CA Foundation · Quantitative Aptitude

Permutations and Combinations: formula sheet

Full chapter guide

Key formulas

Multiplication rule (rule of product): Total ways = m × n × p × ...; Use when stages are done one after another and every stage must be completed. Stage 2 counts are for each choice of stage 1.
Addition rule (rule of sum): Total ways = m + n + p + ...; Use when you choose one of several mutually exclusive cases. No outcome may belong to two cases.
Words that signal the rule: AND / then / followed by → × | OR / either / alternatively → +; A guide only. Always check the actual structure of the job.
Digits with no restriction on repetition: Number of r-digit strings from n digits = n^r; Follows from the multiplication rule when repetition is allowed. If 0 cannot lead a number, the first place has one fewer choice.
Digits without repetition: n × (n − 1) × (n − 2) × ... for r places; Each place has one fewer choice than the previous one. Fill the restricted place first.
Definition: n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1; The product formula is for whole numbers n ≥ 1. 0! = 1 is defined separately, so factorial is defined for all whole numbers n ≥ 0.
Zero factorial: 0! = 1; A definition that keeps n! = n × (n − 1)! true for n = 1. Also 1! = 1.
Recursive property: n! = n × (n − 1)! = n × (n − 1) × (n − 2)!; Use it to expand a large factorial down to a smaller one.
Ratio of factorials: n! ÷ r! = n × (n − 1) × ... × (r + 1), for whole numbers n > r; Only n − r factors remain after cancelling.
Reverse step: (n − 1)! = n! ÷ n; Useful for getting 0! from 1! or checking small values.
Values to memorise: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5,040, 8! = 40,320; Knowing these saves time and lets you match options quickly.
Permutation of r out of n distinct objects: nPr = n! ÷ (n − r)! = n(n − 1)(n − 2)...(n − r + 1); Valid for 0 ≤ r ≤ n, with objects all different and no repetition.
Arranging all n distinct objects: nPn = n!; Uses 0! = 1.
Special values: nP0 = 1, nP1 = n; nP1 = n because one place can be filled in n ways.
Always together: (n − k + 1)! × k!; k specified objects out of n, all arranged in a row. Block counts as one object; k! is the inside order.
Not all together (k specified objects): n! − (n − k + 1)! × k!; Total arrangements minus the arrangements where all k are together. For k = 2 this is the same as 'the two are never together'. For k > 2 it does not mean no two are adjacent.
Two objects never together (shortcut): (n − 2)! × (n − 1)P2; Arrange the other n − 2 objects, then choose 2 of the n − 1 gaps in order. Gives the same value as n! − 2 × (n − 1)!.
No two of k objects adjacent (gap method): (n − k)! × (n − k + 1)Pk; Arrange the other n − k objects, which creates n − k + 1 gaps. Place the k objects in different gaps, in order. For k = 2 it matches the shortcut above.
Recurrence: nPr = n × (n − 1)P(r − 1); Useful for quick checks.
Arrangements with identical objects: n! ÷ (p! × q! × r! ...); n is the total number of objects. p, q, r are the sizes of the groups of alike objects. Letters that appear once contribute 1! and can be ignored.
Arrangements with repetition allowed: n^r; r places, each filled in n ways. Use it for codes, PINs, and numbers where digits can repeat.
Arrangements of n objects taken r at a time, repetition allowed: n^r; Same rule. It is not nPr, because nPr does not allow reuse.
Alike items kept together: Treat the group as one block, then arrange the blocks with the formula for identical objects; Example: if two E's must be together, count them as one unit, so the unit is not divided by 2! again.
Rank of a word in a dictionary: Rank = (sum of words starting with smaller letters at each position) + 1; At each position, count words with each smaller unused letter placed there. The remaining letters are arranged using n! ÷ (p! q! ...).
Numbers from digits with repetition allowed: Choices for first place (non-zero) × choices for other places; If 0 is among the digits, the first place of a multi-digit number cannot be 0.
Circular arrangement of n distinct objects (clockwise and anticlockwise different): (n − 1)!; Use for round tables, people in a ring, and similar cases.
Necklace or garland (clockwise and anticlockwise same): (n − 1)! ÷ 2; Use for n ≥ 3 distinct beads or flowers that can be flipped.
Selecting r from n and arranging in a circle: nCr × (r − 1)! = nPr ÷ r; Dividing nPr by r removes the r rotations.
Two particular persons always together: (n − 2)! × 2!; Treat the pair as one unit: n − 1 units give (n − 2)! circular ways, times 2! for the pair's order.
Two particular persons not together: (n − 1)! − (n − 2)! × 2!; Total minus together.
Women or men placed in gaps: (m − 1)! × mPk; Seat m people in a circle first in (m − 1)! ways. This creates m gaps. Placing k others in these gaps, one per gap, gives mPk ways. When k = m (equal groups, alternate seating), mPk becomes m!, so the total is (m − 1)! × m!.
Combination formula: nCr = n! ÷ (r! × (n − r)!); Valid for whole numbers n and r with 0 ≤ r ≤ n.
Link with permutations: nCr = nPr ÷ r!; Divide by r! to remove the order.
Complementary rule: nCr = nC(n − r); Use it when r is more than n ÷ 2 to shrink the calculation.
Special values: nC0 = nCn = 1; nC1 = n; Only one way to choose none or all; n ways to choose one.
Pascal's relation: nCr + nC(r − 1) = (n + 1)Cr; Adds two neighbouring values of the same n to give the next n.
Ratio of consecutive terms: nCr ÷ nC(r − 1) = (n − r + 1) ÷ r; Very useful for equations and for finding n or r from ratios.
Reduction formula: nCr = (n ÷ r) × (n − 1)C(r − 1); Needs r ≥ 1.
Equal combinations: If nCx = nCy, then x = y or x + y = n; Check both cases. Reject values that are not whole numbers or exceed n.
Sum of all combinations: nC0 + nC1 + ... + nCn = 2ⁿ; Selecting at least one object from n distinct objects gives 2ⁿ − 1 ways.
Combination: nCr = n! ÷ (r! × (n − r)!); Defined for whole numbers with 0 ≤ r ≤ n.
Symmetry: nCr = nC(n − r); Use the smaller of r and n − r to save time. If nCx = nCy, then x = y or x + y = n.
Selecting from groups: Ways = (aCp) × (bCq); Multiply when you must choose p from group A and q from group B together.
Separate cases: Total = Case 1 + Case 2 + ...; Add only when the cases cannot happen together.
At least one: At least one = Total − None; Total ways to choose with no restriction minus ways with none of the wanted type.
Selection of any number of items: Select at least one from n different items = 2ⁿ − 1; Each item is in or out, so 2ⁿ ways, minus the case of choosing nothing.
Lines from points: nC2; For n points with no three collinear. If m points are collinear, lines = nC2 − mC2 + 1.
Triangles from points: nC3; For n points with no three collinear. If m points are collinear, triangles = nC3 − mC3.
Diagonals of a polygon: nC2 − n = n(n − 3) ÷ 2; For a convex polygon with n sides. Subtract the n sides from all pairs of vertices.

Quick revision

Multiplication rule: if one task can be done in m ways and another in n ways, both in sequence can be done in m × n ways.
Addition rule: if tasks are alternatives and cannot happen together, add the ways.
n! = n × (n − 1) × ... × 1, with 0! = 1.
nPr = n! ÷ (n − r)!, the number of ordered arrangements of r objects from n distinct objects.
Arranging n distinct objects in a row: n! ways.
Arrangements of n objects where p are identical of one kind and q of another: n! ÷ (p! × q!).
Circular arrangement of n distinct objects: (n − 1)! when rotations are treated as the same.
nCr = n! ÷ [r! × (n − r)!], with nPr = r! × nCr.
nCr = nC(n − r); nC0 = nCn = 1; nC1 = n.
If nCx = nCy, then x = y or x + y = n.
nCr + nC(r − 1) = (n + 1)Cr.
For 'together' conditions, treat the group as one object, then multiply by the arrangements inside the group.

Common mistakes

Adding when the job has stages, or multiplying when the job has alternatives. Fix: Ask: do I need to complete every step (multiply) or pick only one path (add)?
Counting the unrestricted place first and ignoring the condition. Fix: Fill the restricted place first. For an even number, choose the units digit first. For no leading zero, handle the first place carefully.
Writing 0! = 0. Fix: Remember 0! = 1 by the rule 1! = 1 × 0!. Check the value every time 0! appears in a combination formula.
Treating (a + b)! as a! + b!, or (a − b)! as a! − b!. Fix: Factorial does not distribute. Compute a + b first, then take its factorial. Example: (2 + 3)! = 120, but 2! + 3! = 8.
Using nCr instead of nPr. Fix: Swap two items mentally. If the outcome changes, use nPr. Remember nPr = nCr × r!.
Forgetting the inside arrangement of the block in 'always together'. Fix: Always multiply by k! for the objects inside the block.
Using n! and forgetting to divide by the factorial of alike letters. Fix: Before using n!, scan the word for repeated letters. Write the counts of each letter next to the word.
Dividing by the number of repeated letters instead of its factorial, such as dividing by 4 instead of 4!. Fix: Alike items in a group of p can be shuffled in p! ways. Divide by p!, never by p.
Using n! for a round table. Fix: Fix one person first. Use (n − 1)!.
Forgetting to divide by 2 for necklaces and garlands. Fix: Read the object. If it can be turned over, use (n − 1)! ÷ 2.

Exam tips

Look for the words 'and' and 'or' in the question, but confirm by picturing the job, since the wording can mislead.
In number-formation questions, always check the leading zero and the even/odd condition before you multiply.
Work out the structure of the job and the count first. Use the options only as a sanity check afterwards, and do not remove options by estimating their size.
Mixed questions need both rules. Split into cases, multiply inside each case, then add the totals.
If a question needs long case-work and time is short, skip it and return later, since a wrong answer costs 0.25 marks.
Most questions are either 'simplify' or 'find n'. In both, the first move is to expand the larger factorial down to the smaller one and cancel.
Learn factorials up to 8! by heart. Options are often built from these values, so you can match an answer without full working.
If a 'find n' question gives numeric options, substituting small values like 3, 4, 5 is often faster than solving the quadratic.