2 Counting Techniques

Learn how to count finite outcomes systematically and use addition, multiplication, permutations, combinations, and equally likely probability models to solve counting problems.

The Counting Framework

Counting techniques replace exhaustive listing with structured methods. The central task is to decide what constitutes one distinct outcome and then choose a counting rule that matches the structure of the process.

A useful first question is whether the problem describes alternatives, successive stages, ordered selections, or unordered selections.

Use the for mutually exclusive alternatives.
Use the for successive stages.
Use a when order matters.
Use a when order does not matter.

For an equally likely finite experiment, counting connects directly to probability:

P(A)=\frac{|A|}{|S|}.

Here, $A$ is the event of interest and $S$ is the .

Takeaway: Define the outcome before selecting a formula; the correct method follows from the structure of the outcome.

Alternatives and Overlap

The applies when a choice can be made in one of several mutually exclusive ways. If one alternative has $m$ possibilities and another has $n$ possibilities, the total is

m+n.

For example, choosing one elective from $4$ art courses or $3$ music courses gives

4+3=7

choices, assuming no course belongs to both groups.

When alternatives overlap, adding them directly counts the shared outcomes twice. Correct the count with

|A\cup B|=|A|+|B|-|A\cap B|.

Thus, if $18$ students study French, $12$ study Spanish, and $5$ study both, the number studying at least one language is $18+12-5=25$ .

Takeaway: Add mutually exclusive cases, and subtract overlaps when cases are not disjoint.

Successive Stages

The applies when a procedure has successive stages. If stage $1$ has $n_1$ choices, stage $2$ has $n_2$ choices, and so on, the total is

n_1n_2\cdots n_k.

For a password consisting of two letters followed by three digits, with repetition allowed, the count is

26\cdot26\cdot10\cdot10\cdot10=26^2\cdot10^3=676{,}000.

If repetition is not allowed, the available choices decrease as symbols are used. The count becomes

26\cdot25\cdot10\cdot9\cdot8.

The also combines independent selections within a favorable case. For example, selecting exactly two aces and three non-aces uses

\binom{4}{2}\binom{48}{3}.

Takeaway: Break a process into stages, count the choices at each stage, and multiply them.

Ordered Arrangements

A records the number of ways to arrange distinct objects. For a positive integer $n$ ,

n!=n(n-1)(n-2)\cdots2\cdot1,

and $0!=1$ . For instance,

5!=5\cdot4\cdot3\cdot2\cdot1=120.

A counts an ordered selection. Selecting and assigning $r$ objects from $n$ distinct objects gives

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

Six people competing for president, vice president, and treasurer produce

P(6,3)=\frac{6!}{3!}=6\cdot5\cdot4=120

assignments because the offices are distinct.

When all $n$ objects are arranged, the formula becomes $P(n,n)=n!$ . If objects repeat, divide by the of each repetition count. For the letters in LEVEL, there are two Ls and two Es, so the number of distinct arrangements is

\frac{5!}{2!2!}=30.

Takeaway: Permutations distinguish arrangements that differ in position, and repeated objects require removing duplicate arrangements.

Unordered Selections

A counts a selection in which order does not matter:

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

Choosing a three-person committee from ten people gives

\binom{10}{3}=\frac{10!}{3!7!}=120.

The same three people form one committee regardless of the order in which they are listed.

A practical test is to ask whether switching two selected objects creates a new outcome. If it does, use a ; if it does not, use a . Their relationship is

P(n,r)=\binom{n}{r}r!.

This works because one first chooses the $r$ objects and then arranges them in $r!$ orders.

Takeaway: Use combinations for groups and permutations for roles, positions, or rankings.

Counting and Finite Probability

For equally likely finite outcomes, probability is the ratio of favorable outcomes to total outcomes:

P(A)=\frac{|A|}{|S|}.

The numerator and denominator must describe the same kind of outcome. A five-card hand is an unordered selection, so the total number of hands is

\binom{52}{5}=2{,}598{,}960.

To count hands containing exactly two aces, choose two of the four aces and three of the forty-eight non-aces:

\binom{4}{2}\binom{48}{3}.

Therefore,

P(\text{exactly two aces})=\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}.

For sampling without replacement, a box with $8$ phones, including $3$ defective phones, has $5$ good phones. The probability that two selected phones are both good is

\frac{\binom{5}{2}}{\binom{8}{2}}.

Takeaway: Count all outcomes and favorable outcomes using the same assumptions about order and repetition, then divide.

A Problem-Solving Checklist

A reliable solution process keeps the model, formula, and calculation aligned.

Describe one outcome clearly.
Decide whether order matters.
Separate successive stages from alternative cases.
Check whether repetition is allowed.
Count the total outcomes.
Count the favorable outcomes.
For equally likely outcomes, compute $P(A)=\frac{|A|}{|S|}$ .
Check that the probability is between $0$ and $1$ , inclusive.

Common errors include adding when stages should be multiplied, using a for a committee, ignoring restrictions on repetition, double-counting overlapping cases, and choosing a denominator that represents a different type of outcome than the numerator.

The deciding question remains simple: Does order matter? If yes, begin with permutations; if no, begin with combinations. Then use addition or multiplication to reflect how the cases or stages fit together.