Permutations and Combinations for CA Foundation: Study Guide

What this chapter covers

Permutations and Combinations: topics in the order to study them

1. 1Fundamental Principles of CountingEvery later formula is built from the multiplication and addition rules, so learn them first.
2. 2Factorial NotationYou need to simplify n!, and expressions like n! ÷ (n − 2)!, before using any formula.
3. 3Permutations of Distinct ObjectsThis gives nPr and the basic idea of ordered arrangement, including cases with conditions.
4. 4Permutations with Repetition and Identical ObjectsIt extends arrangements to words with repeated letters, using division by factorials of the repeats.
5. 5Circular PermutationsIt builds on linear arrangements; you fix one position to remove rotations.
6. 6Combinations (nCr) and Its PropertiesOnce permutations are clear, nCr follows by dividing out the order, and its properties speed up calculation.
7. 7Applications of Combinations: Selection ProblemsThis is the final test: mixing selection conditions, committees and cases, using everything before it.

How to prepare Permutations and Combinations

1. Learn the multiplication and addition rules with small examples. Say aloud whether tasks happen together (multiply) or as alternatives (add).
2. Practise factorial simplification until you can cancel terms without writing everything out. For example, 8! ÷ 6! = 8 × 7 = 56.
3. Write the formulas nPr = n! ÷ (n − r)! and nCr = n! ÷ [r! × (n − r)!] and note that nPr = r! × nCr. Use this link to move between the two.
4. For every question, first write one line: order matters or not. Only then pick the formula.
5. Solve word-arrangement questions in three groups: all letters distinct, repeated letters, and conditions such as letters together or fixed positions. For letters together, treat the group as one unit, then arrange inside it.
6. Practise selection questions with conditions such as at least one, specific people included or excluded. Handle each as separate cases or use total minus unwanted.
7. Do timed MCQ sets. Use option elimination, such as checking whether the answer should be divisible by a factor. If a question needs a long case-by-case count, mark it and return later.

Common mistakes in Permutations and Combinations

• Using nPr when order does not matter, or nCr when it does.

  Fix: Ask first: if I swap two chosen items, is it a new outcome? If yes, use permutations. If no, use combinations.

• Forgetting to divide by factorials for repeated letters.

  Fix: Count each letter before starting. Divide n! by the factorial of every repeat count.

• Using n! instead of (n − 1)! for circular arrangements.

  Fix: Check for the words around a table or in a circle. Fix one person's position, then arrange the rest.

• Handling 'together' conditions incompletely.

  Fix: Multiply the outer arrangement by the inner arrangement. For example, for 5 people in a row with 3 specified people together, treat the 3 as one unit: (5 − 3 + 1)! × 3! = 3! × 3! = 6 × 6 = 36.

• Adding where the rule requires multiplying, or the reverse.

  Fix: Use multiply for steps done one after another, and add for separate cases that cannot occur together.

• Missing cases in 'at least' selection questions.

  Fix: Use total minus the unwanted case where possible. For at least one from n items, count all selections and subtract the empty one.

Last-day revision: Permutations and Combinations

• ✓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.

Permutations and Combinations practice questions

• A project manager needs to select 3 engineers from a team of 8 engineers to work on a critical task. In how many ways can this selection be …
• In how many ways can 5 people, Aarav, Bhavna, Chirag, Divya and Esha, be seated in a row if Aarav and Bhavna must sit next to each other?
• Six directors of a company in Pune are to be seated around a circular table for a board meeting. In how many distinct ways can they be seate…
• A surveyor marks 10 points on a plot map, of which exactly 4 lie on one straight line and no other three are collinear. How many triangles c…
• A password must contain exactly 8 characters: a mix of digits (0–9) and uppercase English letters (A–Z). The first character must be a lette…
• A shopkeeper in Jaipur distributes 10 identical laddus among 3 children so that each child gets at least one laddu. In how many ways can thi…
• How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 where no digit is repeated?
• How many distinct arrangements of the letters of the word COMMITTEE are there in which all the vowels stay together?