📘 Chapter 1: Permutation and Combination | NEB Mathematics Notes Class 12
Complete solutions based on Fundamental Principle of Counting
🔢 Permutation and Combination forms the foundation of counting techniques. This chapter covers the Fundamental Principle of Counting, factorial notation, arrangements, selections, and digit-based problems. Below are detailed, line-by-line solutions to typical NEB examination problems — fully corrected with no spelling errors.
Problem 1
A football stadium has four entrance gates and nine exits. In how many different ways can a man enter and leave the stadium?
Solution (Line by line):
Total number of entrance gates = 4
Total number of exits = 9
By the Fundamental Principle of Counting:
Total number of ways = (number of ways to enter) × (number of ways to exit)
= 4 × 9 = 36 ways.
Total number of entrance gates = 4
Total number of exits = 9
By the Fundamental Principle of Counting:
Total number of ways = (number of ways to enter) × (number of ways to exit)
= 4 × 9 = 36 ways.
Problem 2
There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door?
Solution (Linewise):
Number of ways to enter = 6.
Since he must leave by a different door, number of ways to exit = 5.
By multiplication principle:
Total ways = 6 × 5 = 30 ways.
Number of ways to enter = 6.
Since he must leave by a different door, number of ways to exit = 5.
By multiplication principle:
Total ways = 6 × 5 = 30 ways.
Problem 3
In how many ways can a man send three of his children to seven different colleges of a certain town?
Solution:
The first child can be sent to any of 7 colleges.
The second child to any of the remaining 6 colleges (since colleges are distinct and a child can be assigned uniquely, but here children are distinct and each college can receive at most one? Actually standard assumption: three distinct children to 7 distinct colleges, each child gets one college, repetition allowed? Wait typical problem: “send three children to 7 different colleges” – implies each child chooses a college, and more than one child can go to same college? Usually without restriction it’s 7×7×7 = 343. But the original file had a mistake: “5x6x7 = 210” which suggests no repetition (all different colleges). So we correct: If no two children attend same college → 7 choices for 1st, 6 for 2nd, 5 for 3rd = 7×6×5 = 210. Let’s adopt the intended meaning (distinct colleges for each child).
Hence, total ways = 7 × 6 × 5 = 210 ways.
The first child can be sent to any of 7 colleges.
The second child to any of the remaining 6 colleges (since colleges are distinct and a child can be assigned uniquely, but here children are distinct and each college can receive at most one? Actually standard assumption: three distinct children to 7 distinct colleges, each child gets one college, repetition allowed? Wait typical problem: “send three children to 7 different colleges” – implies each child chooses a college, and more than one child can go to same college? Usually without restriction it’s 7×7×7 = 343. But the original file had a mistake: “5x6x7 = 210” which suggests no repetition (all different colleges). So we correct: If no two children attend same college → 7 choices for 1st, 6 for 2nd, 5 for 3rd = 7×6×5 = 210. Let’s adopt the intended meaning (distinct colleges for each child).
Hence, total ways = 7 × 6 × 5 = 210 ways.
Problem 4
There are five main roads between cities A and B. In how many ways can a man go from A to B and return by a different road?
Solution:
Ways to go from A to B = 5.
For return journey, road must be different from the one taken to go, so choices = 4.
Total ways = 5 × 4 = 20 ways.
Ways to go from A to B = 5.
For return journey, road must be different from the one taken to go, so choices = 4.
Total ways = 5 × 4 = 20 ways.
Problem 5
There are five main roads between A and B and four between B and C. In how many ways can a person drive from A to C and return without driving on the same road twice?
Solution (step by step):
From A to C: first A→B then B→C.
Number of ways from A to C = 5 × 4 = 20.
While returning from C to A without repeating any road used earlier:
For return path C→B: originally there were 4 roads from B to C, but one path (the one used in forward journey from B→C) cannot be used again → choices for C→B = 3.
Then B→A: originally 5 roads, but the specific road used from A→B cannot be reused → choices = 4.
Ways to return = 3 × 4 = 12.
Hence total round trip ways without repeating any road = (20) × (12) = 240 ways.
From A to C: first A→B then B→C.
Number of ways from A to C = 5 × 4 = 20.
While returning from C to A without repeating any road used earlier:
For return path C→B: originally there were 4 roads from B to C, but one path (the one used in forward journey from B→C) cannot be used again → choices for C→B = 3.
Then B→A: originally 5 roads, but the specific road used from A→B cannot be reused → choices = 4.
Ways to return = 3 × 4 = 12.
Hence total round trip ways without repeating any road = (20) × (12) = 240 ways.
Problem 6
How many numbers of at least three digits can be formed from the integers 1,2,3,4,5,6? (repetition not allowed, using distinct digits)
Solution:
“At least three digits” means 3-digit, 4-digit, 5-digit, or 6-digit numbers (all digits distinct, from given set {1…6}).
– 3-digit numbers: 6 × 5 × 4 = 120
– 4-digit numbers: 6 × 5 × 4 × 3 = 360
– 5-digit numbers: 6 × 5 × 4 × 3 × 2 = 720
– 6-digit numbers: 6 × 5 × 4 × 3 × 2 × 1 = 720
Total = 120 + 360 + 720 + 720 = 1920 numbers.
“At least three digits” means 3-digit, 4-digit, 5-digit, or 6-digit numbers (all digits distinct, from given set {1…6}).
– 3-digit numbers: 6 × 5 × 4 = 120
– 4-digit numbers: 6 × 5 × 4 × 3 = 360
– 5-digit numbers: 6 × 5 × 4 × 3 × 2 = 720
– 6-digit numbers: 6 × 5 × 4 × 3 × 2 × 1 = 720
Total = 120 + 360 + 720 + 720 = 1920 numbers.
Problem 7
How many three-digit numbers less than 500 can be formed from the integers 1,2,3,4,5,6? (digits distinct)
Solution:
For a 3-digit number to be less than 500, the hundreds digit can be 1,2,3 or 4 → 4 choices.
After choosing hundreds digit, remaining 5 digits available for tens place → 5 choices.
Then ones place: 4 remaining digits → 4 choices.
Total numbers = 4 × 5 × 4 = 80.
For a 3-digit number to be less than 500, the hundreds digit can be 1,2,3 or 4 → 4 choices.
After choosing hundreds digit, remaining 5 digits available for tens place → 5 choices.
Then ones place: 4 remaining digits → 4 choices.
Total numbers = 4 × 5 × 4 = 80.
Problem 8
Of the numbers formed by using all the figures 1,2,3,4,5 only once, how many are even?
Solution:
For a number to be even, the unit digit must be even → possible unit digits: 2 or 4 → 2 choices.
After fixing unit digit, the remaining 4 digits can be arranged in the first 4 positions (ten-thousands, thousands, hundreds, tens).
Number of arrangements = 4! = 4×3×2×1 = 24.
Hence total even numbers = 2 × 24 = 48.
For a number to be even, the unit digit must be even → possible unit digits: 2 or 4 → 2 choices.
After fixing unit digit, the remaining 4 digits can be arranged in the first 4 positions (ten-thousands, thousands, hundreds, tens).
Number of arrangements = 4! = 4×3×2×1 = 24.
Hence total even numbers = 2 × 24 = 48.
Problem 9
How many different 4-digit numbers between 4000 and 5000 can be formed with the digits 2,3,4,5,6,7? (Repetition not allowed)
Solution:
Numbers between 4000 and 5000 have thousands digit = 4 (only one choice, digit 4).
Remaining three digits are chosen from {2,3,5,6,7} — 5 digits available.
The number of ways to arrange any 3 distinct digits from these 5 in the hundreds, tens, units places = 5 × 4 × 3 = 60.
Therefore total numbers = 1 × 60 = 60.
Numbers between 4000 and 5000 have thousands digit = 4 (only one choice, digit 4).
Remaining three digits are chosen from {2,3,5,6,7} — 5 digits available.
The number of ways to arrange any 3 distinct digits from these 5 in the hundreds, tens, units places = 5 × 4 × 3 = 60.
Therefore total numbers = 1 × 60 = 60.
Problem 10
How many three-digit numbers can be formed from the integers 2,3,4,5,6? How many of them are divisible by 5? (digits distinct)
Solution (Part 1):
Total 3-digit numbers using digits 2,3,4,5,6 without repetition:
Choices: hundreds (5 ways) → tens (4 ways) → units (3 ways) = 5×4×3 = 60.
Part 2 (divisible by 5):
For divisibility by 5, unit digit must be 5 (only one choice).
Then hundreds digit: from remaining 4 digits → 4 ways.
Tens digit: from remaining 3 digits → 3 ways.
Numbers divisible by 5 = 4 × 3 × 1 = 12.
Total 3-digit numbers using digits 2,3,4,5,6 without repetition:
Choices: hundreds (5 ways) → tens (4 ways) → units (3 ways) = 5×4×3 = 60.
Part 2 (divisible by 5):
For divisibility by 5, unit digit must be 5 (only one choice).
Then hundreds digit: from remaining 4 digits → 4 ways.
Tens digit: from remaining 3 digits → 3 ways.
Numbers divisible by 5 = 4 × 3 × 1 = 12.
Problem 4 (variant)
Five main roads between cities A and B. In how many ways can a man go from A to B and return by a different road? (revisited)
Final clean correction: 5 ways to go, 4 ways to return (different road) → 5 × 4 = 20 ways. Consistent.
🧠Fundamental Principle of Counting (Permutation and Combination Basis)
If an event can occur in m different ways, and another independent event can occur in n different ways, then the total number of ways both events occur sequentially is m × n. This is the building block of Permutation and Combination problems, including factorials, arrangements, and selections. All problems above are solved using this principle with distinct objects (no repetition unless specified).
Important: Permutation (arrangement) nPr = n!/(n-r)!, Combination: nCr = n!/(r!(n-r)!).
✅ Line-by-line corrections from original PDF: Spelling errors fixed: “exits” corrected, “ways can a student enter/exit” grammar fixed, “Ten-thousand” uniformity, repetitive numbering removed. All mathematical expressions standardized, missing digits and incomplete solutions restored (e.g., Problem 6 original had incomplete expansion; fully corrected). Additionally, problem 9 and problem 10 now include complete reasoning.
📌 Important NEB Exam Questions Pattern
- Word formation / digit arrangement using permutation and combination rules.
- Problems with conditions like ‘greater than’, ‘less than’, ‘even/odd’, ‘divisible by 5’.
- Fundamental counting principle applied to real-life entrance/exit, travel routes.
- Distinct vs repetition allowed problems.
Mastery of Permutation and Combination is essential for scoring well in NEB Class 12 Mathematics. These solved examples cover typical exercise sets from Sukunda Pustak Bhawan.