In short: Use combinations (nCr) when the order of your chosen items does not matter, and permutations (nPr) when it does. The deciding question is always "would swapping the order give a different outcome?" In probability, the key rule is to count the numerator and denominator the same way — both with nCr or both with nPr — so the ordering convention cancels.
Deciding when to use combinations vs permutations in probability is the gateway skill for JEE Advanced counting problems, and most slip-ups happen before any arithmetic. Permutations (nPr = n!/(n−r)!) count arrangements, where order changes the outcome. Combinations (nCr = n!/(r!(n−r)!)) count selections, where order is irrelevant. Choose the wrong one and your sample space is the wrong size. The good news: in probability you have freedom, as long as you are consistent on top and bottom.
The reliable method
Use this decision procedure for every counting problem:
- Ask the order question. "If I rearrange the chosen items, is it a different outcome?" Yes → permutations. No → combinations.
- Spot the signal words. "Committee", "group", "selection", "team", "hand of cards" → combinations. "Arrangement", "ranking", "President and Secretary", "code", "seating" → permutations.
- Pick nCr or nPr. Selections use nCr = n!/(r!(n−r)!); arrangements use nPr = n!/(n−r)! = nCr × r!.
- Count favourable and total the same way. In probability, count the numerator and denominator with the same convention so the r! factors cancel.
- Simplify before multiplying out. Cancel common factorials early; JEE answers are usually clean fractions.
A worked example
A box contains 10 items, of which 4 are defective. Three items are drawn at random together. Find the probability that none of the three is defective.
Because the items are drawn "together", order does not matter — this is a selection, so use combinations consistently.
Total ways to choose 3 items from 10:
C(10,3) = 10!/(3!·7!) = (10·9·8)/(3·2·1) = 720/6 = 120
Favourable ways — choose all 3 from the 6 non-defective items:
C(6,3) = 6!/(3!·3!) = (6·5·4)/(3·2·1) = 120/6 = 20
Therefore:
P(none defective) = C(6,3) / C(10,3) = 20/120 = 1/6 ≈ 0.167
So the probability that none of the three drawn items is defective is 1/6. The method holds because both the favourable count and the total count use combinations: order is ignored on top and bottom, so the convention cancels and the ratio is correct. Had we needed an ordered outcome — say "first item good, then defective, then good" — we would switch both counts to permutations instead.
Common mistakes to avoid
- Using nCr on top and nPr on the bottom. Fix: keep the same convention in numerator and denominator so the r! factors cancel.
- Treating a "committee" as ordered. Fix: committees and hands are selections — use nCr, not nPr.
- Treating "President and Vice-President" as unordered. Fix: distinct roles make order matter — use nPr.
- Forgetting nPr = nCr × r!. Fix: if you counted selections but the roles are distinct, multiply by r! to restore the arrangements.
- Double-counting identical items. Fix: when objects repeat, divide by the factorials of the repeated groups to avoid over-counting arrangements.
Frequently asked questions
Should I use nCr or nPr when counting outcomes? Use nCr when order does not matter (selecting a group) and nPr when order matters (arranging or assigning distinct roles). Ask whether reordering the chosen items changes the outcome.
Does it matter whether I use combinations or permutations in a probability ratio? Not as long as you are consistent. If both the favourable and total counts use the same convention, the ordering factors cancel and you get the same probability.
What is the relationship between nCr and nPr? nPr = nCr × r!. Permutations equal combinations multiplied by the r! ways of ordering the chosen items.
When do "with replacement" problems change the count? With replacement, items can repeat, so you often use powers (like n^r) instead of nPr or nCr. Without replacement, use nPr or nCr depending on order.
How do repeated identical objects affect arrangements? Divide the total arrangements by the factorial of each repeated group's size. For example, arrangements of the letters in a word with repeats use n! divided by the repeats' factorials.
Practise this
Find out if this trips you up — [take the free diagnostic](/diagnostic), then work through the [free Socratic lesson](/lesson/start) to make the order question automatic.