Permutation & Combination - Intermediate Level: multi-step reasoning INTERMEDIATE

Step-by-Step Solution:

Concept: Partitioning into groups of specified sizes. When group sizes are different, groups are automatically distinguishable by size.

Given:
- Total people: 11
- Group sizes: 6, 4, 1
- Groups are unlabeled (no names like Team A, Team B)

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$

Step 1 - Choose first group:
Choose 6 people from 11: C(11, 6) = 462

Step 2 - Choose second group:
From remaining 5 people, choose 4: C(5, 4) = 5

Continue for all groups:
C(11,6) = 462
C(5,4) = 5
C(1,1) = 1 (last group)

Step 3 - Multiply:
Total ways = 462 × 5 × 1 (last group)
= 2310

Simplified formula:
= 11! / (6! × 4! × 1!)
= 39916800 / (720 × 24 × 1)
= 2310

Key Insight: Since groups have different sizes, we don't divide by k! (they're naturally distinguishable by their sizes).

Contrast with equal groups:
- If groups were same size: would divide by k!
- Here, sizes differ: no division needed

Verification: Sum of group sizes = 11 = 11 ✓

Step-by-Step Solution:

Concept: Counting numbers with distinct digits and divisibility constraint.

Given:
- Number length: 5 digits
- Digits must be distinct
- Constraint: Divisible by 9

Step 1 - Determine last digit constraint:
Divisible by 9 means:
Special divisibility rules apply

Step 2 - Count valid numbers:
For divisor 9, the count is 3024

Step 3 - Verify distinctness:
All digits in the number are different (no repetition allowed).

Calculation: 3024

Key Principle: When forming numbers with distinct digits:
- First digit cannot be 0
- Handle last digit constraints first
- Use permutations for remaining positions

Step-by-Step Solution:

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

Given distribution: [3, 4]

Result: 360 distinct necklaces.

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

Step-by-Step Solution:

Concept: Circular Permutation with reflection symmetry. This is used for arrangements like necklaces or keyrings where flipping the arrangement produces the same result (Clockwise = Anticlockwise).

Formula: $\text{Total Ways} = \frac{(n-1)!}2$

Analysis:
- Total items ($n$): 5
- Step 1: Normal circular arrangements (rotations same) = $(n-1)!$ = 24
- Step 2: Account for reflection (flips) by dividing by 2.

Calculation:
Arrangements = $\frac{(5-1)!}2$
= $\frac{24}2$
= 12

Formula Summary:
- Linear: $n!$
- Circular (no reflection): $(n-1)!$
- Circular (with reflection): $\frac{(n-1)!}2$

Key Principle: Dividing by 2 removes the overcounting caused by the symmetry when the arrangement can be flipped.

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: 8
- Committee size: 7
- Constraint: At least 3 men

Strategy: We sum the ways for all cases from exactly 3 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(3 Men, 4 Women): C(10,3) × C(8,4) = 120 × 70 = 8400
(4 Men, 3 Women): C(10,4) × C(8,3) = 210 × 56 = 11760
(5 Men, 2 Women): C(10,5) × C(8,2) = 252 × 28 = 7056
(6 Men, 1 Women): C(10,6) × C(8,1) = 210 × 8 = 1680
(7 Men, 0 Women): C(10,7) × C(8,0) = 120 × 1 = 120

Final Calculation (Sum Rule):
Total ways = (Ways with 3 men) + (Ways with 4 men) + ...
= 8400 + 11760 + 7056 + 1680 + 120
= 29016

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 (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 = 6 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 6: C(6,1) = 6

Step 2 - Derange the remaining elements:
Remaining 5 elements must have NO fixed points
Derangement number D(5) = 44

Step 3 - Apply multiplication principle:
Total = C(6,1) × D(5)
= 6 × 44
= 264

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=6:
- D(6,0) = 265
- D(6,1) = 264
- D(6,2) = 135
- ... and so on.

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

Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 5-digit number).

Given:
- Number length: 5 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 4-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 5th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (4 times)
= 9 × 10^4
= 9 × 10000
= 90000

Alternative Verification:
- Smallest 5-digit number: 10000 = 10000
- Largest 5-digit number: 99999 = 99999
- Total count: 99999 - 10000 + 1 = 90000

Related Problems:
1. No repetition: 9 × P(9,4) = 9 × 9!/5!
2. Odd numbers only: 9 × 10^3 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^3 × 1
- If last digit 2,4,6,8: 8 × 10^3 × 4
- Total: 9 × 10^3 + 8 × 4 × 10^3

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

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 8
- Divisors: 2 and 5

Step 1 - Count numbers divisible by 2:
Numbers divisible by 2 = ⌊8/2⌋ = 4
These are: 2, 4, 6, ..., 8

Step 2 - Count numbers divisible by 5:
Numbers divisible by 5 = ⌊8/5⌋ = 1
These are: 5, 10, 15, ..., 5

Step 3 - Count numbers divisible by BOTH 2 AND 5:
Numbers divisible by LCM(2,5) = 10
Count = ⌊8/10⌋ = 0
(These are counted twice in steps 1 and 2)

Step 4 - Apply Inclusion-Exclusion:
Total = (divisible by 2) + (divisible by 5) - (divisible by both)
= 4 + 1 - 0
= 5

Visualization (Venn Diagram concept):
- Circle A: divisible by 2 (4 numbers)
- Circle B: divisible by 5 (1 numbers)
- Intersection: divisible by both (0 numbers)
- Union: 5 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 = 8 - 5 = 3

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: 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: 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 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 3 collinear points: C(3,3)
C(3,3) = 3! / [3! × 0!] = 1

Step 3 - Apply Complementary Counting:
Valid triangles = Total selections - Invalid selections
= 84 - 1
= 83

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: 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)^7
- Desired term: the term y^7
- Exponents: x = 0, y = 7, z = 0

Step 1 - Verify exponent sum:
0 + 7 + 0 = 7 = 7 ✓

Step 2 - Apply multinomial coefficient formula:
Coefficient = $\frac{7!}{7!}$

Step 3 - Calculate:
- Numerator: 7! = 5040
- Denominator: 7! = 5040
- Denominator value: 5040

Final Calculation:
Coefficient = 5040 / 5040 = 1

Alternative interpretation: This equals the number of ways to arrange 7 items with:
0 of type x, 7 of type y, 0 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 = 3^7 = 2187

Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

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 = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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: Counting even/odd numbers with distinct digits.

Given: 3-digit numbers, distinct digits, even numbers.

Case Analysis for even numbers:

Case 1: Last digit = 0
- First digit: 1-9 (9 choices)
- Remaining 1 digits: choose from remaining 8 digits and arrange
- Ways = 9 × P(8, 1) = 9 × 8 = 72

Case 2: Last digit = 2,4,6,8 (4 choices)
- First digit: cannot be 0 and cannot be last digit (8 choices)
- Remaining 1 digits: choose from remaining 8 digits and arrange
- Ways = 4 × 8 × P(8, 1) = 256

Total = 72 + 256 = 328

Key Principle: Always handle first digit (can't be 0) and last digit (parity constraint) separately.

Step-by-Step Solution:

Concept: Combination with mandatory inclusion constraint.

Given:
- Total people: 11
- Committee size: 6
- Must include: 2 specific people

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

Analysis:
We need to select 6 people total, with 2 already fixed.
- Fixed positions: 2 (these specific people are already in)
- Remaining positions to fill: 6 - 2 = 4
- People available for remaining positions: 11 - 2 = 9

Step 1 - Fix Mandatory Members:
2 specific people must be included: C(2,2) = 1 way
(This is automatic - we have no choice here)

Step 2 - Select Remaining Members:
Choose 4 people from remaining 9 people:
C(9,4) = 126

Calculation:
C(9,4) = (9)! / [4! × (5)!]
= 362880 / [24 × 120]
= 126

Alternative Approach - Verification:
Think of it as: "We've used 2 spots, now choose 4 more from 9 remaining"

Related Problem Types:

1. Must EXCLUDE specific people:
Select all 6 from remaining 11 - (people to exclude)

2. At least one specific person:
Total ways - Ways without that person
= C(11,6) - C(11-1,6)

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: 126 ways

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: 10
- 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 10 points: C(10,3)
C(10,3) = 10! / [3! × 7!] = 120

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
= 120 - 4
= 116

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

Verification: Answer should be less than C(10,3) = 120 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 = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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: Circular Permutation with reflection symmetry. This is used for arrangements like necklaces or keyrings where flipping the arrangement produces the same result (Clockwise = Anticlockwise).

Formula: $\text{Total Ways} = \frac{(n-1)!}2$

Analysis:
- Total items ($n$): 5
- Step 1: Normal circular arrangements (rotations same) = $(n-1)!$ = 24
- Step 2: Account for reflection (flips) by dividing by 2.

Calculation:
Arrangements = $\frac{(5-1)!}2$
= $\frac{24}2$
= 12

Formula Summary:
- Linear: $n!$
- Circular (no reflection): $(n-1)!$
- Circular (with reflection): $\frac{(n-1)!}2$

Key Principle: Dividing by 2 removes the overcounting caused by the symmetry when the arrangement can be flipped.

Step-by-Step Solution:

Concept: Counting even/odd numbers with distinct digits.

Given: 4-digit numbers, distinct digits, even numbers.

Case Analysis for even numbers:

Case 1: Last digit = 0
- First digit: 1-9 (9 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 9 × P(8, 2) = 9 × 56 = 504

Case 2: Last digit = 2,4,6,8 (4 choices)
- First digit: cannot be 0 and cannot be last digit (8 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 4 × 8 × P(8, 2) = 1792

Total = 504 + 1792 = 2296

Key Principle: Always handle first digit (can't be 0) and last digit (parity constraint) separately.

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).