Choosing a Counting Plan
A mixed counting problem is not mainly a test of memorizing formulas. It is a test of matching the count to the outcomes the situation treats as distinct. Changing the order of letters usually creates a different code; changing the order in which committee members are listed does not create a different committee. In a random draw, the physical objects selected matter, but their order may not.
A reliable solution begins by describing the sample space: what counts as one outcome, and how outcomes are generated. Then count all possible outcomes and the favorable outcomes using the same convention. When the favorable outcomes meet a requirement in different ways, separate them into cases that do not overlap. This prevents a favorable outcome from being counted more than once.
The counting notation used here includes \(P(n,r)=n!/(n-r)!\) for ordered selections of \(r\) distinct objects from \(n\), and \(\binom{n}{r}=n!/[r!(n-r)!]\) for unordered selections. These formulas answer different questions: use a permutation count when order distinguishes outcomes and a combination count when it does not. If outcomes are not equally likely, the ratio of counts alone is not a probability; the examples below specify uniform random generation or equally likely physical draws.
Worked Example: A Code With a Letter Requirement
Worked Example: A Code With a Letter Requirement
A code is made by choosing three distinct letters in order from A, B, C, D, E, followed by two distinct digits in order from 0, 1, 2, 3, 4. A code is selected uniformly from all codes of this format. What is the probability that its three-letter part contains at least one vowel? In this set of letters, A and E are vowels.
Define one outcome. A code is the complete ordered string. Thus ABC04 and BAC04 are different outcomes. The three letters must be distinct, and the two digits must be distinct; a letter may have the same symbol as a digit only if it belongs to a different part, which cannot occur here.
Count the whole sample space. There are \(P(5,3)\) possible ordered letter parts and \(P(5,2)\) possible ordered digit parts. Each letter part can be paired with each digit part, so
Count favorable letter parts in disjoint cases. The number of vowels can be exactly one or exactly two; three vowels is impossible because there are only two available vowels. If there is exactly one vowel, choose its position in 3 ways, choose A or E in 2 ways, and fill the other two positions with an ordered selection of two of the three consonants:
If there are exactly two vowels, choose their two positions in \(\binom{3}{2}=3\) ways, arrange A and E in those positions in \(P(2,2)=2\) ways, and choose one of the three consonants for the remaining position. This gives
The cases are disjoint because a letter part cannot have both exactly one and exactly two vowels. Therefore there are \(36+18=54\) favorable letter parts. Each can be paired with any of the 20 digit parts, giving \(54\cdot20=1080\) favorable codes. Uniform selection from the 1200 valid codes gives
So the probability is 0.900, or 90.0%. The code format and uniform selection matter: this calculation would not describe a generator that favored some valid strings over others.
Committees: Count Membership, Not Arrangements
A committee is an unordered selection. For example, the same five people form the same committee regardless of the order in which their names are written. A combination count is therefore appropriate. When a committee must satisfy requirements involving different groups, a useful strategy is to classify each committee by how many members it has from one group. The count in one case is then multiplied across groups using combinations.
Worked Example: A Committee With Minimum Representation
A committee of five is selected from four women and five men. Each five-person committee is equally likely. Find the probability that the committee contains at least two women and at least two men.
Count all committees. There are nine people, and order does not matter, so the sample space has
committees. Every five-person subset is equally likely under the stated selection rule.
Partition the favorable committees. Let \(k\) be the number of women. The requirement of at least two women gives \(k\geq2\). Since the committee has five members, the requirement of at least two men gives \(5-k\geq2\), or \(k\leq3\). Thus the only possible cases are \(k=2\) and \(k=3\). These cases are disjoint: one committee cannot contain both two and three women.
With two women, choose two of the four women and three of the five men. With three women, choose three women and two men. The total favorable count is
Each product counts a unique pairing of a selected group of women with a selected group of men. No committee is duplicated between the cases because its number of women determines which case it belongs to. Therefore,
The probability is about 0.794, or 79.4%. A count such as \(P(9,5)\) would be inappropriate for the sample space: it treats different orderings of the same committee as different outcomes, even though the selection process defines a committee only by its members.
Unordered Draws: Confirm What Counts as an Outcome
A draw may happen in sequence, but the question may concern only which objects were selected. If four distinct objects are drawn without replacement and order is ignored, each possible four-object subset is one outcome. When all ordered draw sequences are equally likely, every such subset is equally likely too: a fixed subset of four objects can appear in \(4!\) different orders, and this number is the same for every subset. Thus the total count can be made with combinations, provided the event is also counted in terms of subsets.
Worked Example: A Draw With a Color Requirement
A container holds five red, four blue, and three green tokens. Four tokens are drawn without replacement, and every four-token subset is equally likely. Find the probability of drawing exactly two red tokens and at least one blue token.
Treat the 12 tokens as distinct physical objects, even when they have the same color. The total number of unordered four-token draws is
To count favorable draws, first choose exactly two of the five red tokens. There are then two additional tokens to choose. For those two, the requirement “at least one blue” allows two disjoint cases: exactly one blue and one green, or exactly two blue. There cannot be any red among these remaining tokens because the event requires exactly two red in total.
In the first case, choose one of four blue tokens and one of three green tokens, giving \(\binom{4}{1}\binom{3}{1}=12\) possibilities. In the second case, choose two of the four blue tokens, giving \(\binom{4}{2}=6\) possibilities. The cases cannot overlap because the first contains one blue token and the second contains two. Therefore,
The favorable count is a count of distinct subsets: first a pair of red tokens, then one of the allowed two-token selections. The total and favorable counts use the same unordered-outcome convention. Hence
The probability is about 0.364, or 36.4%. Counting ordered sequences in the numerator but subsets in the denominator would mix two different sample spaces and produce an invalid ratio.
A Short Justification for Adding Case Counts
The case method works because disjoint cases divide an event into separate pieces. Suppose a favorable event \(A\) is the union of cases \(A_1,\ldots,A_m\), and no outcome belongs to two different cases. Then each outcome in \(A\) appears in exactly one case count. Consequently,
If the sample space contains \(N\) equally likely outcomes, then \(P(A)=|A|/N\). Substituting the disjoint-case count gives
This is not a special shortcut that makes overlap harmless. If a committee satisfies two descriptions that overlap, adding both case counts counts that committee twice. One must either redefine the cases so they are disjoint or account for the overlap separately. In the examples, the case labels—number of vowels, number of women, or number of blue tokens—assign every favorable outcome to exactly one case.
Checks That Prevent Counting Errors
Before calculating, write down what a single outcome is. This one choice determines whether order matters, which formula fits, and whether all outcomes can reasonably be treated as equally likely. Then check the event count independently: identify the permitted case values, verify that no case is impossible or omitted, and ask whether two cases could describe the same outcome.
- Codes: Positions generally matter, so a different ordering is a different code. Check whether repetition is allowed and whether each position has the same available symbols.
- Committees: Membership matters, not the order of names. Count possible group sizes that satisfy every requirement before evaluating the cases.
- Draws: If order is ignored, count subsets in both numerator and denominator. Treat same-colored objects as distinct when they are physically different tokens.
- Probabilities from counts: Use favorable count divided by total count only when the sample space outcomes are equally likely.
Common mistake: A student counts committees with two women and three men, then adds the count for three women and two men, but says the cases might overlap because both describe a five-person committee. That reasoning confuses the total committee size with the case-defining feature. The cases cannot overlap: a given committee has a fixed number of women. Conversely, counting “at least one blue” once for each blue token selected would overlap, because a draw with two blue tokens would be counted twice. The correction is to define cases by an exact count—such as exactly one blue or exactly two blue—or to count the event by a method that handles overlap explicitly.
Check Your Understanding
For each question, identify the outcome convention and explain how you would organize the count.
- A five-character code uses distinct symbols chosen from eight available symbols. Does changing the order create a different code, and which counting notation matches the total?
- A four-person committee is selected from six researchers and five clinicians. How can you count committees with at least one member from each group using disjoint cases?
- A bag has three yellow and five purple tokens. Three tokens are drawn without replacement, and order is ignored. What is the total number of possible outcomes if every token subset is equally likely?
- In a draw problem, an event is split into “at least one green” and “at least two green.” Are these cases disjoint? Explain the double-counting issue.
- Why is it not valid to divide an ordered favorable-sequence count by an unordered total-subset count?