Permutation, Combination & Probability — IBPS PO Quant

This is the topic that scares people out of proportion to its difficulty. In IBPS the questions are simple and formulaic — 1–3 in Mains — and the entire skill is answering one question first: does order matter? If yes, it's a permutation (arrangement); if no, a combination (selection). Probability is then just favourable outcomes ÷ total outcomes, where both are combination counts. Learn the handful of templates and the "order?" test, and this becomes reliable marks; over-think it and you lose time you don't have.

1. What IBPS actually asks

• Selection: "In how many ways can 3 men be chosen from 7?" (order doesn't matter → combination).
• Arrangement: "How many words from the letters of MONDAY?" (order matters → permutation).
• Probability: "A bag has 4 red, 5 blue balls; probability 2 drawn are red?"
• Common contexts: committees, teams, seating in a line, forming numbers/words, and balls from a bag.

2. The one decision: does order matter?

${}^n{P}_r=\frac{n!}{(n-r)!}(\text{order matters}),{}^n{C}_r=\frac{n!}{r!(n-r)!}(\text{order doesn’t})$

• Combination (selection): committees, teams, choosing balls — order irrelevant. ⁷C₃ = 35.
• Permutation (arrangement): seating, ranking, forming words/numbers — order matters. ⁷P₃ = 210.
• Relationship: ⁿPᵣ = ⁿCᵣ × r! (choose, then arrange).

Quick nCr values: ⁿC₀ = ⁿCₙ = 1, ⁿC₁ = n, ⁿC₂ = n(n−1)/2, and ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ (choosing r to include = choosing n−r to leave out — use the smaller).

3. Selection templates (most common)

• "At least / at most": add the valid cases. "At least 2 red from 4 red, 3 blue, choose 3": (2R1B) + (3R0B) = ⁴C₂·³C₁ + ⁴C₃ = 18 + 4 = 22.
• "Both/all from a group": multiply the group choices. "2 men from 5 AND 1 woman from 4": ⁵C₂ × ⁴C₁ = 40.
• "Either/or (mutually exclusive)": add the cases.

The rule of thumb: AND ⇒ multiply, OR ⇒ add.

4. Arrangement templates

• All distinct letters: n! ways. MONDAY (6 distinct) → 720.
• Repeated letters: divide by the factorial of each repeat. BANANA = 6!/(3!·2!) (3 A's, 2 N's) = 60.
• Circular arrangement: (n−1)! ways (fix one to break rotation).
• Together/not-together: treat "together" items as one block, arrange, then multiply by internal arrangements; "not together" = total − together.

5. Probability — favourable ÷ total

$P(E)=\frac{\text{favourable outcomes}}{\text{total outcomes}},0\le P\le 1$

• Both counts are usually combinations. "2 red from a bag of 4 red + 5 blue (9 balls)": P = ⁴C₂ / ⁹C₂ = 6/36 = 1/6.
• Complement: P(at least one) = 1 − P(none) — often far faster than summing cases.
• Independent events: multiply (P(A and B) = P(A)·P(B)). Mutually exclusive: add.
• Dice/coins: total outcomes = 6ⁿ / 2ⁿ; count favourable directly for small n.

7. Common traps

• Using permutation for a selection (or vice versa) — always answer "does order matter?" first.
• Forgetting to divide by repeats in word arrangements.
• Adding when you should multiply — AND ⇒ multiply, OR ⇒ add.
• Over-counting in "at least" — sum disjoint cases, or use the complement.
• Circular vs linear — circular is (n−1)!, not n!.

8. The protocol

1. Ask "does order matter?" → permutation (arrange) or combination (select).
2. AND ⇒ multiply, OR ⇒ add the case counts.
3. Word arrangements: n! ÷ (repeats!); circular = (n−1)!.
4. Probability = favourable ÷ total, both as combinations; use the complement for "at least one".
5. Keep it simple — IBPS PnC is basic; if the arithmetic explodes, you've mis-modelled the "order" question.