Modern Math (Permutations, Combinations, Probability)

Syllabus Unit: Quantitative Ability (QA)

Why it matters

Chapter overview

Permutations, combinations, probability and set theory form CAT's most conceptually distinct Quant cluster — the skill tested is not calculation but correctly modelling a counting or chance scenario before any arithmetic happens, since a single wrong model (permutation vs combination, with vs without replacement) invalidates the entire answer.

Management Aptitude Syllabus Quantitative Ability (QA) 34% weightage

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Decide: does order matter? (permutation vs combination)

Handle any 'together' groups as a single clubbed unit

Apply the repetition formula if items repeat

Multiply/subtract to combine sub-counts into the final answer

1Permutations & Combinations

Permutations count arrangements where order matters; combinations count selections where it doesn't — get this choice right first.

Learning Objectives

Apply nPr = n!/(n−r)! for arrangements and nCr = n!/(r!(n−r)!) for selections.
Handle arrangements with repeated items using n!/(p!q!r!...) where p,q,r are repetition counts.
Solve 'items together' problems by clubbing them into one unit, and 'items never together' by subtracting from total.
Distinguish circular permutations ((n−1)!) from linear permutations (n!) for arranging n distinct items.

Key Terms

Permutation (nPr)The number of ways to arrange r items out of n distinct items where order matters: nPr = n!/(n−r)!.

Combination (nCr)The number of ways to select r items out of n distinct items where order does not matter: nCr = n!/(r!(n−r)!).

Factorial (n!)The product of all positive integers up to n (n! = n x (n−1) x ... x 1); 0! is defined as 1.

Repetition FormulaFor arranging n items where some are identical (p of one kind, q of another...), the count is n!/(p! x q! x ...).

The single most important decision in any counting problem is whether **order matters**. If it does, use permutations (nPr); if only the group composition matters, use combinations (nCr).

nPr = n! / (n−r)! nCr = n! / (r!(n−r)!) nCr = nPr / r!

Together / never-together problems

'2 people always together' in a row of 5: club them as 1 unit → 4 units arrange in 4! ways, and the pair internally swaps in 2! ways → 4! x 2! = 48.

'2 people never together': total arrangements (5!) minus 'always together' arrangements (48) = 120 − 48 = 72.

Repeated letters and circular arrangements

Arrangements of the word 'BANANA' (6 letters, A x3, N x2): 6!/(3! x 2!) = 720/12 = 60, since identical letters swapping produce indistinguishable words.

Exam Tipn distinct people around a circular table: (n−1)! arrangements, not n! — because rotating the whole circle produces the same arrangement, removing one degree of freedom.