Permutation & Combination - Beginner-Intermediate Level: exam-oriented practice BEGINNER-INTERMEDIATE

Step-by-Step Solution:

Concept: Geometrical combination with constraint. Three collinear points cannot form a triangle.

Strategy: Use complementary counting:
Total valid triangles = All possible triangles - Invalid triangles

Given:
- Total points: 12
- Collinear points: 4

Triangle Formation Rule: We need exactly 3 non-collinear points to form a triangle.

Step 1 - Calculate Total Possible Selections:
Selecting any 3 points from 12 points: C(12,3)
C(12,3) = 12! / [3! × 9!] = 220

Step 2 - Calculate Invalid Triangles:
3 collinear points don't form a triangle.
Selecting 3 points from 4 collinear points: C(4,3)
C(4,3) = 4! / [3! × 1!] = 4

Step 3 - Apply Complementary Counting:
Valid triangles = Total selections - Invalid selections
= 220 - 4
= 216

Key Technique: Complementary counting is often easier than direct counting when dealing with restrictions.

Verification: Answer should be less than C(12,3) = 220 since we have a constraint.

Related Concepts:
- For lines from n points: C(n,2) - (collinear points consideration)
- For quadrilaterals: C(n,4) with appropriate constraints

Step-by-Step Solution:

Concept: Fixing a specific digit at a specific position.

Given:
- Number length: 5
- Digit 2 fixed at position 4
- All digits distinct

Step 1 - Handle position 4:

Position 4 (not first): Fixed as 2 (1 choice)
- First digit: cannot be 0 and cannot be 2 (8 choices)
- Remaining 3 positions: choose from remaining 8 digits and arrange
- Ways = 8 × P(8, 3) = 8 × 336 = 2688


Calculation: 2688

Key Point: When fixing a digit in first position, it cannot be 0.

Step-by-Step Solution:

Concept: Inclusion-Exclusion Principle - when counting elements in the union of sets, add individual counts and subtract overcounted intersections.

Formula: |A ∪ B| = |A| + |B| - |A ∩ B|

Given:
- Range: 1 to 9
- Divisors: 3 and 5

Step 1 - Count numbers divisible by 3:
Numbers divisible by 3 = ⌊9/3⌋ = 3
These are: 3, 6, 9, ..., 9

Step 2 - Count numbers divisible by 5:
Numbers divisible by 5 = ⌊9/5⌋ = 1
These are: 5, 10, 15, ..., 5

Step 3 - Count numbers divisible by BOTH 3 AND 5:
Numbers divisible by LCM(3,5) = 15
Count = ⌊9/15⌋ = 0
(These are counted twice in steps 1 and 2)

Step 4 - Apply Inclusion-Exclusion:
Total = (divisible by 3) + (divisible by 5) - (divisible by both)
= 3 + 1 - 0
= 4

Visualization (Venn Diagram concept):
- Circle A: divisible by 3 (3 numbers)
- Circle B: divisible by 5 (1 numbers)
- Intersection: divisible by both (0 numbers)
- Union: 4 numbers

Key Principle: Subtract the intersection to avoid double counting.

Extension to Three Sets:
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|

Common Application: Finding numbers NOT divisible by either = 9 - 4 = 5

Step-by-Step Solution:

Concept: Combination with constraints - selection from multiple groups with specific requirements.

Given:
- Men available: 6
- Women available: 5
- Committee size: 5
- Required: 2 men and 3 women

Strategy: Select from each group independently, then multiply (Multiplication Principle).

Step 1 - Select Men:
Choose 2 men from 6 men = C(6,2)
C(6,2) = 6! / [2! × 4!] = 15

Step 2 - Select Women:
Choose 3 women from 5 women = C(5,3)
C(5,3) = 5! / [3! × 2!] = 10

Step 3 - Apply Multiplication Principle:
Total ways = C(6,2) × C(5,3)
= 15 × 10
= 150

Key Principle: When selecting from different independent groups with specific requirements from each:
- Calculate selections from each group separately
- Multiply the results

Common Error: Don't add the combinations - multiply them! Each selection from one group can be paired with each selection from the other.

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y+z)^6
- Desired term: the term y^5 * z^1
- Exponents: x = 0, y = 5, z = 1

Step 1 - Verify exponent sum:
0 + 5 + 1 = 6 = 6 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{6!}{5! × 1!}$

Step 3 - Calculate:
- Numerator: 6! = 720
- Denominator: 5! × 1! = 120 × 1
- Denominator value: 120

Final Calculation:
Coefficient = 720 / 120 = 6

Alternative interpretation: This equals the number of ways to arrange 6 items with:
0 of type x, 5 of type y, 1 of type z

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 3^6 = 729

Step-by-Step Solution (Sum Rule):

Concept: 'At least' problems require finding the sum of ways for all valid, mutually exclusive cases.

Given:
- Men available: 10
- Women available: 7
- Committee size: 6
- Constraint: At least 4 men

Strategy: We sum the ways for all cases from exactly 4 men up to the maximum possible number of men (6).

Valid Cases (Men, Women) and Calculation:

(4 Men, 2 Women): C(10,4) × C(7,2) = 210 × 21 = 4410
(5 Men, 1 Women): C(10,5) × C(7,1) = 252 × 7 = 1764
(6 Men, 0 Women): C(10,6) × C(7,0) = 210 × 1 = 210

Final Calculation (Sum Rule):
Total ways = (Ways with 4 men) + (Ways with 5 men) + ...
= 4410 + 1764 + 210
= 6384

Key Principle: Use the Sum Rule (addition) because the cases are mutually exclusive (you cannot simultaneously select exactly $k$ men and exactly $j$ men, where $k
e j$).

Step-by-Step Solution:

Concept: Geometrical combination with constraint. Three collinear points cannot form a triangle.

Strategy: Use complementary counting:
Total valid triangles = All possible triangles - Invalid triangles

Given:
- Total points: 9
- Collinear points: 4

Triangle Formation Rule: We need exactly 3 non-collinear points to form a triangle.

Step 1 - Calculate Total Possible Selections:
Selecting any 3 points from 9 points: C(9,3)
C(9,3) = 9! / [3! × 6!] = 84

Step 2 - Calculate Invalid Triangles:
3 collinear points don't form a triangle.
Selecting 3 points from 4 collinear points: C(4,3)
C(4,3) = 4! / [3! × 1!] = 4

Step 3 - Apply Complementary Counting:
Valid triangles = Total selections - Invalid selections
= 84 - 4
= 80

Key Technique: Complementary counting is often easier than direct counting when dealing with restrictions.

Verification: Answer should be less than C(9,3) = 84 since we have a constraint.

Related Concepts:
- For lines from n points: C(n,2) - (collinear points consideration)
- For quadrilaterals: C(n,4) with appropriate constraints

Step-by-Step Solution:

Concept: Derangement - permutation where no element appears in its original position. Denoted as D(n) or !n.

Given:
- Number of items: 5
- Constraint: No item in its correct position

Derangement Formula:
D(n) = n! × [1 - 1/1! + 1/2! - 1/3! + ... + (-1)ⁿ/n!]

Or recursively: D(n) = (n-1)[D(n-1) + D(n-2)]

Step-by-Step Calculation:
D(0) = 1 (convention)
D(1) = 0 (one item must be in its position)
D(2) = 1 (only one swap possible)

For n ≥ 3, we use: D(n) = (n-1)[D(n-1) + D(n-2)]

Let's calculate:
D(0) = 1
D(1) = 0
D(2) = 1
D(3) = (3-1)[D(2) + D(1)] = 2 × [1 + 0] = 2
D(4) = (4-1)[D(3) + D(2)] = 3 × [2 + 1] = 9
D(5) = (5-1)[D(4) + D(3)] = 4 × [9 + 2] = 44

Answer: D(5) = 44

Intuitive Understanding:
Total arrangements = 5! = 120
Derangements = 44
Probability of derangement ≈ 1/e ≈ 0.368 (for large n)

Real-World Application:
- Secret Santa where no one gets their own name
- Permutation ciphers in cryptography
- Hat-check problem in probability

Key Formula Memory Aid:
D(n) ≈ n!/e for large n (within 1 unit)
For 5: 120/e ≈ 44 ≈ 44

Common Error: Don't subtract n! - n, that's not derangement count.

Step-by-Step Solution:

Concept: Geometrical combination with constraint. Three collinear points cannot form a triangle.

Strategy: Use complementary counting:
Total valid triangles = All possible triangles - Invalid triangles

Given:
- Total points: 8
- Collinear points: 3

Triangle Formation Rule: We need exactly 3 non-collinear points to form a triangle.

Step 1 - Calculate Total Possible Selections:
Selecting any 3 points from 8 points: C(8,3)
C(8,3) = 8! / [3! × 5!] = 56

Step 2 - Calculate Invalid Triangles:
3 collinear points don't form a triangle.
Selecting 3 points from 3 collinear points: C(3,3)
C(3,3) = 3! / [3! × 0!] = 1

Step 3 - Apply Complementary Counting:
Valid triangles = Total selections - Invalid selections
= 56 - 1
= 55

Key Technique: Complementary counting is often easier than direct counting when dealing with restrictions.

Verification: Answer should be less than C(8,3) = 56 since we have a constraint.

Related Concepts:
- For lines from n points: C(n,2) - (collinear points consideration)
- For quadrilaterals: C(n,4) with appropriate constraints

Step-by-Step Solution (Rencontres Numbers):

Concept: This is a partial derangement problem. We want exactly k fixed points, with the remaining n-k elements deranged.

Formula:
$$D(n, k) = \binom{n}{k} \cdot !(n-k)$$
where $!(m)$ is the derangement number.

Given:
- n = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

Step 2 - Derange the remaining elements:
Remaining 4 elements must have NO fixed points
Derangement number D(4) = 9

Step 3 - Apply multiplication principle:
Total = C(5,1) × D(4)
= 5 × 9
= 45

Verification:
- When k=0: D(n,0) = derangement(n) ✓
- When k=n-1: D(n,n-1) = 0 (can't have all but one fixed) ✓
- When k=n: D(n,n) = 1 (identity permutation) ✓

Rencontres numbers table for n=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Step-by-Step Solution (Rencontres Numbers):

Concept: This is a partial derangement problem. We want exactly k fixed points, with the remaining n-k elements deranged.

Formula:
$$D(n, k) = \binom{n}{k} \cdot !(n-k)$$
where $!(m)$ is the derangement number.

Given:
- n = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

Step 2 - Derange the remaining elements:
Remaining 4 elements must have NO fixed points
Derangement number D(4) = 9

Step 3 - Apply multiplication principle:
Total = C(5,1) × D(4)
= 5 × 9
= 45

Verification:
- When k=0: D(n,0) = derangement(n) ✓
- When k=n-1: D(n,n-1) = 0 (can't have all but one fixed) ✓
- When k=n: D(n,n) = 1 (identity permutation) ✓

Rencontres numbers table for n=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Step-by-Step Solution:

Concept: Derangement - permutation where no element appears in its original position. Denoted as D(n) or !n.

Given:
- Number of items: 5
- Constraint: No item in its correct position

Derangement Formula:
D(n) = n! × [1 - 1/1! + 1/2! - 1/3! + ... + (-1)ⁿ/n!]

Or recursively: D(n) = (n-1)[D(n-1) + D(n-2)]

Step-by-Step Calculation:
D(0) = 1 (convention)
D(1) = 0 (one item must be in its position)
D(2) = 1 (only one swap possible)

For n ≥ 3, we use: D(n) = (n-1)[D(n-1) + D(n-2)]

Let's calculate:
D(0) = 1
D(1) = 0
D(2) = 1
D(3) = (3-1)[D(2) + D(1)] = 2 × [1 + 0] = 2
D(4) = (4-1)[D(3) + D(2)] = 3 × [2 + 1] = 9
D(5) = (5-1)[D(4) + D(3)] = 4 × [9 + 2] = 44

Answer: D(5) = 44

Intuitive Understanding:
Total arrangements = 5! = 120
Derangements = 44
Probability of derangement ≈ 1/e ≈ 0.368 (for large n)

Real-World Application:
- Secret Santa where no one gets their own name
- Permutation ciphers in cryptography
- Hat-check problem in probability

Key Formula Memory Aid:
D(n) ≈ n!/e for large n (within 1 unit)
For 5: 120/e ≈ 44 ≈ 44

Common Error: Don't subtract n! - n, that's not derangement count.

Step-by-Step Solution:

Concept: For necklaces with identical beads, we use Burnside's Lemma.

Given distribution: [4, 2, 4]

Result: 181440 distinct necklaces.

Note: Full calculation requires summing over all rotation and reflection symmetries, which is complex for general cases.

Step-by-Step Solution (Rencontres Numbers):

Concept: This is a partial derangement problem. We want exactly k fixed points, with the remaining n-k elements deranged.

Formula:
$$D(n, k) = \binom{n}{k} \cdot !(n-k)$$
where $!(m)$ is the derangement number.

Given:
- n = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

Step 2 - Derange the remaining elements:
Remaining 4 elements must have NO fixed points
Derangement number D(4) = 9

Step 3 - Apply multiplication principle:
Total = C(5,1) × D(4)
= 5 × 9
= 45

Verification:
- When k=0: D(n,0) = derangement(n) ✓
- When k=n-1: D(n,n-1) = 0 (can't have all but one fixed) ✓
- When k=n: D(n,n) = 1 (identity permutation) ✓

Rencontres numbers table for n=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Step-by-Step Solution:

Concept: Combination with mandatory inclusion constraint.

Given:
- Total people: 10
- Committee size: 6
- Must include: 3 specific people

Strategy: Fix the mandatory selections first, then choose remaining from available pool.

Analysis:
We need to select 6 people total, with 3 already fixed.
- Fixed positions: 3 (these specific people are already in)
- Remaining positions to fill: 6 - 3 = 3
- People available for remaining positions: 10 - 3 = 7

Step 1 - Fix Mandatory Members:
3 specific people must be included: C(3,3) = 1 way
(This is automatic - we have no choice here)

Step 2 - Select Remaining Members:
Choose 3 people from remaining 7 people:
C(7,3) = 35

Calculation:
C(7,3) = (7)! / [3! × (4)!]
= 5040 / [6 × 24]
= 35

Alternative Approach - Verification:
Think of it as: "We've used 3 spots, now choose 3 more from 7 remaining"

Related Problem Types:

1. Must EXCLUDE specific people:
Select all 6 from remaining 10 - (people to exclude)

2. At least one specific person:
Total ways - Ways without that person
= C(10,6) - C(10-1,6)

3. Exactly k from group A, rest from group B:
C(|A|,k) × C(|B|,6-k)

Common Error: Don't forget to reduce both the total pool and the selection size by the number of mandatory inclusions.

Answer: 35 ways

Step-by-Step Solution:

Concept: Multinomial theorem expansion:
$$(x_1 + x_2 + ... + x_k)^n = \sum_{a_1+...+a_k=n} \frac{n!}{a_1! a_2! ... a_k!} x_1^{a_1} x_2^{a_2} ... x_k^{a_k}$$

Given:
- Expression: (x+y)^4
- Desired term: the term x^3 * y^1
- Exponents: x = 3, y = 1

Step 1 - Verify exponent sum:
3 + 1 = 4 = 4 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{4!}{3! × 1!}$

Step 3 - Calculate:
- Numerator: 4! = 24
- Denominator: 3! × 1! = 6 × 1
- Denominator value: 6

Final Calculation:
Coefficient = 24 / 6 = 4

Alternative interpretation: This equals the number of ways to arrange 4 items with:
3 of type x, 1 of type y

Key Principle: Multinomial coefficients generalize binomial coefficients:
- Binomial: C(n, k) = n!/(k!(n-k)!)
- Multinomial: n!/(a! b! c! ...) where a+b+c+... = n

Quick Check: The sum of all multinomial coefficients for given n is k^n = 2^4 = 16

Step-by-Step Solution:

Concept: This problem uses the Fundamental Counting Principle (Multiplication Rule). When making sequential independent choices, multiply the number of options at each step.

Analysis:
- First choice: 4 options
- Second choice: 6 options
- Third choice: 3 options

Calculation:
Total ways = 4 × 6 × 3 = 72

Key Principle: For independent sequential events, multiply the number of choices at each stage.

Verification: Each of the 4 first choices can be paired with each of the 6 second choices (4×6=24), and each of these can be combined with any of the 3 third choices.

Step-by-Step Solution:

Concept: Group division where groups are indistinguishable (no labels like Team A, Team B).

Given:
- Total people: 8
- Group size: 4
- Number of groups: 2

Key Question: Are the groups distinguishable?
- If groups have labels (Team A, Team B): Groups are distinguishable
- If groups have no labels: Groups are indistinguishable (our case)

Strategy for Indistinguishable Groups:

Step 1 - Select first group:
Choose 4 people from 8: C(8,4)
C(8,4) = 8!/[4! × 4!] = 70

Step 2 - Remaining people form second group:
Remaining 4 people automatically form the other group: C(4,4) = 1

Step 3 - Remove overcounting:
Since groups are indistinguishable, we've counted each division twice.
(Selecting ABCD for group 1 and EFGH for group 2 is same as selecting EFGH for group 1 and ABCD for group 2)

Divide by 2! = 2

Calculation:
Total ways = C(8,4) / 2!
= 70 / 2
= 35

General Formula:
For dividing n items into k equal groups of size m each (where n = k×m):
Ways = C(n,m) × C(n-m,m) × ... × C(m,m) / k!
= n! / [(m!)^k × k!]

For our case:
= 8! / [(4!)^2 × 2!]
= 40320 / [24^2 × 2]
= 40320 / [576 × 2]
= 35

Contrast:
- Distinguishable groups (labeled teams): 70 ways
- Indistinguishable groups (unlabeled): 35 ways

Common Error: Forgetting to divide by k! when groups are indistinguishable.

Step-by-Step Solution:

Concept: Permutation with restriction - specific position must have certain type of letter.

Strategy: Fix the restricted position first, then arrange the remaining letters.

Analysis of 'EQUATION':
- Total letters: 8
- Vowels: E, U, A, I, O = 5 vowels
- First position must be a vowel

Step 1 - Fix First Position:
Choose a vowel for first position: 5 choices

Step 2 - Arrange Remaining:
Remaining 7 letters can be arranged in 7! ways
7! = 5040

Calculation:
Total arrangements = 5 × 5040 = 10080

Key Strategy: When dealing with restrictions:
1. Handle the restriction first (fix the constrained position)
2. Arrange the remaining elements freely
3. Multiply the results

Verification: This should be less than the total arrangements (8! = 40320) since we've added a constraint.

Step-by-Step Solution:

Concept: Derangement - permutation where no element appears in its original position. Denoted as D(n) or !n.

Given:
- Number of items: 4
- Constraint: No item in its correct position

Derangement Formula:
D(n) = n! × [1 - 1/1! + 1/2! - 1/3! + ... + (-1)ⁿ/n!]

Or recursively: D(n) = (n-1)[D(n-1) + D(n-2)]

Step-by-Step Calculation:
D(0) = 1 (convention)
D(1) = 0 (one item must be in its position)
D(2) = 1 (only one swap possible)

For n ≥ 3, we use: D(n) = (n-1)[D(n-1) + D(n-2)]

Let's calculate:
D(0) = 1
D(1) = 0
D(2) = 1
D(3) = (3-1)[D(2) + D(1)] = 2 × [1 + 0] = 2
D(4) = (4-1)[D(3) + D(2)] = 3 × [2 + 1] = 9

Answer: D(4) = 9

Intuitive Understanding:
Total arrangements = 4! = 24
Derangements = 9
Probability of derangement ≈ 1/e ≈ 0.368 (for large n)

Real-World Application:
- Secret Santa where no one gets their own name
- Permutation ciphers in cryptography
- Hat-check problem in probability

Key Formula Memory Aid:
D(n) ≈ n!/e for large n (within 1 unit)
For 4: 24/e ≈ 9 ≈ 9

Common Error: Don't subtract n! - n, that's not derangement count.