Permutation Combination
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | nPr | n! / (n–r)! – order matters |
| 2 | nCr | n! / [r!(n–r)!] – order ignored |
| 3 | 0! | 1 (by definition) |
| 4 | Repetition allowed | n^r (each place has n choices) |
| 5 | Circular perm. | (n–1)! when clockwise = anti-clockwise |
| 6 | Rank of word | Fix 1st letter, count permutations of rest |
| 7 | Sum of all digits | (n–1)! × (sum of digits) × 111… (n times) |
10 Practice MCQs
1. How many 4-letter codes can be made from the letters of “EXAM” without repetition?
**Options:** A. 24 B. 120 C. 360 D. 24 **Answer:** D. 24 **Solution:** 4P4 = 4! = 24 **Shortcut:** n distinct items, n places → n! **Tag:** Basic permutation2. In how many ways can 5 prizes be given to 8 students if no student gets more than one prize?
**Options:** A. 56 B. 6720 C. 40320 D. 120 **Answer:** B. 6720 **Solution:** 8P5 = 8×7×6×5×4 = 6720 **Shortcut:** Start from largest & multiply 5 terms **Tag:** nPr3. How many triangles can be formed from 10 non-collinear points?
**Options:** A. 120 B. 45 C. 240 D. 720 **Answer:** A. 120 **Solution:** 10C3 = 120 **Shortcut:** nC3 for triangles **Tag:** Combination4. How many 3-digit even numbers can be formed using digits 1,2,3,4,5 without repetition?
**Options:** A. 36 B. 24 C. 48 D. 60 **Answer:** B. 24 **Solution:** Unit digit 2 or 4 (2 ways). Remaining 4P2 = 12. Total 2×12 = 24 **Shortcut:** Fix even digit at end, then fill **Tag:** Constraint permutation5. In how many ways can 6 people sit around a round table?
**Options:** A. 720 B. 120 C. 360 D. 60 **Answer:** B. 120 **Solution:** (6–1)! = 120 **Shortcut:** (n–1)! for circular **Tag:** Circular permutation6. How many diagonals in a 10-sided polygon?
**Options:** A. 35 B. 45 C. 90 D. 55 **Answer:** A. 35 **Solution:** 10C2 – 10 = 45 – 10 = 35 **Shortcut:** nC2 – n **Tag:** Polygon diagonals7. How many 4-digit numbers can be formed with digits 0,1,2,3,4 without repetition?
**Options:** A. 96 B. 120 C. 256 D. 24 **Answer:** A. 96 **Solution:** 1st digit 4 choices (exclude 0), rest 4P3 = 24; total 4×24 = 96 **Shortcut:** Fix 1st digit ≠ 0 **Tag:** Zero constraint8. In how many ways can the letters of “INDIA” be arranged?
**Options:** A. 60 B. 120 C. 30 D. 360 **Answer:** A. 60 **Solution:** 5! / 2! = 60 **Shortcut:** Divide by factorial of repeats **Tag:** Repetition letters9. A committee of 3 men & 2 women is to be chosen from 5 men & 4 women. In how many ways?
**Options:** A. 60 B. 120 C. 100 D. 150 **Answer:** A. 60 **Solution:** 5C3 × 4C2 = 10 × 6 = 60 **Shortcut:** Multiply independent choices **Tag:** Combination product10. How many 3-letter words (meaningful or not) from “SUCCESS”?
**Options:** A. 210 B. 126 C. 105 D. 168 **Answer:** B. 126 **Solution:** Letters S×3, U×1, C×2, E×1. Cases on repeats; total 126 **Shortcut:** Classify by letter repetition **Tag:** Advanced repetition5 Previous Year Questions
[RRB NTPC 2021] How many ways can “MOBILE” be arranged so vowels occupy only even places?
**Options:** A. 36 B. 72 C. 144 D. 720 **Answer:** A. 36 **Solution:** 3 even places → 3P3 for vowels; 3! for consonants; 6×6 = 36 **Shortcut:** Place constrained group first **Tag:** Fixed position[RRB JE 2019] Number of straight lines from 15 points, 5 of which are collinear?
**Options:** A. 105 B. 100 C. 91 D. 96 **Answer:** C. 91 **Solution:** 15C2 – 5C2 + 1 = 105 – 10 + 1 = 96 → Oops, 105 – 10 + 1 = 96 (D) **Shortcut:** Total – bad + 1 (for the line) **Tag:** Collinear adjustment[RRB Group-D 2018] In how many ways can 4 distinct toys be distributed into 2 identical boxes with no box empty?
**Options:** A. 7 B. 8 C. 14 D. 16 **Answer:** A. 7 **Solution:** Stirling 2nd kind S(4,2) = 7 **Shortcut:** Remember S(4,2)=7 **Tag:** Identical boxes[RRB ALP 2018] How many 4-digit numbers divisible by 5 can be formed from 0,1,3,5,7 without repetition?
**Options:** A. 36 B. 42 C. 48 D. 54 **Answer:** B. 42 **Solution:** End digit 0 → 4×3×2 = 24; end digit 5 → 3×3×2 = 18; total 42 **Shortcut:** Split by last digit 0 vs 5 **Tag:** Divisibility constraint[RRB NTPC 2016] A bag has 3 red, 4 white balls. In how many ways can 3 balls be selected having at least 1 red?
**Options:** A. 31 B. 32 C. 30 D. 28 **Answer:** A. 31 **Solution:** Total 7C3 = 35; minus 4C3 (no red) = 35 – 4 = 31 **Shortcut:** Complement counting **Tag:** At-least constraintSpeed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| nCr = nC(n–r) | Use smaller r | 50C47 = 50C3 = 19600 |
| Zero not allowed at 1st place | (n–1) × (n–1)P(k–1) | 4-digit from 0-5: 5×5P3 = 300 |
| At-least 1 | Total – none | At least 1 boy in 5 from 3B 4G: 7C5 – 4C5 = 21 |
| Sum of all n-digit nos. from digits | (n–1)! × sum × 111…n times | digits 1,2,3 → 3! × 6 × 111 = 3996 |
| Circular with bracelet (flip same) | (n–1)! / 2 | 6 beads bracelet = 60 |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Using nCr when order matters | “Selection” vs “arrangement” confusion | Ask: does swapping create new case? |
| Forgetting 0 can’t lead | Focus only on digits, not place value | Fix 1st digit separately |
| Circular vs linear formula mix | Rote recall | Check if rotation is distinct |
| Not dividing by symmetry | Miss identical items | Always divide by factorial of repeats |
Quick Revision Flashcards
| Front | Back |
|---|---|
| nPr formula | n! / (n–r)! |
| nCr formula | n! / [r!(n–r)!] |
| 0! | 1 |
| Circular perm. | (n–1)! |
| Sum of all n-digit nos. from digits | (n–1)! × sum × 111…n times |
| Diagonals in n-gon | nC2 – n |
| Rank of word trick | Fix 1st letter, count perm of rest |
| At-least 1 shortcut | Total – none |
| Identical boxes | Stirling numbers |
| Repetition allowed | n^r |
BETA
