Permutations and Combinations: Formulas, Shortcuts, Worked Examples

Every permutations and combinations problem in a placement aptitude test reduces to one question: does the order of selection matter?

If yes, use nPr. If not, use nCr. Everything else in P&C is arithmetic built on those two decisions.

The Core Distinction: Order Matters vs. Order Doesn’t

A permutation counts arrangements. A combination counts selections.

Take three objects: A, B, C. Select two of them:

• With order mattering (permutations): AB, BA, AC, CA, BC, CB — 6 outcomes
• Without order mattering (combinations): AB, AC, BC — 3 outcomes

Same objects, same selection size, two different counts. The question “is AB different from BA for this problem?” determines which formula to apply.

Passwords and PINs are permutations: 1234 and 4321 are different PINs. Committee memberships are combinations: selecting Ramya, Priya, and Anil produces one committee regardless of the order they were named. Row seating is a permutation: seat 1 and seat 3 are different positions. Team formation is a combination: the team is the same whether you picked Ramya first or last.

Getting this classification right before touching a formula is the step that most wrong answers skip.

The Permutation Formula (nPr)

nPr = n! / (n-r)!

Where n is the total number of distinct objects and r is the number of positions to fill.

The factorial n! means n multiplied by every positive integer below it: 5! = 5 x 4 x 3 x 2 x 1 = 120. Values to know: 4! = 24, 3! = 6, 2! = 2, 1! = 1, 0! = 1.

Worked example: arrange 3 objects from 5 distinct objects (A, B, C, D, E).

• Given: n = 5, r = 3
• n! = 5! = 120
• (n-r)! = (5-3)! = 2! = 2
• 5P3 = 120 / 2 = 60 ordered arrangements

Worked example: arrange all 5 letters in “TIGER” (all distinct).

• nPn where n = 5: 5! / 0! = 120 / 1 = 120 arrangements

Permutations with Identical Items

When objects include repeated items, the identical orderings of those items should not be counted separately. Divide by the factorial of each repeat count:

n! / (p1! x p2! x ... x pk!)

where p1, p2, … pk are the counts of each identical group.

Worked example: arrange letters in “BBC” (3 letters, 2 identical Bs).

• Total letters: n = 3, so 3! = 6
• Count of repeated Bs: 2, so 2! = 2
• Arrangements = 3! / 2! = 6 / 2 = 3
• Verify manually: BBC, BCB, CBB — exactly 3 distinct arrangements ✓

Circular Permutations

Arranging n objects in a circle fixes one object as a reference point, removing rotationally identical arrangements. The count becomes (n-1)! instead of n!.

Worked example: 6 people seated at a round table.

• Formula: (6-1)! = 5! = 120 arrangements

Worked example: same table, Sam and Suhana must not sit adjacent.

• Total circular arrangements: (6-1)! = 120
• Arrangements where Sam and Suhana sit together: treat the pair as one unit, giving 5 units in a circle
  • Arrangements of 5 units in a circle: (5-1)! = 4! = 24
  • Sam and Suhana can swap within their unit: 2 ways
  • Together count: 24 x 2 = 48
• Not adjacent: 120 - 48 = 72

The Combination Formula (nCr)

nCr = n! / (r! x (n-r)!)

The only structural difference from nPr is the extra r! in the denominator. That factor divides out the r! orderings of the selected group that permutations count but combinations treat as duplicates.

Stated differently: nCr = nPr / r!. Permutation divided by the orderings of the chosen items gives combination. This relationship is useful when you need both counts from the same setup.

Worked example: select 3 from 5 distinct objects.

• Given: n = 5, r = 3
• 5! = 120
• r! = 3! = 6; (n-r)! = 2! = 2
• 5C3 = 120 / (6 x 2) = 120 / 12 = 10

Worked example: select 2 students from 10 for a scholarship (order doesn’t matter).

• C(10,2) = 10! / (2! x 8!) = (10 x 9) / (2 x 1) = 90 / 2 = 45

Worked example: form a committee of 1 man from 3 men and 3 women from 4 women.

• Ways to choose 1 man: C(3,1) = 3
• Ways to choose 3 women: C(4,3) = 4! / (3! x 1!) = 24 / (6 x 1) = 4
• Total committees: 3 x 4 = 12

Shortcut Tricks for Faster Calculation

Four shortcuts reduce arithmetic load on placement test problems.

• Mirror property: nCr = nC(n-r). To compute C(10,7), compute C(10,3) instead: (10 x 9 x 8) / (3 x 2 x 1) = 120. Always use the smaller r value.
• Quick C(n,2) formula: C(n,2) = n(n-1)/2. For C(10,2): 10 x 9 / 2 = 45. No factorial chain needed.
• Connect nPr and nCr: nPr = nCr x r!. If you have already computed C(5,3) = 10, then P(5,3) = 10 x 3! = 10 x 6 = 60.
• Boundary values: nC0 = nCn = 1. nC1 = nP1 = n. nPn = n!.

The mirror property is the most useful under time pressure. Any C(n,r) where r is close to n can be recast as a C(n, small), saving a full factorial chain calculation.

What Campus Placement Tests Actually Ask

Five question types cover the bulk of P&C problems across placement aptitude rounds.

1. Letter arrangements: Identify repeated letters and apply the divided-factorial formula. “TIGER” has 5 distinct letters: 120 arrangements. “BBC” has 2 identical Bs: 3 arrangements.
2. Group selections: Committees, teams, and scholarship problems use nCr. Conditions such as “1 man from 3, 3 women from 4” multiply the nCr calls.
3. Circular seating: Apply (n-1)! for the total, then subtract cases that violate the stated constraint.
4. At-least-one problems: Compute total minus zero-from-that-group. For example, total committees minus committees with no women is faster than summing individual “at least 1, at least 2” cases.
5. Repetition problems: C(n+r-1, r) when the same item can be chosen more than once.

The TCS NQT Numerical Ability section lists permutations and combinations as part of its standard quantitative aptitude syllabus. The AMCAT Quantitative Ability module covers the same five types with a similar exam-style format.

The decision rule for every problem: if the problem describes an arrangement, sequence, ranking, or row, use nPr. If it describes a group, team, committee, or selection where the named order doesn’t change the outcome, use nCr.

FACE Prep’s Campus Placement Evaluation Test benchmarks your current aptitude percentile across the full placement syllabus. P&C is one of the tested topics; it’s a direct measure of where this article’s formula coverage sits in your preparation.

For the same formula-first approach applied to work-rate and LCM problems, FACE Prep’s time and work guide covers the five question types that appear on TCS, Infosys, and Wipro aptitude papers.

Analytical roles at companies like Mu Sigma test combinatorial thinking heavily in their aptitude rounds. FACE Prep’s Mu Sigma aptitude guide covers the MuApt test pattern and question types where P&C and probability overlap.

The permutation and combination framework in this article (classify the problem, match it to the correct formula, compute step by step) is the same structured decomposition that applies when evaluating any system’s outputs analytically. TinkerLLM is an LLM playground where you can practice that kind of systematic evaluation for ₹299; the nCr mindset of “how many ways can these elements be selected” translates directly into structured prompt testing.