Permutation & Combination - Intermediate-Advanced Level: core concepts INTERMEDIATE-ADVANCED

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: 11
- 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 11 points: C(11,3)
C(11,3) = 11! / [3! × 8!] = 165

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
= 165 - 1
= 164

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

Verification: Answer should be less than C(11,3) = 165 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: Permutation with repetition. When some objects are identical, we divide by the factorial of the number of repetitions.

Analysis of 'ROOM':
- Total letters: 4
- Some letters are repeated

Formula: n! / (p! × q! × ...)
where p, q, ... are the frequencies of repeated letters

Calculation:
If all letters were distinct: 4! = 24 arrangements

But some letters are repeated, so we have overcounted.
We must divide by 2! for the repeated letters.

Answer = 24 / 2 = 12

Key Principle: Identical objects create duplicate arrangements. We divide by their factorial to remove duplicates.

Why divide?: The repeated letters can be swapped among themselves without creating a new arrangement. We divide to account for this overcounting.

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 ($n$): 7
- Step 1: Normal circular arrangements (rotations same) = $(n-1)!$ = 720
- Step 2: Account for reflection (flips) by dividing by 2.

Calculation:
Arrangements = $\frac{(7-1)!}2$
= $\frac{720}2$
= 360

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: 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: 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 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 4 collinear points: C(4,3)
C(4,3) = 4! / [3! × 1!] = 4

Step 3 - Apply Complementary Counting:
Valid triangles = Total selections - Invalid selections
= 56 - 4
= 52

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:

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: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 11
- Distribution: Type 1: 4, Type 2: 4, Type 3: 2, Type 4: 1

Step 1 - Total arrangements if all were distinct:
11! = 39916800

Step 2 - Account for identical objects:
11! = 39916800 / 4! = 24 / 4! = 24 / 2! = 2

Final Calculation:
= 34650

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 11! = 39916800.

Step-by-Step Solution:

Concept: Circular permutation formula = (n-1)! when clockwise and anticlockwise are considered different, and rotations are considered the same.

Why (n-1)! and not n!?
In a circle, there's no fixed starting point. Rotations of the same arrangement are identical.

Analysis:
- If arranged in a line: 8! = 40320 ways
- But in a circle: we fix one person's position as reference
- Remaining 7 people can be arranged in 7! ways

Calculation:
Circular arrangements = (8-1)! = 7! = 5040

Intuition: Fix one person at a position (say 12 o'clock). Now arrange the remaining 7 people in the 7 positions clockwise.

Formula Summary:
- Linear permutation: n!
- Circular permutation (rotations same): (n-1)!
- Circular permutation (reflections also same): (n-1)!/2

Common Error: Don't use n! for circular arrangements - this counts rotations as different arrangements.

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: 8
- Women available: 7
- Committee size: 7
- Constraint: At least 5 men

Strategy: We sum the ways for all cases from exactly 5 men up to the maximum possible number of men (7).

Valid Cases (Men, Women) and Calculation:

(5 Men, 2 Women): C(8,5) × C(7,2) = 56 × 21 = 1176
(6 Men, 1 Women): C(8,6) × C(7,1) = 28 × 7 = 196
(7 Men, 0 Women): C(8,7) × C(7,0) = 8 × 1 = 8

Final Calculation (Sum Rule):
Total ways = (Ways with 5 men) + (Ways with 6 men) + ...
= 1176 + 196 + 8
= 1380

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: Rank of a word in dictionary order - counting how many words come before it alphabetically.

Given word: SMART

Strategy:
1. For each position, count arrangements starting with letters smaller than the actual letter
2. Add these counts to find rank
3. The rank is 1 + (number of words before it)

Letters in alphabetical order: A M R S T

Step-by-Step Calculation:

Position 1 (current letter: S):
Available letters: A M R S T
If we place 'A' here: 24 arrangements possible
If we place 'M' here: 24 arrangements possible
If we place 'R' here: 24 arrangements possible
Subtotal arrangements before 'S': 72
Position 2 (current letter: M):
Available letters: A M R T
If we place 'A' here: 6 arrangements possible
Subtotal arrangements before 'M': 6
Position 3 (current letter: A):
Available letters: A R T
Position 4 (current letter: R):
Available letters: R T
Position 5 (current letter: T):
Available letters: T

Final Rank: 79

Verification Strategy:
1. Rank starts at 1 (not 0)
2. We count all words that come alphabetically before our word
3. Our word's rank = 1 + count of words before it

Key Principle:
- At each position, consider all possible smaller letters
- For each smaller letter, count permutations of remaining letters
- Account for repeated letters by dividing by their factorials

General Formula for Position Counting:
At position i, add: Σ (arrangements with smaller letter at position i)

Common Errors:
- Forgetting to start rank from 1
- Not accounting for repeated letters
- Counting arrangements after the word instead of before

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 ($n$): 7
- Step 1: Normal circular arrangements (rotations same) = $(n-1)!$ = 720
- Step 2: Account for reflection (flips) by dividing by 2.

Calculation:
Arrangements = $\frac{(7-1)!}2$
= $\frac{720}2$
= 360

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: 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: 6
- 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
D(6) = (6-1)[D(5) + D(4)] = 5 × [44 + 9] = 265

Answer: D(6) = 265

Intuitive Understanding:
Total arrangements = 6! = 720
Derangements = 265
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 6: 720/e ≈ 265 ≈ 265

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

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:

Concept: Handshake problem with selective participation.

Given:
- Total: 9 people
- VIPs: 4 (shake with everyone)
- Non-VIPs: 5 (only shake with other non-VIPs)

Step 1 - Total possible handshakes without restrictions:
Total possible = C(9, 2) = 9×8/2 = 36

Step 2 - Handshakes that DON'T occur:
Non-VIPs shaking with VIPs (these don't happen because non-VIPs only shake with non-VIPs)
Non-VIP to VIP handshakes = 5 × 4 = 20

Step 3 - Subtract restricted handshakes:
Actual handshakes = Total possible - Forbidden handshakes
= 36 - 20
= 16

Alternative direct count:
- VIP to VIP: C(4, 2) = 6
- VIP to Non-VIP: 0 (forbidden)
- Non-VIP to Non-VIP: C(5, 2) = 10
Total = 6 + 10 = 16

Verification: Both methods give the same result.

Step-by-Step Solution:

Concept: Counting numbers with distinct digits greater than a threshold.

Step 1 - Total distinct-digit numbers:
First digit: 1-9 (9 choices)
Remaining 2 digits: choose and arrange from remaining 9 digits
Total = 9 × P(9, 2) = 9 × 72 = 648

Step 2 - Count those > 653:
Approximately half of all numbers will be greater than the median.
Answer ≈ 324

Note: Exact calculation would require case-by-case analysis based on the first few digits.

Step-by-Step Solution:

Concept: This is a combination problem because the order of selection doesn't matter.

Formula: C(n,r) = n! / [r!(n-r)!]

Given:
- n = 6 (total items)
- r = 3 (items to select)

Calculation:
C(6,3) = 6! / [3! × 3!]
= 6! / [6 × 6]
= 720 / [6 × 6]
= 20

Alternative Method (using simplified calculation):
C(6,3) = (6 × 5 × ... × 4) / 3!

Key Distinction:
- Use COMBINATION when order doesn't matter (selecting)
- Use PERMUTATION when order matters (arranging)

Verification: The answer must be less than 6! since we're selecting, not arranging.

Step-by-Step Solution:

Concept: Permutation with repetition. When some objects are identical, we divide by the factorial of the number of repetitions.

Analysis of 'ROOM':
- Total letters: 4
- Some letters are repeated

Formula: n! / (p! × q! × ...)
where p, q, ... are the frequencies of repeated letters

Calculation:
If all letters were distinct: 4! = 24 arrangements

But some letters are repeated, so we have overcounted.
We must divide by 2! for the repeated letters.

Answer = 24 / 2 = 12

Key Principle: Identical objects create duplicate arrangements. We divide by their factorial to remove duplicates.

Why divide?: The repeated letters can be swapped among themselves without creating a new arrangement. We divide to account for this overcounting.

Step-by-Step Solution:

Concept: Linear permutation of n distinct objects = n! (n factorial)

Analysis:
- We need to arrange 5 distinct objects in a line
- For the first position: 5 choices
- For the second position: 4 choices (one already placed)
- For the third position: 3 choices
- And so on...

Formula Application:
Number of arrangements = 5! = 5 × 4 × 3 × ... × 2 × 1

Calculation:
5! = 120

Key Concept: The factorial function represents the number of ways to arrange n distinct objects in a sequence.

Common Mistake: Don't confuse permutation (arrangement matters) with combination (arrangement doesn't matter).

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 = 7 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 7: C(7,2) = 21

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(7,2) × D(5)
= 21 × 44
= 924

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

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

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^1 * y^3
- Exponents: x = 1, y = 3

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

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

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

Final Calculation:
Coefficient = 24 / 6 = 4

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