Permutation & Combination - Flashcards
Master the fundamental concepts of permutations and combinations with these comprehensive flashcards. These cards cover key formulas, problem-solving strategies, and real-world applications essential for Philippine entrance exams like UPCAT, CSE, and other major assessments. Focus on understanding when to use permutations vs combinations and how to apply the formulas correctly.
Flashcards
What is the key difference between permutation and combination?
Answer
Permutation considers the ORDER of arrangement (arrangement matters), while combination does NOT consider order (selection only matters). Example: ABC and BAC are different in permutation but same in combination.
What is the formula for permutation nPr?
Answer
nPr = n! / (n-r)!
Where:
- n = total number of objects
- r = number of objects selected
- ! = factorial
Example: 5P3 = 5! / (5-3)! = 5! / 2! = 60
What is the formula for combination nCr?
Answer
nCr = n! / (r!(n-r)!)
Where:
- n = total number of objects
- r = number of objects selected
- ! = factorial
Example: 5C3 = 5! / (3!×2!) = 120 / (6×2) = 10
What is factorial and how do you calculate it?
Answer
Factorial (n!) is the product of all positive integers from 1 to n.
Examples:
- 3! = 3 × 2 × 1 = 6
- 5! = 5 × 4 × 3 × 2 × 1 = 120
- 0! = 1 (by definition)
Used in both permutation and combination formulas.
When should you use PERMUTATION in a problem?
Answer
Use permutation when:
- ORDER matters
- Arrangement is important
- Keywords: arrange, order, sequence, ranking
Example: 'In how many ways can 5 students line up?' - Order matters, so use permutation.
When should you use COMBINATION in a problem?
Answer
Use combination when:
- ORDER does NOT matter
- Only selection is important
- Keywords: choose, select, pick, group
Example: 'Choose 3 students from 10 for a committee' - Order doesn't matter, so use combination.
How many ways can a director choose 3 of their 10 talented artists to take part in an art class?
Answer
This is a COMBINATION problem (choosing/selecting).
Given: n = 10, r = 3
10C3 = 10! / (3!(10-3)!)
= 10! / (3! × 7!)
= (10 × 9 × 8) / (3 × 2 × 1)
= 720 / 6 = 120 ways
In how many ways can the letters of 'ALPHA' be arranged?
Answer
This is a PERMUTATION problem (arrangement).
Since all letters are different:
5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
Alternatively, since we're using all letters, it's simply 5! = 120
What are consecutive numbers and give examples?
Answer
Consecutive numbers follow each other in order without skipping.
Examples:
- Consecutive integers: 1, 2, 3, 4, 5
- Consecutive even: 2, 4, 6, 8, 10
- Consecutive odd: 1, 3, 5, 7, 9
Pattern: If first number is n, then n, n+1, n+2, n+3...
For even/odd: n, n+2, n+4, n+6...
What is the difference between consecutive even and consecutive odd numbers?
Answer
Both consecutive even and consecutive odd numbers have a difference of 2 between adjacent numbers.
Consecutive even: 22, 24 → difference = 2
Consecutive odd: 11, 13 → difference = 2
For any two consecutive even numbers: difference = 2
For any two consecutive odd numbers: difference = 2
If the sum of two consecutive numbers is 9, find the smaller number.
Answer
Let x = smaller number, x+1 = larger number
Equation: x + (x+1) = 9
2x + 1 = 9
2x = 8
x = 4
Therefore, the smaller number is 4.
Check: 4 + 5 = 9 ✓
A security system uses a 4-letter password with no repeated letters. How many possible passwords are there?
Answer
This is a PERMUTATION problem (order matters for passwords).
Total letters in alphabet = 26
Selecting 4 letters without repetition:
26P4 = 26! / (26-4)! = 26! / 22!
= 26 × 25 × 24 × 23 = 358,800 possible passwords
Compare nPr and nCr. Which is larger and why?
Answer
nPr is ALWAYS larger than or equal to nCr for the same values of n and r.
Reason: nPr = n!/(n-r)! while nCr = n!/(r!(n-r)!)
Since nCr has an additional r! in the denominator, it's smaller.
Relationship: nPr = nCr × r!
Example: 5P3 = 60, 5C3 = 10, and 60 = 10 × 3!
In a national election, voters select 12 senators from 15 candidates. How many ways can this be done?
Answer
This is a COMBINATION problem (selection, order doesn't matter).
Given: n = 15, r = 12
15C12 = 15! / (12!(15-12)!) = 15! / (12! × 3!)
= (15 × 14 × 13) / (3 × 2 × 1)
= 2730 / 6 = 455 ways
What keywords indicate a COMBINATION problem?
Answer
Keywords for COMBINATION:
- Choose
- Select
- Pick
- Group
- Committee
- Team formation
Remember: These suggest that ORDER DOESN'T MATTER, only the selection itself matters.
What keywords indicate a PERMUTATION problem?
Answer
Keywords for PERMUTATION:
- Arrange
- Order
- Sequence
- Ranking
- Position
- Line up
- Password/codes
Remember: These suggest that ORDER MATTERS in the arrangement.
Sum of three consecutive numbers is 63. Find the highest number.
Answer
Let x = highest number
Then: (x-2) + (x-1) + x = 63
Solving: 3x - 3 = 63
3x = 66
x = 22
Highest number = 22
Check: 20 + 21 + 22 = 63 ✓
Five candidates run for president, vice president, and secretary. How many different combinations can fill these three positions?
Answer
This is a PERMUTATION problem because positions are different (order matters).
We're selecting 3 positions from 5 candidates:
5P3 = 5! / (5-3)! = 5! / 2!
= (5 × 4 × 3 × 2 × 1) / (2 × 1)
= 120 / 2 = 60 different combinations
What is the sum property of two consecutive numbers?
Answer
The sum of two consecutive numbers is ALWAYS an odd number.
Examples:
- 1 + 2 = 3 (odd)
- 5 + 6 = 11 (odd)
- 20 + 21 = 41 (odd)
This happens because: even + odd = odd, and consecutive numbers are always one even and one odd.
A company has 5 vacancies for junior auditor positions from 7 candidates. How many groups of five can be chosen?
Answer
This is a COMBINATION problem (choosing groups, order doesn't matter).
Given: n = 7, r = 5
7C5 = 7! / (5!(7-5)!) = 7! / (5! × 2!)
= (7 × 6) / (2 × 1)
= 42 / 2 = 21 different groups
Diagrams
Permutation & Combination Concept Overview
This mind map shows the complete structure of permutation and combination concepts, including formulas, when to use each method, keywords, and applications.
Problem-Solving Decision Tree
This flowchart guides students through the decision-making process of determining whether to use permutation or combination, including helpful keywords and formula application steps.