Most aspirants prepare permutations and combinations by memorising a sheet of formulas, then freeze when a GATE question matches none of them. The problem is not the formulas. Counting questions are decision problems, and you need a decision procedure, not a lookup table.
This post teaches one procedure, three moves (choose, arrange, subtract), and applies it to 20 GATE-style counting problems.
Permutations and combinations: why one method beats twenty formulas
Every counting question asks how many ways a task can be completed. The task breaks into stages, and the subject reduces to three moves.
Choose. Pick r objects out of n when order does not matter: C(n, r) = n! / (r! (n - r)!). A committee, a subset.
Arrange. Put k distinct objects in order: k!. Arranging r objects picked from n gives P(n, r) = C(n, r) × r!, so a permutation is literally a choice followed by an arrangement.
Subtract. When the question says "at least one" or "never together", count the total, count the violating cases, and subtract. The complement is almost always easier to count.
Every standard formula is one of these moves, or two multiplied together. Seen that way, the formula sheet becomes receipts, not memory work.
The decision procedure to run before every counting question
Run these four questions, in this order, on every problem.
What are the stages? Split the task into independent decisions and multiply their counts. This is the product rule.
Does order matter? Distinguishable slots or a sequence: arrange. A group where shuffling changes nothing: choose.
Is repetition allowed? Repeats mean every slot keeps all n options, so n^r. No repeats means shrinking options, so factorials.
Is there a restriction? Build the restricted part first, or count everything and subtract the violations, whichever is smaller.
Product rule problems: multiply the stages (Problems 1 to 5)
Problem 1. How many passwords have 3 lowercase letters followed by 2 digits? Five independent slots: 26 × 26 × 26 × 10 × 10 = 1,757,600.
Problem 2. How many binary strings of length 8 exist? Each position independently picks 0 or 1, so 2^8 = 256.
Problem 3. How many functions exist from a set A with 3 elements to a set B with 5 elements? Each element of A independently chooses an image in B: 5^3 = 125.
Problem 4. How many of those functions are one-to-one? Repetition is now banned, so options shrink: 5 × 4 × 3 = 60, exactly P(5, 3).
Problem 5. How many 3-digit even numbers have all distinct digits? Build the constrained slot first. Last digit 0: 9 × 8 = 72. Last digit 2, 4, 6 or 8: the first digit has 8 options (not zero, not the last digit), the middle has 8, so 4 × 8 × 8 = 256. Total 328.
Permutation problems: arrange when order matters (Problems 6 to 10)
Problem 6. In how many ways can 6 distinct books be placed in a row on a shelf? Arrange all six: 6! = 720.
Problem 7. How many 4-letter strings over the English alphabet have no repeated letter? Shrinking options: 26 × 25 × 24 × 23 = 358,800, i.e. P(26, 4).
Problem 8. How many distinct arrangements does the word ENGINE have? Six letters, E twice, N twice. Arrange all, divide out the shuffles of identical letters: 6! / (2! × 2!) = 180.
Problem 9. In how many ways can 7 people sit around a circular table? Fix one person to kill the rotations, arrange the rest: (7 - 1)! = 720.
Problem 10. In how many ways can 5 boys and 5 girls sit in a row, alternating? The pattern starts with a boy or a girl (2 ways); arrange boys in their seats and girls in theirs: 2 × 5! × 5! = 28,800.
Combination problems: choose when order does not matter (Problems 11 to 15)
Problem 11. How many 3-member committees can be formed from 10 people? A committee has no order: C(10, 3) = 120.
Problem 12. If 12 people all shake hands once, how many handshakes happen? Each handshake is an unordered pair: C(12, 2) = 66, the same count as edges in a complete graph on 12 vertices.
Problem 13. How many binary strings of length 10 contain exactly three 1s? Choosing which 3 positions hold the 1s fully determines the string: C(10, 3) = 120.
Problem 14. How many non-negative integer solutions does x + y + z = 10 have? Line up 10 identical units and 2 dividers and choose where the 2 dividers go: C(12, 2) = 66. Stars and bars is a choose move.
Problem 15. From 4 vowels and 5 consonants, how many ways can you select 2 vowels and 3 consonants? Two independent choices multiplied: C(4, 2) × C(5, 3) = 6 × 10 = 60.
Choose then arrange: the compound count GATE actually asks (Problems 16 to 18)
Problem 16. Continue Problem 15: how many 5-letter words can be formed from those selections? Choose gave 60 letter sets; arrange each set of 5 distinct letters: 60 × 5! = 7,200. Stopping after the choose step is the most common trap in the topic.
Problem 17. A committee of 5 is formed from 6 men and 4 women with exactly 2 women. Choose the women, choose the men: C(4, 2) × C(6, 3) = 6 × 20 = 120. No arrange step, a committee is unordered.
Problem 18. From 8 students, in how many ways can you pick 3 and seat them in a row of 3 chairs? Choose then arrange: C(8, 3) × 3! = 56 × 6 = 336, equal to P(8, 3). That is the receipt for the P(n, r) formula itself.
Subtract: complement counting for "at least" and "never" (Problems 19 and 20)
Problem 19. How many binary strings of length 8 contain at least one 1? Total minus the all-zero string: 2^8 - 1 = 255. One subtraction replaces eight cases.
Problem 20. How many 5-person committees from 6 men and 4 women contain at least one woman? Total minus all-male committees: C(10, 5) - C(6, 5) = 252 - 6 = 246. Direct counting needs four cases; the complement, one.
The same move handles "two people never sit together": glue the pair into one block to count the together cases, then subtract from the total.
Permutations and combinations in the GATE CS paper: where counting shows up
In GATE CS, combinatorics sits inside the Engineering Mathematics portion of the syllabus, and it rarely stays there. Counting leaks into the core subjects: functions and relations in discrete mathematics, edges and labelled graphs in graph theory, strings of a given length in automata questions. All of them reduce to the patterns above.
Many counting questions appear as NAT (numerical answer type), where you key in an integer with no options to eliminate. A half-trusted formula is dangerous there; a re-derivable procedure is not. For the exact section weightage in your cycle, check the official GATE information brochure, and for how Engineering Mathematics stacks up against the core subjects see our GATE CS subject-wise weightage breakdown. The current official GATE portal is the source for that cycle's brochure and paper rules.
The short version, and where to practise
Run the four questions on every problem: stages, order, repetition, restriction. Then execute choose, arrange, subtract as the answers dictate. That procedure solved all 20 problems here, and it will solve the variants the paper throws at you. Drill until the decision step is automatic.
Work the topic on the Permutation and Combination learn module, nearly 70 lessons in the Engineering Mathematics track, backed by hundreds of published permutation and combination questions in our bank.
For the full syllabus with this sequencing built in, GATE Guidance by Sanchit Sir plans practice around a target score, flags syllabus hotspots, and tracks topic-wise performance.
Still comparing options? Start from the GATE category page.
Learn the three moves, run the four questions, and counting stops being a formula lottery and becomes the most predictable marks in the mathematics section.
