Permutation and Combination
Overview
Permutation and Combination forms the foundation of counting problems in competitive exams.
The core distinction is simple: Permutation counts arrangements where order matters (who stands first vs second), while Combination counts selections where order doesn't matter (choosing 3 people from 5). SBI Clerk focuses on basic applications—committee formation, word arrangements, and selection problems—rather than complex scenarios involving restrictions.
Students who grasp the fundamental formulas and practice recognizing whether a problem needs "arrangement" or "selection" can quickly secure these marks. The calculations involved are straightforward once you identify the correct approach.
Key Concepts
- Fundamental Counting Principle: If task A can be done in 'm' ways and task B in 'n' ways, both tasks together can be done in m × n ways (for AND) or m + n ways (for OR).
- Factorial (n!): Product of all positive integers from 1 to n. Example: 5! = 5 × 4 × 3 × 2 × 1 = 120. Special case: 0! = 1.
- Permutation (nPr): Number of ways to arrange 'r' items from 'n' distinct items. Order matters—ABC is different from BAC.
- Combination (nCr): Number of ways to select 'r' items from 'n' distinct items. Order doesn't matter—selecting A, B, C is same as selecting B, C, A.
- Key relationship: nPr = nCr × r! (Every combination can be arranged in r! ways to get permutations).
- Complementary property: nCr = nC(n-r). Choosing 3 items from 10 equals choosing 7 items to leave out.
- Repeated items in permutation: When arranging n items where some are identical, divide by factorials of repeated counts.
Formulas / Key Facts
Permutation formula: nPr = n! / (n-r)!
Combination formula: nCr = n! / [r! × (n-r)!]
Permutation with repetition allowed: n^r (each of r positions can be filled in n ways)
Arrangement of n items with repetitions: n! / (p! × q! × r!...) where p, q, r are counts of identical items
Quick values to memorize:
- 5! = 120
- 6! = 720
- 7! = 5040
- 5C2 = 10
- 6C2 = 15
- 6C3 = 20
- 10C2 = 45
- 10C3 = 120
Selection with "at least one" condition: Total selections − selections with none = selections with at least one
Worked Examples
Example 1: Basic Permutation How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
Step 1: We need to fill 3 positions using 5 digits, and order matters (123 ≠ 321). Step 2: This is permutation: 5P3 = 5!/(5-3)! = 5!/2! = 120/2 = 60
Answer: 60
Example 2: Basic Combination In how many ways can a committee of 3 members be formed from 8 people?
Step 1: We're selecting 3 from 8. Order doesn't matter (selecting A, B, C same as B, A, C). Step 2: This is combination: 8C3 = 8!/(3! × 5!) Step 3: = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56
Answer: 56
Example 3: Arrangement with Repeated Letters How many different words can be formed using all letters of "BANANA"?
Step 1: Total letters = 6 (B=1, A=3, N=2) Step 2: If all were distinct: 6! arrangements Step 3: Divide by repetitions: 6!/(3! × 2! × 1!) = 720/(6 × 2 × 1) = 720/12 = 60
Answer: 60
Example 4: Selection with Conditions From 5 men and 4 women, a committee of 4 is to be formed with at least 2 women. How many ways?
Step 1: "At least 2 women" means 2W+2M or 3W+1M or 4W+0M Step 2: Case 1 (2W, 2M): 4C2 × 5C2 = 6 × 10 = 60 Step 3: Case 2 (3W, 1M): 4C3 × 5C1 = 4 × 5 = 20 Step 4: Case 3 (4W, 0M): 4C4 × 5C0 = 1 × 1 = 1 Step 5: Total = 60 + 20 + 1 = 81
Answer: 81
Common Mistakes
Wrong: Using permutation when combination is needed. Fix: Ask yourself—does rearranging the same items create a different outcome? If selecting people for a committee, order doesn't matter (use nCr). If arranging people in a line, order matters (use nPr).
Wrong: Forgetting to account for repeated elements when calculating arrangements. Fix: When letters/digits repeat, always divide total factorial by the factorial of each repeated element's count.
Wrong: Calculating nCr by expanding full factorials. Fix: Use shortcut: for 10C3, write (10 × 9 × 8)/(3 × 2 × 1). Take only 'r' terms in numerator starting from n.
Wrong: Adding cases when you should multiply (or vice versa). Fix: "AND" situations = multiply. "OR" situations = add. Forming a committee needs member 1 AND member 2 (multiply). Having 2 women OR 3 women means either works (add).
Wrong: Double-counting in problems with restrictions. Fix: When items must be together, treat them as a single unit first, then arrange within the unit.
Quick Reference
- Order matters → Permutation (nPr); Order doesn't matter → Combination (nCr)
- nPr = n!/(n-r)! and nCr = n!/[r!(n-r)!]
- Repeated letters: Divide n! by factorial of each repetition count
- nCr = nC(n-r) — use smaller value for faster calculation
- At least one = Total − None
- AND = Multiply, OR = Add