! ÷ (5 − 3)! = 5! ÷ 2! = (5 × 4 × 3 × 2 × 1) ÷ (2 × 1) = 120 ÷ 2 = 60 ways. You can also reason it directly: 5 choices for 1st place, then 4 remaining choices for 2nd, then 3 remaining choices for 3rd: 5 × 4 × 3 = 60.

Worked Example 3: Combination — Order Does Not Matter

From a group of 6 volunteers, a committee of 2 will be chosen to represent the group at an event, with no distinction between the two committee roles. How many different committees are possible?

Because the two committee spots are identical (order doesn't matter — choosing Sam-then-Alex is the same committee as Alex-then-Sam), this is a combination: 6C2 = 6! ÷ [2! × (6 − 2)!] = 6! ÷ (2! × 4!) = 720 ÷ (2 × 24) = 720 ÷ 48 = 15 committees.

Core Terms: Probability

Probability of an event is the number of favorable outcomes divided by the total number of possible outcomes, always expressed as a value from 0 (impossible) to 1 (certain), and often written as a fraction, decimal, or percent. A simple event involves one outcome (rolling a single die and getting a 4). A compound event combines two or more simple events, and can be independent (the outcome of one event does not affect the other, like two separate coin flips) or dependent (the outcome of the first event changes the possibilities for the second, like drawing cards from a deck without putting the first one back).

For independent compound events joined by "and," multiply the individual probabilities. For dependent events (sampling without replacement), recalculate the total outcomes for the second event based on what was removed. The complement of an event is "the event does not happen," and P(not A) = 1 − P(A) — this is often the fastest route to "at least one" problems.

Worked Example 4: Independent Compound Probability

A fair coin is flipped twice. What is the probability of getting heads both times?

Each flip is independent, and each has a probability of 1/2 for heads. Multiply: 1/2 × 1/2 = 1/4.

Worked Example 5: Dependent Compound Probability

A bag contains 4 red marbles and 6 blue marbles (10 total). Two marbles are drawn without replacement. What is the probability both are red?

First draw: 4 red out of 10 total, so P(first red) = 4/10. Because the marble is not replaced, the second draw has only 9 marbles left, with 3 red remaining: P(second red) = 3/9. Multiply: 4/10 × 3/9 = 12/90 = 2/15. The trap here is using 4/10 again for the second draw, which would incorrectly treat the event as independent ("with replacement") when the problem specifies otherwise.

Worked Example 6: Complementary Probability

The probability that it rains tomorrow is 0.35. What is the probability that it does NOT rain?

P(not rain) = 1 − P(rain) = 1 − 0.35 = 0.65.

Common Traps Checklist

• Applying the combination formula to a problem where order actually matters (or vice versa) — always ask "does swapping the order create a different outcome?"
• Using the same denominator for both draws in a "without replacement" problem, instead of shrinking the total for the second draw.
• Adding probabilities for "and" situations instead of multiplying (multiplication is for independent/joint events; addition is for "or" with mutually exclusive events).
• Forgetting the complement shortcut on "at least one" problems, leading to unnecessarily long calculations.
• Miscounting factorial terms — remember 0! = 1, and nPr always has a larger value than nCr for the same n and r because order is being counted separately.

Takeaways: use the Fundamental Counting Principle to multiply choices across independent stages, decide permutation-versus-combination by asking whether order changes the outcome, compute basic probability as favorable ÷ total, multiply probabilities for independent "and" events, adjust the total for dependent events sampled without replacement, and reach for the complement (1 minus the opposite event) whenever a problem asks for "at least one."