GMAT Focus Edition · Advanced Quantitative Aptitude
Permutations, Combinations and Probability
Counting principles, arrangements, selections and basic probability.
Six concepts cover counting and probability for CAT — arrangements, selections, circular seating, repeated letters, equally likely outcomes and restriction tricks. Each card states the rule, shows where CAT wording hides it, and ends in a ledger you can copy.
- GMAT Focus Edition
- Hard level
- 6 concepts
- 21 practice questions
1Permutations versus combinations
A permutation counts arrangements: ^nP_r = \frac{n!}{(n-r)!} fills r labelled positions from n distinct objects. A combination counts selections: ^nC_r = \frac{n!}{r!(n-r)!} picks an r-subset and ignores who sits where inside it. The link ^nP_r = r!\,^nC_r is exactly the r! ways to reorder the same subset.
CAT wording hides the symbol. "Gold and silver medals", "president and vice-president", "a password with distinct digits in order" are permutations because the positions carry different names. "A committee of two", "choose any two cards", "pick a pair of friends" are combinations because swapping the two names does not create a new outcome.
Figure. Permutations keep order (AB \neq BA); combinations collapse them to one set \{A,B\}. Ask whether rearranging the selection creates a new outcome before picking P or C.
How to choose
- Name the positionsIf the slots are labelled differently — 1st, 2nd, president, digit-1 — order counts.
- Swap testIf exchanging the two chosen objects gives the same outcome, treat it as a selection.
- Pick the formulaArrangement \rightarrow ^nP_r; selection \rightarrow ^nC_r. Probability without replacement usually uses combinations once the sample space is defined as unordered draws.
| Wording cue | Symbol | Order |
|---|---|---|
| Medals, ranked seats, passwords with distinct digits | ^nP_r | Matters |
| Committee, hand of cards, unordered pair | ^nC_r | Does not |
| Same r objects, all orders | r!\,^nC_r | Equals ^nP_r |
Medals and a committee from five runners
Five distinct runners. (a) In how many ways can gold and silver medals be awarded? (b) In how many ways can a two-person committee be chosen?
- Medals: positions labelled, so ^5P_2 = 5 \times 420
- Committee: order irrelevant, so ^5C_2 = \frac{5!}{2!\,3!}10
- Check: 20 \div 10 = 2 = 2!^nP_r = r!\,^nC_r
Pro tip. Before multiplying, ask whether swapping the two chosen objects creates a new outcome. If not, divide by r! — or start with ^nC_r and stop. Probability questions that draw without replacement almost always count unordered outcomes in the sample space, so combinations beat permutations there even when the story mentions "drawing two balls".
From 6 students, a president and a vice-president are to be chosen. The count is
- ^6C_2 = 15
- ^6P_2 = 30
- 6! = 720
The two offices are differently labelled, so order counts: ^6P_2 = 6 \times 5 = 30. Treating it as ^6C_2 = 15 counts unordered pairs and ignores which student holds which office. 6! arranges all six students in a line, which is not the two-office selection.
2Fundamental counting and linear arrangements
If one task can be done in m ways and an independent second task in n ways, together they can be done in m \times n ways. Independence is the keyword: the choice for the first task does not shrink the options for the second. Chain three or more stages by multiplying their counts.
When all n objects are distinct and every position in a row is filled, the count is n! = n \times (n-1) \times \cdots \times 1. That is the special case ^nP_n — every object gets a seat and order matters because the seats are labelled left to right.
Figure. Independent successive choices multiply: 3 shirts and 2 pants give 3\times 2=6 complete outfits. Draw the tree when a CAT stem hides a dependency.
How it works
- List the stagesBreak the task into sequential choices — shirt then pants, digit-1 then digit-2, and so on.
- Count each stageHow many options remain at that stage? If earlier choices remove objects, the count drops; if not, it stays fixed.
- MultiplyIndependent stages multiply. A full row of n distinct objects is one stage per seat, giving n!.
Outfits and a bookshelf
A wardrobe has 4 shirts and 3 trousers, all distinct. (a) How many shirt–trouser outfits? (b) In how many orders can 5 different books be placed on a shelf?
- Outfits: independent choices, 4 \times 312
- Books: all five distinct, 5! = 5 \times 4 \times 3 \times 2 \times 1120
- Check: 5! = {}^5P_5120
Pro tip. Do not add counts from independent stages — that is the error when a question lists two separate menus and asks for a pair. Only when the choices are mutually exclusive alternatives (pick shirt A or shirt B, not both) do you add.
A meal offers 3 starters, 4 mains and 2 desserts, and you pick one of each. The number of complete meals is
- 9
- 24
- 12
Three independent stages multiply: 3 \times 4 \times 2 = 24. Adding gives 9, which treats the courses as alternatives rather than a sequence of choices.
3Circular arrangements
n distinct people around a round table can be seated in (n-1)! ways, not n!. Rotating everyone one seat clockwise gives the same arrangement — there is no distinguished "first chair" unless the question names one (a head of table, a marked seat).
Adjacency in a circle uses the block trick from linear work, but the outer count stays (n-1)!. Treat the pair as one block to get (n-2)! arrangements of blocks, then multiply by 2 for the internal order of the two friends inside the block.
Figure. Rotations of the same seating are identical, so fix one person and arrange the rest: (n-1)!. Divide by 2 only when the problem treats reflections as the same (necklace / bracelet).
Round-table method
- Fix one personBreak rotational symmetry by pinning one person; arrange the remaining n-1 relative to them.
- Count relative seatsThe count is (n-1)! for all distinct people with no further restriction.
- AdjacencyGlue the required neighbours into one block, count (n-2)! block arrangements, multiply by internal orders (2 for a pair).
Two friends together at a round table
Five friends sit around a round table. What is the probability that two particular friends sit next to each other?
- Total circular arrangements = (5-1)! = 4!24
- Pair as one block: (4-1)! \times 2! = 3! \times 212
- Probability = \frac{12}{24}\frac12
Pro tip. Using 5! = 120 for the denominator overcounts by a factor of 5 — every circular seating appears five times as a different linear rotation. The (n-1)! formula is already rotation-correct; do not divide again unless you started from n!.
Six distinct trophies are arranged in a circle on a display stand. The number of distinct arrangements is
- 6! = 720
- (6-1)! = 120
- \frac{6!}{2} = 360
A circle has no first position unless one is marked, so the count is (6-1)! = 5! = 120. Using 6! treats some rotation of the same display as a new arrangement.
4Arrangements with repeated letters
When some objects are identical, n! overcounts: swapping two identical S's does not produce a new arrangement. Divide by the factorial of each repeat count: \frac{n!}{p!\,q!\,r!} when the multiset has p alike of one kind, q of another, and so on.
The division removes indistinguishable permutations inside each group. The numerator still counts all positions; each denominator factor collapses the p! orders of the repeated letter into one.
Figure. Start from n! orderings of the letters, then divide by n_i! for each letter that repeats n_i times — those swaps do not create a new word.
Repeated-letter recipe
- Count the multisetTotal letters n, and how many of each letter — M\times1, I\times4, S\times4, P\times2 for MISSISSIPPI.
- Write the denominatorOne factorial factor per repeated letter: divide by p! for each group of p identical items.
- SimplifyCancel factorials before multiplying the remaining integers — CAT numbers are sized to cancel cleanly.
Distinct arrangements of MISSISSIPPI
In how many distinct ways can the letters of the word MISSISSIPPI be arranged?
- Eleven letters: M\times1, I\times4, S\times4, P\times2n = 11
- \frac{11!}{4!\,4!\,2!} = \frac{39916800}{24 \times 24 \times 2}\frac{39916800}{1152}
- Final count34650
Pro tip. A letter that appears only once needs no denominator factor — only repeats divide. Forgetting one repeated group (here, either the four S's or the four I's) leaves the answer too large by that factorial.
The number of distinct arrangements of the letters in BANANA is
- 6! = 720
- \frac{6!}{3!\,2!} = 60
- \frac{6!}{2!\,2!} = 180
Six letters with three A's and two N's: \frac{6!}{3!\,2!} = \frac{720}{6 \times 2} = 60. Dividing only by 2! twice ignores that the three A's contribute a 3! collapse.
5Probability of equally likely outcomes
When every outcome in the sample space is equally likely, P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}}. Build the sample space with combinations when the draw order does not matter — the same unordered pair of balls should not appear twice with the colours swapped.
For "at least one" events, compute 1 - P(\text{none}) instead of summing many disjoint cases. The complement rule P(E') = 1 - P(E) also applies when "not both red", "not all heads" or "at least one defect" is easier via its opposite.
Figure. When outcomes are equally likely, probability is favourable count over total count. Here two favourable faces out of six give 2/6=1/3 — simplify, and never invent unequal weights unless the stem does.
Probability from counting
- Define the sample spaceList what counts as one outcome. Unordered draws \rightarrow combinations; ordered rankings \rightarrow permutations.
- Count favourable outcomesUse the same counting rule as the denominator so the ratio is apples to apples.
- Complement when easier"At least one" often means 1 - P(\text{zero}) — one subtraction beats several cases.
| Rule | Formula | When to use |
|---|---|---|
| Equally likely | P(E) = \frac{\text{favourable}}{\text{total}} | Every outcome equally likely |
| Union | P(A \cup B) = P(A) + P(B) - P(A \cap B) | Either A or B (or both) |
| Complement | P(E') = 1 - P(E) | "At least one", "not all", "at least one fails" |
Probability of drawing two red balls
A bag contains 4 red and 3 blue balls. Two are drawn at random without replacement. What is the probability that both are red?
- Total ways to pick 2 of 7: ^7C_2 = \frac{7 \times 6}{2}21
- Favourable (both red): ^4C_2 = \frac{4 \times 3}{2}6
- Probability = \frac{6}{21}\frac{2}{7}
Pro tip. Using ^7P_2 = 42 and 4 \times 3 = 12 gives the same \frac{2}{7} only if both numerator and denominator count ordered draws. Mixing ordered favourable with unordered total — or vice versa — is the silent error; pick one convention and keep it.
A fair coin is tossed three times. The probability of at least one head is
- \frac{3}{8}
- \frac{7}{8}
- \frac{1}{2}
Complement: P(\text{no heads}) = P(\text{all tails}) = \left(\frac12\right)^3 = \frac{1}{8}, so P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}. Summing one-head and two-head and three-head cases also works but is slower on longer strings.
6Restrictions: block, gap and stars-and-bars
When certain items must stay together in a line, treat them as one block, arrange the blocks, then multiply by the internal arrangements of the block. Five people with Alice and Bob adjacent: 4! \times 2! — four blocks (the AB block plus three others) times the two orders inside the block.
When certain items must not be together, place the unrestricted ones first and slot the restricted ones into the gaps between them. For "no two girls together" with m boys, there are m+1 gaps; choose and fill enough gaps for the girls. Identical items into distinct boxes use stars and bars: ^{n+r-1}C_{r-1} ways to distribute n identical objects into r distinct groups (each group may receive zero).
Figure. Three restriction templates: glue people who must sit together (block), seat the unrestricted first and drop constrained items into gaps, or map non-negative integer solutions to stars and bars.
Three restriction moves
- Together (block)Glue the cluster into one super-item, count block arrangements, multiply by internal permutations.
- Not together (gap)Line up the free items, count the gaps on both sides, assign restricted items to distinct gaps.
- Identical into distinct boxesStars and bars: n identical items, r distinct recipients \rightarrow ^{n+r-1}C_{r-1}.
Block in a row and identical chocolates
(a) Five students sit in a row; Alice and Bob must sit together. How many seatings? (b) Ten identical chocolates are distributed among four children (each may get zero). How many distributions?
- Alice–Bob block plus three others: 4! \times 2!48
- Stars and bars: ^{10+4-1}C_{4-1} = {}^{13}C_3\frac{13 \times 12 \times 11}{3 \times 2 \times 1}
- Distinct distributions286
Pro tip. Block and gap are opposites — do not mix them on the same pair. "A and B together" is a block; "A and B not together" is a gap after placing everyone else. Stars and bars counts distributions, not seatings: the chocolates are identical, so only the split among children matters.
Four boys stand in a row. Two identical trophies must be placed in the gaps (including the ends) so that no two trophies occupy the same gap. The number of placements is
- ^5C_2 = 10
- ^5P_2 = 20
- 4! = 24
Four boys create five gaps; the trophies are identical, so choose two gaps: ^5C_2 = 10. Using ^5P_2 treats the trophies as distinct; 4! counts boy arrangements and ignores the trophies entirely.
Notes
- Fundamental counting principle: if one task can be done in m ways and an independent second task in n ways, together they can be done in m\times n ways.
- Permutations versus combinations: use ^nP_r when order matters (arrangements) and ^nC_r when it does not (selections).
- Circular arrangements: n distinct objects around a circle can be seated in (n-1)! ways because rotations of the same arrangement coincide.
- Probability of equally likely events: P(E)=\frac{\text{favourable outcomes}}{\text{total outcomes}}, and every probability lies between 0 and 1.
- Addition and complement rules: P(A\cup B)=P(A)+P(B)-P(A\cap B), and P(\text{not }E)=1-P(E).
Formulas
- Permutations: ^nP_r=\frac{n!}{(n-r)!}
- Combinations: ^nC_r=\frac{n!}{r!(n-r)!}
- Circular arrangement: (n-1)!
- Arrangements with repeats: \frac{n!}{p!\,q!\,r!}
- Conditional probability: P(A\mid B)=\frac{P(A\cap B)}{P(B)}
- Complement: P(E')=1-P(E)
Exam traps & shortcuts
- For an 'at least one' probability, compute 1-P(\text{none}) rather than summing many separate cases.
- Treat items that must stay together as one block, arrange the blocks, then multiply by the internal arrangements of the block.
- For 'no two together', place the unrestricted items first and slot the restricted ones into the gaps between them.
- Distributing n identical items into r distinct groups uses stars and bars: ^{n+r-1}C_{r-1}.
Reference tables
| Situation | Count | Probability hook |
|---|---|---|
| Arrange r from n distinct, order matters | ^nP_r = \frac{n!}{(n-r)!} | Ordered sample space |
| Choose r from n distinct, order ignored | ^nC_r = \frac{n!}{r!(n-r)!} | Unordered draws |
| n distinct around a circle | (n-1)! | Adjacency: block then (n-2)! \times r! |
| n letters with repeats | \frac{n!}{p!\,q!\,r!} | Favourable outcomes in word problems |
| n identical into r distinct boxes | ^{n+r-1}C_{r-1} | Distribution, not seating |
| Conditional probability | P(A \mid B) = \frac{P(A \cap B)}{P(B)} | Restrict the sample space to B first |
Recap
Read only this the night before.
- Order test
- Labelled positions \rightarrow ^nP_r. Swapping names changes nothing \rightarrow ^nC_r. Probability draws are usually unordered.
- Multiply stages
- Independent choices multiply. A full row of n distinct objects is n!.
- Circle
- (n-1)! seats a circle. Adjacency: glue the pair, count (n-2)!, multiply internal orders.
- Repeats
- Divide n! by each repeated letter's factorial. Single occurrences need no factor.
- Probability
- Equally likely \rightarrow favourable over total. At least one \rightarrow 1 - P(\text{none}).
- Restrictions
- Together \rightarrow block. Apart \rightarrow gaps after the free items. Identical into distinct boxes \rightarrow stars and bars.
Practise Permutations, Combinations and Probability
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