8 questions from actual SSC CGL papers · all shown free · click option to reveal solution
• Permutation = Arrangement (ORDER matters) • Combination = Selection (ORDER does NOT matter) --- KEY RULES & PROPERTIES 1
n! = n × (n−1) × (n−2) × ... × 1 Example: 5! = 5 × 4 × 3 × 2 × 1 = 120 Special rule: 0! = 1 (memorise this — it is always asked) 2
nPr = n! / (n−r)! This gives arrangements of r items from n items. 3
nCr = n! / (r! × (n−r)!) This gives selections of r items from n items. 4
nPr = nCr × r
Permutation = Combination × arrangements of selected items 5
nCr = nC(n−r) Example: 10C3 = 10C7 — Use this to reduce calculations! 6. nC0 = nCn = 1 (selecting none or selecting all = 1 way) 7. nC1 = n (selecting 1 from n = n ways) ---
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— WHAT GETS ASKED • How many ways to arrange letters of a word (with/without repetition) • Selecting a committee of X people from Y people • Forming numbers using given digits • Sitting arrangements in a row or circle • Handshakes / matches in a tournament (always nC2) Circular Arrangement Formula: (n−1)! for n people sitting in a circle --- SHORTCUTS & TRICKS Trick 1 — Handshakes/Matches Shortcut: If n people shake hands with each other, total handshakes = nC2 = n(n−1)/2 Example: 10 people = 10×9/2 = 45 handshakes. Fast and clean. Trick 2 — Vowel/Consonant Grouping: To keep vowels together in a word arrangement — treat all vowels as ONE unit. Then arrange (remaining letters + 1 unit) and multiply by arrangements of vowels within the unit. Trick 3 — Complement Method: When 'at least one' is mentioned: At least 1 = Total ways − Ways with NONE selected This avoids long addition and saves time in exam. ---
Q: In how many ways can 5 students be arranged in a row? Solution: This is a pure arrangement — use Permutation. 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 Answer: 120 ways WORKED EXAMPLE 2 Q: A committee of 3 is to be selected from 6 people. How many ways? Solution: Order does not matter in a committee — use Combination. 6C3 = 6! / (3! × 3!) = (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20 Answer: 20 ways WORKED EXAMPLE 3 Q: How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 without repetition? Solution: First digit: 5 choices Second digit: 4 choices (one used) Third digit: 3 choices Total = 5 × 4 × 3 = 60 OR use 5P3 = 5!/(5−3)! = 120/2 = 60 Answer: 60 numbers ---
— #1 TRAP Students mix up
Permutation vs Combination. The golden test: Ask yourself — Does changing the ORDER give a NEW outcome? Yes → Permutation. No → Combination. Example: Selecting a President AND Secretary from 5 people = ORDER matters (different roles) = Permutation = 5P2 = 20. Selecting 2 members for a team = ORDER does NOT matter = Combination = 5C2 = 10. Many students use Combination here and get 10 instead of 20.
This is the most common wrong answer in SSC CGL.
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