Counting outcomes for probability constantly asks one question first: does the order the items come out in matter? Getting that one question right decides whether a problem needs a permutation or a combination. Everything else is the same formula, applied correctly or not.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Discrete Mathematics I — Logic, Proof, and Structure
A 5-card poker hand is dealt, then sorted by rank afterward for display. Does the later sorting step make the deal itself a permutation?
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Why
A processing step performed on an already-counted group never retroactively adds order to that group's own count.
From Discrete Mathematics I — Logic, Proof, and Structure
Assigning a starting pitcher and a closer from the same 12-player roster is a...
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Why
Starting pitcher and closer are distinct roles -- swapping the two players changes who does what, a permutation.
From Discrete Mathematics I — Logic, Proof, and Structure
From a 9-person pool, how many more ways are there to rank the top 3 (1st, 2nd, 3rd) than to simply choose an unranked top-3 group?
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Why
P(9,3) = 504 and C(9,3) = 84. 504 minus 84 equals 420 more ordered outcomes.
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Formula
Permutations of n items taken r at a time
also known as: P(n,r) or nPr
Recalled from College Algebra · Counting, permutations, combinations, and probability — first taught there.
The number of ways to arrange r items chosen from n distinct items, where order matters, is P(n,r) = \frac{n!}{(n-r)!}.
This is just the counting principle, run r times in a row. The first position in the arrangement can be filled by any of the n items. The second position can be filled by any of the n - 1 items that remain (one has been used). The third by any of the n-2 remaining, and so on, down to the rth position, which has n - (r-1) = n - r + 1 choices left. Multiply across all r positions: P(n,r) = n \cdot (n-1) \cdot (n-2) \cdots (n - r + 1). That product has exactly r factors, starting at n and counting down. To see it equals \frac{n!}{(n-r)!}, write n! in full and notice the tail end cancels: n! = \bigl[n(n-1)\cdots(n-r+1)\bigr] \cdot (n-r)!, so dividing both sides by (n-r)! leaves exactly the product above.
Formula
Combinations of n items taken r at a time
also known as: C(n,r), nCr, or n choose r
Recalled from College Algebra · Counting, permutations, combinations, and probability — first taught there.
The number of ways to choose r items from n distinct items, where order does not matter, is C(n,r) = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}.
Start from P(n,r), which counts every arrangement of r chosen items. Each unordered group of r items corresponds to many of those arrangements — specifically, every one of the r! ways to reorder that same group is a different permutation but the exact same combination. So P(n,r) overcounts each combination by a factor of r!, and dividing it out gives the true count: C(n,r) = \frac{P(n,r)}{r!} = \frac{n!/(n-r)!}{r!} = \frac{n!}{r!\,(n-r)!}.
Picking 3 runners for gold, silver, and bronze out of 8 racers is a permutation problem. Swapping who gets gold and who gets silver changes the result completely, so order matters. Picking 3 runners to send to a different meet, with no medals attached, is a combination problem instead. The same 3 names in a different order are still the same trio. Same 8 racers, same "choose 3," but the counts come out wildly different, because one job cares who's first and the other doesn't.
The number of ways to arrange r items chosen from n distinct items, where order matters, is P(n,r) = \dfrac{n!}{(n-r)!}. The number of ways to choose r items from n, where order does not matter, is C(n,r) = \binom{n}{r} = \dfrac{n!}{r!\,(n-r)!} = \dfrac{P(n,r)}{r!}. Every combination corresponds to exactly r! permutations: the same r items rearranged into every possible order. That r! is exactly the factor C(n,r) divides out of P(n,r).
Order matters, or doesn't
Click 3 names, then toggle 'order matters.' Watch the count switch between a permutation and a combination.
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Worked example
Permutation: awarding medals
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8 racers finish a race. In how many ways can gold, silver, and bronze be awarded?
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Order matters (gold \neq silver), so this is P(8,3).
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3\boxed{P(8,3) = \frac{8!}{5!} = 8 \times 7 \times 6 = 336}
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Worked example
Combination: choosing a delegation
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From the same 8 racers, a delegation of 3 (no roles, no order) is sent to a meeting. How many delegations are possible?
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Order does not matter here. Divide the permutation count by 3! to remove the overcounted orderings of each group.
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3\boxed{C(8,3) = \frac{336}{3!} = \frac{336}{6} = 56}
Worked example
Spotting which one a word problem needs
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A 5-card hand deals 2 cards face up, in a specific left-right order, to one player. It also deals 2 cards face down, with no order recorded, to a discard pile, both from the same 5-card set with no overlap. Which formula does each group need?
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The face-up cards have a recorded position (left vs. right), so that group is a permutation. The face-down discard pile has no recorded order, so that group is a combination.
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3\boxed{\text{Face-up: } P(5,2)=20, \qquad \text{Discard: } C(3,2)=3 \text{ (from the remaining 3 cards)}}
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Where this goes next
Chapter 2 (Counting for probability) and Chapter 5 (Discrete distributions)
The binomial distribution's entire formula is built from \binom{n}{k}. That factor is literally a combination count, standing for "how many ways can k successes land among n trials." Getting comfortable with permutations and combinations now is what makes that formula legible later instead of a symbol string to memorize.
Check your understanding (counts toward your progress)
In how many ways can a 3-person subcommittee (no roles) be chosen from the same 9-member club?
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Why
Order doesn't matter for an unranked subcommittee: C(9,3) = 9!/(3!6!) = 84.
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In how many ways can an unranked group of 2 be chosen from 9 distinct people?
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Why
Order doesn't matter, so this is C(9,2) = \binom{9}{2} = 36.
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A 5-card poker hand is dealt from a standard 52-card deck. How many different 5-card hands are possible?
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Why
A hand has no internal order, and cards aren't replaced: C(52,5) = 2,598,960.
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Why
A dealt hand has no ranking among its 5 cards, and a card can't be dealt twice — that's a combination, without replacement.
In how many ways can 3 distinct roles be filled, in order, by 3 of 14 candidates?
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Why
Order matters here, so this is P(14,3) = \frac{14!}{(14-3)!} = 2184.
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Check your understanding (counts toward your progress)
A fruit basket is made by selecting 3 different fruits from 6 available types (order doesn't matter). How many different baskets are possible?
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Why
No order for a basket's contents: C(6,3) = 6!/(3!3!) = 20.
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A 5-person team is chosen from 12 people, but one specific person (Sam) must be on the team. How many different teams are possible?
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Why
With Sam's seat fixed, only the other 4 team spots need filling from the remaining 11 people: C(11,4) = 330.
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Why
Since Sam's spot is already decided, only the remaining team members need choosing from everyone else.
A race has 8 runners. In how many ways can gold, silver, and bronze medals be awarded (one runner per medal, no ties)?
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Why
Three distinct medal positions from 8 runners: P(8,3) = 8×7×6 = 336.
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Why
Order matters here — gold, silver, and bronze are three distinct positions.
In how many ways can 4 distinct trophies be lined up on a shelf?
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Why
Arranging all 4 distinct trophies in a row is 4! = 24.
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Practice: Arranging things in order (5 problems · doesn’t affect your progress)
A relay race assigns 4 runners to positions 1st, 2nd, 3rd, and 4th leg. Does determining the number of possible leg assignments use permutations or combinations?
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The four legs are distinct ordered positions (1st through 4th), not one unordered group.
This counts arrangements into distinct positions, which is a permutation, not a sum-rule situation.
Permutations and combinations give different counts whenever order matters, and it does here.
Why
Each of the 4 legs is a distinct, ordered position, so assigning runners to them is a permutation, not a combination.
Permutations — each runner gets a distinct, ordered position
How many ways can 5 distinct paintings be hung in a row on a wall?
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Why
Arranging all 5 distinct paintings in order is 5! = 120.
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In how many ways can a 3-digit code be formed using the digits 1-9 (each digit used at most once)?
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Why
Order matters and no repeats are allowed: P(9,3) = 9×8×7 = 504.
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Why
No repeats means each position has one fewer available digit than the last.
A club has 7 members. In how many ways can a president and a vice-president be chosen (two DIFFERENT people, distinct roles)?
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Why
Order matters here since the roles are distinct: P(7,2) = 7×6 = 42.
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Why
Two distinct roles from 7 people is a permutation, P(7,2).
6 people, including a married couple, sit in a row of 6 chairs. The married couple insists on sitting next to each other. How many seating arrangements satisfy this?
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Why
Treat the couple as one combined block: now there are 5 units to arrange (5! = 120 ways), and the couple can be ordered 2 ways within their block. Multiply: 120 × 2 = 240.
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Glue the couple together into one 'block' first, then arrange the block along with the other 4 people.
Don't forget the couple can sit in 2 different orders within their own block.
Practice: Choosing unordered groups (5 problems · doesn’t affect your progress)
A pizza shop lets you pick 3 toppings from a list of 8 (no order, no repeats). Does counting the possible topping sets use permutations or combinations?
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The finished pizza doesn't record what order the toppings were added — only which 3 ended up on it.
The product rule counts independent sequential stages, not a single unordered selection.
The sum rule is for disjoint alternatives (this OR that), not selecting a group all at once.
Why
A topping set has no order — only which 3 toppings appear — so this is the unordered case a combination counts.
Combinations — a topping set has no internal order
A hand of 2 cards is dealt from a standard 52-card deck. How many different 2-card hands are possible?
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Why
No order for a hand's two cards: C(52,2) = 1326.
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A study group of 5 students is chosen from a class of 15. How many different groups are possible?
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Why
C(15,5) = 15!/(5!10!) = 3003.
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Why
Order doesn't matter for a study group — use C(n,r), not P(n,r).
A trivia team needs exactly 2 members from a pool of 6 science majors and exactly 2 members from a pool of 5 history majors. How many different 4-person teams are possible?
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Why
Choose 2 of 6 science majors (C(6,2)=15) and, independently, 2 of 5 history majors (C(5,2)=10). Multiply the two independent stages: 15×10=150.
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Choose from each pool separately (two combinations), then multiply since both choices happen together.
A jar has 8 distinct marbles, split into two unlabeled piles of 4 marbles each (there's no way to tell 'pile 1' from 'pile 2' — swapping which pile is which doesn't create a new split). How many distinct splits are possible?
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Why
Choosing 4 marbles for a labeled 'first pile' gives C(8,4) = 70 ways, but each real split shows up twice this way — once for each pile you could call 'first.' Since the piles are actually unlabeled, divide by 2: 70/2 = 35.
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First count C(8,4) as if the piles WERE labeled, then account for the fact that each actual split got counted twice (once with each pile called 'first').
Dividing by 2 removes exactly the double-count from swapping the two piles' labels.
Practice: Permutation or combination? Deciding first (4 problems · doesn’t affect your progress)
A teacher selects 3 students from a class of 15 to form an unranked study group. Does order matter?
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Being SELECTED in some sequence doesn't mean the final group is ordered — what matters is whether swapping two chosen students changes the outcome.
Whether order matters is a property of the TASK's structure, not of which specific names end up chosen.
This is exactly the combination formula's use case — a group with no ranking.
Why
Swapping which student is named first, second, or third among the 3 chosen doesn't change who's in the study group — the outcome is unordered, so this is a combination.
No — a combination, since the 3 students form one unranked group either way
8 runners finish a race. In how many ways can gold, silver, and bronze specifically be awarded to 3 of them?
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Swapping which runner gets gold versus bronze is a genuinely DIFFERENT outcome — the medal ranking makes order matter.
Combination would apply only if the 3 runners were an undifferentiated group with no distinct roles like gold/silver/bronze.
Repeats aren't possible — the same runner can't win two medals.
Why
Gold, silver, and bronze are three distinct roles, so which runner lands in which specific role matters — a permutation, P(8,3)=336.
P(8,3) = 336 — a permutation, since swapping medal positions changes who gets which medal
Explain why dividing P(n,r) by r! gives C(n,r), instead of dividing by some other quantity.
AI grading isn't configured on this deployment.
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Why
Every unordered group of r items corresponds to exactly r! different ordered arrangements of those same r items — dividing the ordered count P(n,r) by that r! removes exactly the overcounting caused by treating each group's own internal orderings as distinct outcomes.
A 10-person club selects an unordered 4-person committee, which then separately elects one of its own 4 members as chair. Which combination of tools counts the total number of outcomes?
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Committee selection alone (P(10,4)) would overcount — the committee itself has no internal order, only the later chair choice singles someone out.
This only counts the committee itself and ignores the second, separate stage of choosing a chair from within it.
The two stages happen together (form committee AND THEN pick a chair), so they multiply — adding them would treat the stages as alternatives instead.
Why
Forming the committee is a combination (C(10,4), since it's unordered), and THEN choosing one of the 4 committee members as chair is a separate stage (4 ways) — the two stages multiply: C(10,4)×4.
C(10,4) for the committee, then multiply by 4 for choosing the chair from within it