Permutations & Combinations is the chapter where 1st PUC students first have to think about counting rather than calculating — and it’s genuinely one of the most conceptually different chapters in the whole syllabus. There’s no algebraic manipulation to lean on here, no formula you can blindly substitute into without first understanding what the question is actually asking. Daniel Sir has watched the same pattern repeat every year since 1993: students who memorise "when to use nPr" and "when to use nCr" as a checklist, without understanding the underlying distinction, get the two backwards the moment a question is phrased even slightly unfamiliarly.
The One Question That Decides Everything
Before touching either formula, ask: does the order of selection matter? If arranging three people in a queue, ABC and BCA are genuinely different outcomes — order matters, so it’s a permutation. If choosing three people for a committee, {A, B, C} is the same committee regardless of the order they were selected in — order doesn’t matter, so it’s a combination. Every single P&C question in this chapter reduces to correctly answering this one question first.
Permutations — Counting Arrangements
nPr = n!/(n−r)! counts the number of ways to arrange r items chosen from n, where order matters. For example, arranging 3 of 6 books on a shelf: 6P3 = 6!/(6−3)! = 6!/3! = 6×5×4 = 120 — you multiply n, then n−1, then n−2, continuing r times total.
Combinations — Counting Selections
nCr = n!/(r!(n−r)!) counts the number of ways to select r items from n, where order doesn’t matter. Choosing 3 of 6 books to give away (not arranging them, just selecting which three): 6C3 = 6!/(3!3!) = 720/(6×6) = 20. Notice this is smaller than 6P3 — that’s not a coincidence. nCr is always nPr divided by r!, because combinations group together all the different orderings of the same selection into a single count.
Check your own nPr and nCr calculations instantly →
Worked Example — Distinguishing the Two in One Problem
A committee of 4 is to be selected from 7 people, and then a chairperson is to be chosen from those 4. How many total outcomes are possible? This is a two-step problem: first, selecting the committee (order doesn’t matter — combination): 7C4 = 35. Then, choosing a chairperson from the 4 selected (this is really just choosing 1 of 4, order irrelevant, but effectively picking which one — 4C1 = 4, or equivalently just 4 choices). Total: 35 × 4 = 140. Recognising that the first step is a combination and the second is a simple selection — not defaulting to a permutation for the whole problem — is exactly the skill this chapter is built to test.
The Factorial Foundation Underneath Both
Both formulas rest on n! = n×(n−1)×(n−2)×…×2×1, the number of ways to arrange all n items in a row. Understanding this as "n choices for the first position, n−1 for the second, and so on" — rather than a symbol to memorise — makes both nPr and nCr feel like natural extensions rather than two more formulas to keep separate.
Where Marks Are Actually Lost
Mistake 1 — Defaulting to permutation because it "feels more thorough." Students under exam pressure sometimes default to nPr because it produces a bigger, more impressive-looking number — but using it when order genuinely doesn’t matter overcounts every answer.
Mistake 2 — Not recognising multi-step problems need different methods per step. As shown above, a single question can require a combination for one part and something simpler for another — treating the whole problem with one formula throughout is a common structural error.
Mistake 3 — Forgetting to account for restrictions. Questions with conditions ("if two specific people must sit together," "if a particular item must be included") require adjusting the base count — students who apply the unrestricted formula and ignore the stated condition lose marks on an otherwise correct method.
Where This Chapter Leads
P&C is the direct foundation for the Binomial Theorem (the coefficients in a binomial expansion are literally combinations) and for Probability later in 1st PUC. A shaky grasp of nCr specifically makes both of those chapters noticeably harder than they need to be.
Frequently Asked Questions
Q: Is there a quick way to check whether I should use nPr or nCr?
Ask yourself: if I swapped the order of my selected items, would I have a genuinely different outcome? If yes, permutation. If it’s the same outcome either way, combination.
Q: Why is 0! defined as 1, not 0?
By convention and consistency — it’s the number of ways to arrange zero items, which is exactly one way (doing nothing), and defining it this way keeps the nPr and nCr formulas working correctly even at the boundary cases.
Coming Soon — Problem Vault, Permutations & Combinations: a full set of Karnataka board P&C questions, solved with the order-matters reasoning shown explicitly at every step. See what’s already in the Problem Vault →
See how Daniel Classes teaches 1st and 2nd PUC Mathematics →