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Decide whether order matters, enforce restrictions and correct duplicate counts.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
Count a finite set of outcomes exactly once using stages, cases, permutations, combinations or complements.
Use multiplication for successive decisions when each partial choice has a known number of possible continuations. Use addition for alternative cases that are disjoint. If a case can occur in both lists, addition alone counts the overlap twice. A factorial n! multiplies positive integers from one through n; zero factorial is one. Factorials compactly express many arrangement counts, but writing the first few decision counts often explains the problem more clearly than immediately substituting into a formula.
| Term | What it means |
|---|---|
| Permutation | An ordered selection or arrangement. |
| Combination | A selection in which order does not distinguish outcomes. |
| Replacement | Returning a selected object so it can be selected again. |
| Disjoint cases | Cases with no outcomes in common. |
Counting problems begin with an identity question: when are two results considered different? A chair-secretary pair changes when the same two people exchange roles. An ordinary two-person committee does not. A code can begin with zero if the rules allow it, whereas a three-digit positive integer cannot. The arithmetic depends on these definitions, so choosing a formula before defining the object invites overcounting.
A good counting method accounts for every permitted object exactly once. You can build an object in stages, split the set into disjoint cases, or count a larger set and subtract forbidden objects. Whichever route you choose, explain why no valid result is omitted and no result is counted twice. A small version that can be listed by hand is often the fastest check of a proposed general method.
Another way: steps
Define distinct outcomes; identify order and replacement rules; choose stages or disjoint cases; correct overcounting; check a small example.
If a meal chooses one of three soups, one of five mains and one of two drinks, there are 3 times 5 times 2 = 30 meals. The factors describe successive components of one result. The numbers of choices need not be identical across stages, but the simple product assumes each earlier choice leaves the stated number of later choices. If a particular soup removes a drink option, split into cases or count its continuations separately. The multiplication principle does not require probabilistic independence; it requires correct counts of the available continuations.
A four-character code drawn from six symbols with repetition allowed has six choices at every position, giving 6 to the fourth power. Without repetition, the choices decrease: 6 times 5 times 4 times 3. Both are ordered counts because changing character positions changes the code. For a selected group, order may not matter even when replacement is forbidden. Ask these questions separately: can an object be used again, and does swapping positions create a different outcome? Neither answer automatically determines the other.
Choosing three people in order from seven gives 7 times 6 times 5 sequences. Each unordered committee appears 3! = 6 times, once for each ordering of its three distinct members. Dividing by six gives thirty-five committees. This division works because every committee has exactly the same number of ordered representations. If different objects are counted different numbers of times, one universal divisor will not repair the count. State what each counted sequence represents before deciding whether and by how much to divide.
Five distinct letters have 5! arrangements. If two letters are identical, swapping those two labels does not create a new visible arrangement, so divide by 2!. With several groups of repeated letters, divide by each group's factorial. For the letters in MOMENT, there are six positions and two identical Ms, giving 6!/2! = 360 arrangements. This method assumes the only indistinguishability comes from those repeated symbols. Do not divide merely because two different people share a profession; their identities still distinguish a team unless the question says otherwise.
For a four-digit integer formed from digits zero through five without repetition, the first position has five nonzero choices. After choosing it, five digits remain for the second position, including zero, then four and three remain. Choosing the leading digit first makes the special restriction explicit. An alternative is to count all four-position strings and subtract those beginning with zero. Both methods should agree. Treating zero as forbidden in every position undercounts, while allowing it in the first position counts strings that are not four-digit integers.
If a set has n distinct members, each member is either included or excluded, giving 2 to power n subsets. Subtract one to exclude the empty set. If two specified people must sit next to one another in a row, treat them as a block while arranging the block with the others, then account for the two internal orders. If they must not sit together, subtract adjacent arrangements from all arrangements. Be cautious with multiple adjacency conditions: overlapping blocks can require separate cases, and a shortcut that works for one pair need not extend unchanged to several pairs.
For two sets, the union count equals the first count plus the second count minus their intersection. For three sets, add the three individual counts, subtract each pairwise intersection, and add back the triple intersection once. The triple group was initially counted three times and then removed three times, so it needs one restoration. Distinguish at least one category, exactly one category and all categories; they describe different regions. Fill the most specific overlap regions first when constructing a diagram, then compute the nonoverlapping remainder of each set. If a count includes members of all three groups, subtract that group from each pairwise region before labeling a diagram region as exactly two memberships. Confusing an inclusive intersection with an exclusive region can double-count people even when the final addition is arithmetically correct.
A project panel must include two designers and one engineer. First choose the unordered designer pair, then choose the engineer. Multiplying these two counts creates every eligible panel exactly once. Treating all three roles as ordered positions would count the two designers in both orders even though the panel has no first and second designer role. Assigning a chair later would create a new counting stage and a different question.
If a panel is then chosen uniformly from the eligible panels, a probability can be calculated by dividing favorable eligible panels by all eligible panels. Uniform selection from all people followed by rejecting invalid panels can produce the same model only if the sampling procedure gives each eligible panel equal weight; do not assume that from vague wording.
Check your count with a smaller situation, such as three designers and two engineers, where you can list all pairs. A small exhaustive list reveals missing panels and duplicates more clearly than an unexplained factorial expression. Use the list to test the method, then return to the original numbers.
Distinct roles make order matter. Repeated use and ordering are separate properties. A leading zero can be allowed in a code but not in a multi-digit integer. Divide by a factorial only when every object is counted that many times. Adding overlapping cases requires correcting the overlap. A complement must exclude exactly the forbidden outcomes, no more and no less.
Define the outcome.
Choose a captain and deputy from five people, with different people in the roles.
The roles distinguish an ordered pair.
Choose the first role.
There are five possible captains.
Every person is eligible.
Choose the second role.
Four people remain for deputy.
The captain cannot hold both roles.
Multiply and check order.
5 times 4 = 20 appointments.
Exchanging captain and deputy produces a different appointment, so do not divide by two.
Define what stays the same.
Choose a three-person committee from six people.
Reordering the same three members does not create a new committee.
Count ordered selections first.
6 times 5 times 4 = 120.
No person can appear twice.
Count representations of one committee.
Three distinct members have 3! = 6 orders.
Every committee appears the same six times.
Remove the uniform overcount.
120/6 = 20 committees.
Division retains one representative for each group.
Check a complementary count.
Choosing the three excluded people also gives twenty.
Each committee corresponds to exactly one excluded group.
Identify the allowed symbols.
Use 0, 1, 2, 3 and 4 to form a three-digit positive integer without repetition.
A leading zero is forbidden.
Fill the restricted first position.
Four nonzero digits are possible.
Choosing this position first isolates the exceptional rule.
Fill the second position.
Four digits remain, including zero.
Zero is allowed after the leading position.
Fill the third position.
Three unused digits remain.
No repetition is permitted.
Multiply the stage counts.
4 times 4 times 3 = 48.
Each sequence defines one valid integer.
Verify by subtracting invalid strings.
5 times 4 times 3 minus 4 times 3 = 48.
The subtracted strings are exactly those beginning with zero.
Separate the required categories.
Choose two researchers from five and one analyst from three.
The categories contain distinct eligible people.
Count each selection correctly.
The researcher pair has (5 times 4)/2 = 10 possibilities.
Determine the number of analyst choices and combine.
Multiply the category choices.
A menu offers 4 starters and 6 main dishes. How many meals consisting of one starter and one main dish are possible?
How many different orders are possible for 4 distinct books on a shelf?
How many three-letter codes can be formed from A, B, C, D and E if repetition is allowed?
How many three-letter codes can be formed from A, B, C, D and E without repetition?
A committee of 3 people is chosen from 8 people. How many committees are possible?
From 7 people, one chair and one secretary are appointed, with no person holding both roles. How many appointments are possible?
How many distinct arrangements of the letters in LEVEL are possible?
How many four-digit positive integers can be formed using only 0, 1, 2 and 3 if repetition is allowed?
A team of 2 is selected from 5 scientists and 4 artists, with one member from each group. How many teams are possible?
How many subsets of a set of 6 distinct elements contain at least one element?
A project has 6 eligible designers and 5 eligible engineers. A panel must contain 2 designers and 1 engineer. How many panels are possible?
Answer three reserved questions without hints. These cases are separate from lesson practice.
A committee of 2 is selected from 9 people. How many committees are possible?
How many three-digit positive integers use only digits 0, 2, 4 and 6 without repetition?
How many distinct arrangements of the letters in BANANA are possible?
Tomorrow, explain why a committee count divides an ordered selection by a factorial. Then contrast a code that permits repetition with a number whose first digit cannot be zero, and test your method on a small list.
17. Complete a mixed committee count, step 3
10 times 3 = 30 committees.
Only the two-researcher selection needed a correction for internal order.