Factorials, combinations and probability

Define one outcome before counting

A counting problem begins with a description of a single outcome. A list of prize winners in first, second and third place is ordered: swapping first and third produces a different result. A committee of three people is normally unordered: listing the same members in another order does not create a new committee. Write one concrete outcome and swap two entries. If the result changes the object the problem asks for, order matters. This test prevents applying a remembered formula to the wrong situation.

Also decide whether an item can be reused. A code made from digits may allow repetition, while choosing distinct people for roles normally does not. Replacement and order are separate questions. A process can be ordered with or without replacement. Avoid assuming that the number of choices decreases at each stage simply because several stages occur. It decreases only when an earlier choice removes possibilities or another stated restriction changes the available set.

Multiply successive choices

The product rule counts a process with successive stages by multiplying the number of choices available at each stage. Three shirt choices and two trouser choices give six outfits when every shirt can be paired with every pair of trousers. If a restriction removes one specific pairing, count the allowed combinations directly or subtract that forbidden pairing from the unrestricted six. Multiplication is justified by the stage structure, not by the mere presence of two numbers in a question.

For distinct objects arranged without replacement, the choices decrease. Arranging all n distinct objects gives n factorial, written n!, equal to the product of the integers from n down to one. By convention, zero factorial equals one. There is one empty arrangement when no objects need to be placed, which also keeps the usual counting formulas consistent. A factorial is a compact notation for a product, not a rule to multiply the number by itself repeatedly.

Choosing r distinct objects from n for ordered roles gives n! divided by (n−r)!. It represents the product of the r decreasing choice counts. For example, two different roles filled from five candidates give five times four outcomes. The unused three candidates have no internal ordering in the outcome, so there is no reason to keep multiplying by three, two and one after both roles have been assigned.

Remove the order that the question does not distinguish

For an unordered selection of r objects, each selected group has been counted r! times by the ordered calculation. Divide by r! to remove those repeats. The resulting binomial coefficient is written $\binom{n}{r}=n!/[r!(n-r)!]$. A selection of two from five therefore gives twenty ordered pairs divided by two, or ten distinct pairs. The division is justified because each pair appears exactly twice in the ordered list.

Choosing r objects to include is equivalent to choosing n−r objects to leave out, so the two binomial coefficients are equal. This gives a useful check. Selecting four committee members from six should give the same count as selecting the two people excluded. For small values, listing the excluded pairs may be easier than expanding factorials. Cancel factors before multiplying so the arithmetic stays manageable on paper.

Convert counts into a probability only when justified

For equally likely outcomes, probability is the number of favourable outcomes divided by the total number. Use the same definition of outcome in both counts. If the denominator counts ordered draws but the numerator counts unordered groups, the ratio is inconsistent. You can use either representation when it is valid, but the ordering convention must match. With non-uniform outcomes, counting alone is insufficient; probabilities must reflect the different chances.

For sampling without replacement, successive event probabilities change. The chance of drawing two red objects can be calculated by multiplying the initial red fraction by the red fraction remaining after a red draw. Alternatively, count red pairs and divide by all pairs. Agreement between these methods is a strong check on both the arithmetic and the outcome model. Finally ensure that the probability lies between zero and one and that every counted favourable outcome is also included in the denominator.