Permutation, Combination & Probability for GATE Aptitude 2027: Formulas & Solved PYQs

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Quick Summary: Permutation, Combination and Probability give 2 to 4 marks in GATE General Aptitude 2027 across every paper (DA, RA, CS, ME, EE, CE, EC), and probability also reappears in the core Engineering Mathematics section. GATE tests a small toolkit: the counting principle, nPr and nCr, arrangements with repetition, circular arrangements, classical probability, complement, addition and conditional probability. This guide covers that toolkit with 8 solved previous-year and GATE-style questions, a 5-question practice set, a one-page formula sheet and the traps that convert a sure mark into a negative.

🎯 Score 15/15 in General Aptitude. Practise this topic with 100+ exam-level questions, detailed solutions and All-India ranking in the Best GATE Aptitude Test Series 2027 by PiyushAI. Topic-wise tests + full-length GA mocks, valid till GATE 2027.

📋 Table of Contents

Why this topic matters in GATE GA
What the official GA syllabus says
Fundamental counting principle, nPr and nCr
Arrangements with repetition, restrictions and circles
Classical probability, complement and addition
Conditional probability and independence
Shortcuts and traps
Solved PYQs and GATE-style questions
Practice set
One-page formula sheet
Common mistakes
FAQs

Why this topic matters in GATE GA

Counting and probability are the highest-yield Quant chapter in General Aptitude for one reason: the same ideas are tested twice. GATE asks a 1-mark or 2-mark GA question on arrangements or dice, and then asks a 2-mark Engineering Mathematics question on conditional probability or distributions. Learn the GA version properly and the core version becomes half as hard.

The GA questions are short, concrete and almost always have a small integer or a clean fraction as the answer. They look like puzzles (“how many rectangles in this grid”, “probability the socks match”) but every one of them reduces to nCr, nPr or favourable/total. The mark is lost by double counting, by forgetting the complement, or by treating dependent events as independent.

Here is roughly how much this cluster has contributed. Candidates targeting 15/15 in GA should treat these as banked marks.

Year	Marks from this topic (approximate, across papers)	Typical question type
2019	2 to 3	Arrangements with constraints, dice probability
2020	1 to 3	Selection from groups, coin tosses
2021	2 to 3	Seating in a row, probability of at least one
2022	2 to 4	Balls drawn without replacement, number of words
2023	2 to 3	Counting paths on a grid, cards
2024	2 to 3	Committee selection, conditional probability
2025	1 to 3	Arrangements of letters, independent events
2026	2 to 3	Distribution of identical objects, dice

What the official GA syllabus says

Official GATE GA syllabus, Quantitative Aptitude: “Numerical computation and estimation: ratios, percentages, powers, exponents and logarithms, permutations and combinations, and series. Mensuration and geometry. Elementary statistics and probability.”

“Permutations and combinations” and “elementary statistics and probability” are the two phrases this post covers. The complete GA syllabus and weightage are in the GATE General Aptitude 2027 hub. The previous post covered Powers, Exponents, Surds and Number System.

Fundamental counting principle, nPr and nCr

Multiplication principle: if a task has m ways and a second independent task has n ways, both together have m × n ways. Addition principle: if the tasks are alternatives (one or the other), the count is m + n. “And” multiplies, “or” adds.

Permutation (order matters): nPr = n!/(n − r)!. Arranging 3 of 5 books on a shelf: 5P3 = 5 × 4 × 3 = 60.
Combination (order does not matter): nCr = n!/[r!(n − r)!]. Choosing 3 of 5 books to carry: 5C3 = 10.
nCr = nC(n−r); nC0 = nCn = 1; nC1 = n; nC2 = n(n − 1)/2.
nPr = nCr × r!. A permutation is a combination followed by an arrangement.

The decision rule: if swapping two chosen items gives a different outcome (seats, ranks, passwords), it is a permutation. If swapping gives the same outcome (a committee, a hand of cards, a subset), it is a combination.

✅ Solved Example (GATE 2023 style): A committee of 3 is to be formed from 5 men and 4 women. How many committees contain at least one woman?
Solution: Total committees = 9C3 = 84. Committees with no woman = 5C3 = 10. At least one woman = 84 − 10 = 74.

✅ Solved Example (GATE 2023 style): How many diagonals does a convex polygon with 12 sides have?
Solution: Any two vertices give a line: 12C2 = 66. Subtract the 12 sides: 66 − 12 = 54. General formula: n(n − 3)/2.

Arrangements with repetition, restrictions and circles

Repeated letters: arrangements of n objects where p are alike of one kind, q alike of another = n!/(p! q!). MISSISSIPPI: 11!/(4! 4! 2!) = 34,650.
Objects together: glue them into one block, arrange the blocks, then arrange inside the block.
Objects never together: total arrangements minus arrangements with them together.
Circular arrangement: (n − 1)! for people around a table; (n − 1)!/2 for a necklace or garland (flipping gives the same thing).
Selection with repetition allowed (identical objects into distinct boxes): distributing n identical items among r people = (n + r − 1)C(r − 1). With each getting at least one: (n − 1)C(r − 1).
Each of n positions filled from k choices independently: kⁿ. A 4-digit PIN has 10⁴ possibilities; the number of subsets of an n-element set is 2ⁿ.

✅ Solved Example (GATE 2023 style): In how many ways can the letters of COMPUTER be arranged so that the vowels are always together?
Solution: COMPUTER has 8 distinct letters with vowels O, U, E. Treat the three vowels as one block: 6 units to arrange = 6! = 720. The vowels inside the block can be arranged in 3! = 6 ways. Total = 720 × 6 = 4,320.

✅ Solved Example (GATE 2023 style): In how many ways can 10 identical chocolates be given to 3 children so that each child gets at least one?
Solution: Give each child one first; 7 remain to be distributed freely: (7 + 3 − 1)C(3 − 1) = 9C2 = 36. Same as (10 − 1)C(3 − 1) = 9C2.

Classical probability, complement and addition

P(E) = (favourable outcomes)/(total equally likely outcomes). Every probability lies between 0 and 1. The two rules that solve 80% of GA probability questions:

Complement: P(not E) = 1 − P(E). “At least one” is almost always solved as 1 − P(none).
Addition: P(A or B) = P(A) + P(B) − P(A and B). If A and B are mutually exclusive, P(A and B) = 0.

Standard sample spaces: one coin 2, two coins 4, three coins 8, one die 6, two dice 36, a card from a deck 52 (4 suits of 13, 4 aces, 12 face cards, 26 red). Two dice sum table: sum 7 appears 6 times, sums 6 and 8 appear 5 times each, sums 2 and 12 appear once each.

✅ Solved Example (GATE 2023 style): A fair die is thrown twice. What is the probability of getting at least one six?
Solution: P(no six in a throw) = 5/6. P(no six in two throws) = (5/6)² = 25/36. P(at least one six) = 1 − 25/36 = 11/36.

✅ Solved Example (GATE 2023 style): A card is drawn from a standard deck. What is the probability that it is a king or a heart?
Solution: P(king) = 4/52, P(heart) = 13/52, P(king of hearts) = 1/52. P(king or heart) = (4 + 13 − 1)/52 = 16/52 = 4/13.

Conditional probability and independence

P(A | B) = P(A and B)/P(B), the probability of A given that B has happened. Rearranged: P(A and B) = P(B) × P(A | B). This is the “without replacement” rule: the second draw’s probability depends on what the first draw removed.

A and B are independent when P(A and B) = P(A) × P(B), equivalently P(A | B) = P(A). Coin tosses, dice throws and draws with replacement are independent. Draws without replacement are not.

Bayes in one line: P(A | B) = P(B | A) P(A) / P(B). In GA it appears as a “given that the item is defective, which machine made it” question; a 2 × 2 table of counts solves it faster than the formula.

✅ Solved Example (GATE 2023 style): Two cards are drawn without replacement from a deck of 52. What is the probability that both are aces?
Solution: P(first ace) = 4/52. Given that, P(second ace) = 3/51. Product = 12/2652 = 1/221. Combination check: 4C2/52C2 = 6/1326 = 1/221.

Each of these question types has a timed, ranked test in the GATE Aptitude Test Series; the P&C and probability topic test alone has 40 questions with video solutions.

Shortcuts and traps

💡 GATE Tip: “At least one” means complement. Compute P(none) and subtract from 1. Direct computation needs three or four cases and invites a slip.

💡 GATE Tip: Rectangles in an m × n grid = (m + 1)C2 × (n + 1)C2, because a rectangle is a choice of two horizontal and two vertical lines. Squares in an n × n grid = 1² + 2² + … + n².

💡 GATE Tip: Shortest grid paths from one corner of an m × n grid to the opposite corner, moving only right and up = (m + n)C m. You are choosing which m of the m + n steps are “right”.

💡 GATE Tip: Probability by combinations and probability by sequential draws always agree. Use whichever is quicker, but check the answer with the other when time permits; it catches almost every counting error.

❌ Using nPr when order does not matter (committees, hands, subsets). The answer comes out r! times too big.
❌ Double counting in “at least one” by adding P(one), P(two), P(three) without checking overlaps.
❌ Treating draws without replacement as independent.
❌ Forgetting to divide by p! for repeated letters.
❌ Using n! for a circular arrangement instead of (n − 1)!.
❌ Adding P(A) and P(B) for “A or B” when A and B can occur together.
❌ Counting the 12 face cards as including aces (they are J, Q, K only).

Solved previous-year and GATE-style questions

✅ Solved Example (GATE 2016, CS): In a 2 × 4 rectangle grid, each cell is a rectangle. How many rectangles can be observed in the grid?
Solution: A 2 × 4 grid has 3 horizontal and 5 vertical lines. Choose 2 of each: 3C2 × 5C2 = 3 × 10 = 30.

✅ Solved Example (GATE 2017, CS): There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is:
Solution: Same-colour pairs = 3C2 + 4C2 + 3C2 = 3 + 6 + 3 = 12. Total pairs = 10C2 = 45. Probability = 12/45 = 4/15.

✅ Solved Example (GATE 2014, CS): The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ____.
Solution: Inclusion-exclusion. Divisible by 2: 50, by 3: 33, by 5: 20. By 6: 16, by 10: 10, by 15: 6. By 30: 3. Divisible by at least one = 50 + 33 + 20 − 16 − 10 − 6 + 3 = 74. Not divisible = 26. Probability = 0.26.

✅ Solved Example (GATE 2011, CS): A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is:
Solution: Of the 36 outcomes, 6 are ties. The remaining 30 split equally between “second higher” and “first higher”. Favourable = 15. Probability = 15/36 = 5/12.

✅ Solved Example (GATE 2015, CS): A coin is tossed thrice. Let X be the event that head occurs in each of the first two tosses, Y the event that a tail occurs on the third toss, and Z the event that two tails occur in three tosses. Which is TRUE: X and Y are independent, or Y and Z are independent?
Solution: P(X) = 1/4, P(Y) = 1/2, P(X and Y) = P(HHT) = 1/8 = P(X)P(Y), so X and Y are independent. P(Z) = 3/8; P(Y and Z) = P(HTT or THT) = 2/8 = 1/4, but P(Y)P(Z) = 3/16. Not equal. X and Y are independent; Y and Z are not.

✅ Solved Example (GATE-style): What is the probability that a leap year selected at random contains 53 Saturdays?
Solution: A leap year has 366 days = 52 weeks + 2 days. The extra two days are one of 7 consecutive pairs (Sun-Mon, Mon-Tue, …, Sat-Sun). Saturday appears in 2 of them (Fri-Sat and Sat-Sun). Probability = 2/7.

✅ Solved Example (GATE-style): A box contains 3 green and 2 orange balls. Balls are drawn one at a time without replacement. What is the probability that both orange balls are drawn before any green ball?
Solution: The first two draws must both be orange: (2/5) × (1/4) = 1/10. Combination check: of the 5C2 = 10 equally likely position-pairs for the oranges, only positions {1, 2} work.

✅ Solved Example (GATE-style): How many distinct arrangements of the letters of MISSISSIPPI are possible?
Solution: 11 letters: I appears 4 times, S 4 times, P 2 times, M once. Arrangements = 11!/(4! × 4! × 2!) = 39,916,800/1,152 = 34,650.

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Practice set

Attempt these in 7 minutes before opening the key.

The number of diagonals of a decagon is (A) 35 (B) 45 (C) 30 (D) 40
Two fair dice are thrown. The probability that the sum is 7 is (A) 1/12 (B) 1/6 (C) 7/36 (D) 5/36
How many 4-digit numbers with all digits distinct can be formed using the digits 1 to 7? (A) 2401 (B) 840 (C) 35 (D) 5040
A bag has 5 red and 3 blue balls. Two balls are drawn at random. The probability that both are red is (A) 5/14 (B) 25/64 (C) 5/8 (D) 10/56
In how many ways can 8 people be seated around a circular table? (A) 40320 (B) 5040 (C) 2520 (D) 720

Show answer key

(A) 10C2 − 10 = 45 − 10 = 35.
(B) Six pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) out of 36: 6/36 = 1/6.
(B) 7P4 = 7 × 6 × 5 × 4 = 840.
(A) 5C2/8C2 = 10/28 = 5/14.
(B) (8 − 1)! = 7! = 5040.

One-page formula sheet

1. “And” multiplies, “or” adds. k choices for each of n positions: kⁿ.
2. nPr = n!/(n − r)!; nCr = n!/[r!(n − r)!]; nPr = nCr × r!; nCr = nC(n−r).
3. Arrangements with repeats: n!/(p! q! r!). Objects together: glue, arrange blocks, arrange inside.
4. Circular: (n − 1)!; necklace: (n − 1)!/2.
5. n identical objects to r people: (n + r − 1)C(r − 1); each at least one: (n − 1)C(r − 1).
6. Diagonals of an n-gon: n(n − 3)/2. Rectangles in m × n grid: (m + 1)C2 × (n + 1)C2. Grid paths: (m + n)Cm.
7. Subsets of an n-set: 2ⁿ. Handshakes among n people: nC2.
8. P(E) = favourable/total; P(not E) = 1 − P(E); “at least one” = 1 − P(none).
9. P(A or B) = P(A) + P(B) − P(A and B).
10. P(A and B) = P(A) × P(B | A); independent when P(A and B) = P(A)P(B).
11. Bayes: P(A | B) = P(B | A)P(A)/P(B). Use a count table when possible.
12. Two dice: 36 outcomes; sum s has (6 − |s − 7|) outcomes. Leap year has 2 extra weekdays; ordinary year has 1.

Common mistakes

❌ Answering a probability NAT as a fraction when the question asks for a decimal rounded to two places (or the other way around).
❌ Counting (1,6) and (6,1) as one outcome for two dice. They are different outcomes.
❌ Forgetting to subtract the n sides when counting diagonals.
❌ Allowing the digit 0 in the leading position when forming numbers.
❌ Reading “at most two” as “exactly two”.
❌ Treating “mutually exclusive” and “independent” as the same thing. Mutually exclusive events with non-zero probability are never independent.

FAQs

How many P&C and probability questions appear in GATE GA?

Usually 1 to 2 questions worth 2 to 4 marks per paper, and probability is tested again in the core Engineering Mathematics section of most papers, so the chapter is worth more than its GA share.

Do I need Bayes’ theorem for GA?

Occasionally, in a “given that the item is defective” form. A 2 × 2 table of counts answers such questions in under a minute without the formula. Learn the formula for the Engineering Mathematics section.

What is the fastest way to decide between permutation and combination?

Ask whether swapping two chosen items changes the outcome. Seats, ranks and passwords: yes, so permutation. Committees, hands and subsets: no, so combination.

Is the PiyushAI Aptitude Test Series enough for GA?

Yes. The Best GATE Aptitude Test Series 2027 has a topic test on permutation, combination and probability, tests on every other GA chapter, full-length GA mocks in the GATE interface, solved PYQs from 2010 to 2026 and All-India rank analysis. With the free guides on this site, you do not need another GA resource.

Are probability answers in GATE GA usually fractions or decimals?

MCQs give fractions in the options. NAT questions ask for a decimal, typically rounded to two places, and accept a small range. Compute exactly, then convert.

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