What are Permutation and Combination?

The mathematical concepts of permutations and combinations serve as basic tools which enable people to calculate all available methods of item arrangement and item selection. The definition of permutation describes it as an item arrangement which requires specific sequence order to be maintained while combination defines it as a method of selecting items without any regard for their sequence. The study of historical permutation and combination problems requires students to learn various counting techniques which include factorial counting methods and methods to arrange both distinct items and identical items along with circular arrangement techniques and selection methods that use multiple constraints. The common question types require people to reorder the characters in words while they create groups and panels and determine all possible ways to distribute items and solve complex problems that need both selection and organization according to specific rules. The resolution of these problems depends on two essential mathematical formulas, which include nPr and nCr, and require students to develop analytical skills within defined limits.

Permutation and Combination Questions (Basic to Advanced)

Question 1:
A letter lock consists of three rings, each marked with ten different letters. In how many ways can an unsuccessful attempt be made to open the lock?

Solution:
Each ring can be set to any of the 10 letters. Therefore, the total number of possible attempts is:

$$[10 \times 10 \times 10 = 1000]$$

Out of these 1000 attempts, only one attempt is successful.

$$[\text{Number of unsuccessful attempts} = 1000 – 1 = 999]$$

Answer: 999

Question 2:
Find the total number of positive integral solutions for (x, y, z) such that:

$$[x \times y \times z = 24]$$

Solution:
The prime factorization of 24 is:

$$[24 = 2^3 \times 3^1]$$

We need to distribute the powers of 2 and 3 among (x, y, z).

• The number of ways to distribute three 2’s among three variables:

$$[(3 + 3 – 1)C(3 – 1) = 5C2 = 10]$$

• The number of ways to distribute one 3 among three variables:

$$[(1 + 3 – 1)C(3 – 1) = 3C2 = 3]$$

Multiplying these together gives the total number of positive solutions:

$$[10 \times 3 = 30]$$

$$Answer: 30$$

Question 3:
Find the total number of signals that can be made using five differently colored flags when any number of flags may be used in a signal.

Solution:

• Case 1: Using 1 flag: (5P1 = 5)
• Case 2: Using 2 flags: $$(5P2 = 5 \times 4 = 20)$$
• Case 3: Using 3 flags: $$(5P3 = 5 \times 4 \times 3 = 60)$$
• Case 4: Using 4 flags: $$(5P4 = 5 \times 4 \times 3 \times 2 = 120)$$
• Case 5: Using all 5 flags: $$(5P5 = 5! = 120)$$

$$[\text{Total signals} = 5 + 20 + 60 + 120 + 120 = 325]$$

Answer: 325

Question 4:
Two numbers are chosen from the sequence (1, 3, 5, 7, \dots, 149, 151). How many ways exist such that their product is a multiple of 5?

Solution:

• Numbers divisible by 5 in the sequence: (5, 15, 25, \dots, 145). This forms an arithmetic progression with:
  $$[a = 5, d = 10, T_n = 145 \Rightarrow n = 15]$$
• Total numbers in the sequence:
  $$[1 + (m-1) \times 2 = 151 \Rightarrow m = 76]$$
• Ways to select numbers giving a product divisible by 5:
  $$[(\text{both numbers divisible by 5}) + (\text{one divisible by 5, one not})
  = 15C2 + 15C1 \times (76-15)C1 = 105 + 915 = 1020]$$

Answer: 1020

Question 5:

A five-digit number divisible by 3 has to be formed using the digits (0, 1, 2, 3, 4, 5) without repetition. How many such numbers can be formed?

Solution:

• A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of all digits (0 + 1 + 2 + 3 + 4 + 5 = 15), which is divisible by 3.
• We must exclude either 0 or 3 while forming the five-digit number to satisfy the criteria.

Case 1: Excluding 0 → Digits used: 1, 2, 3, 4, 5
$$[\text{Number of arrangements} = 5P5 = 120]$$

Case 2: Excluding 3 → Digits used: 0, 1, 2, 4, 5

• The first digit cannot be 0 → 4 choices for the first digit
• Remaining digits can be arranged in 4! ways
  $$[\text{Arrangements} = 4 \times 4! = 4 \times 24 = 96]$$

Total numbers:
$$[120 + 96 = 216]$$

Answer: 216

Permutation and Combination JEE Mains Questions

Level 1: Basic/Conceptual

1. Question 1: In how many ways can 5 books be arranged on a shelf?
   Solution hint: This is simple permutation of 5 distinct objects:
   $$[5! = 120 \text{ ways}]$$
2. Question 2: From a group of 7 students, a team of 3 is to be selected. In how many ways can this be done?
   Solution hint: Combination of 7 taken 3 at a time:
   $$[{}^7C_3 = \frac{7!}{3!4!} = 35]$$

Level 2: Moderate

Question 3: A password consists of 3 letters followed by 2 digits. How many such passwords are possible if letters and digits can be repeated?
Solution hint:

• Letters: 26 choices each → $$(26^3)$$
• Digits: 10 choices each → $$(10^2)$$
  $$[\text{Total} = 26^3 \times 10^2 = 175,760]$$

Question 4: In how many ways can 4 boys and 3 girls sit in a row such that no two girls sit together?
Solution hint:

• Arrange 4 boys: (4! = 24) ways
• Place 3 girls in gaps between boys → 5 gaps → choose 3 gaps:$$({}^5C_3 = 10) ways$$
• Arrange 3 girls: (3! = 6)
  $$[\text{Total} = 24 \times 10 \times 6 = 1,440]$$

Level 3: Challenging

5. Question 5: From 10 different books, 4 are to be selected and arranged on a shelf such that 2 particular books are not together.
   Solution hint:
   • Total ways to select and arrange 4 books: $$( {}^{10}C_4 \times 4! = 210 \times 24 = 5,040 )$$
   • Treat 2 particular books as 1 unit (they are together) → choose 2 more from remaining $$8: ( {}^8C_2 = 28 )$$
   • Arrange 3 units: $$(3! = 6)$$
   • Arrange the 2 particular books within unit: $$(2! = 2)$$
   • Total arrangements where they are together: $$(28 \times 6 \times 2 = 336)$$
   • Total arrangements where they are not together: $$(5,040 – 336 = 4,704)$$
6. Question 6 (JEE Type Classic): There are 5 men and 4 women. In how many ways can they be seated in a row so that no two women are together?
   Solution hint:
   • Arrange 5 men: $$(5! = 120)$$
   • Place 4 women in 6 gaps →$$ ({}^6C_4 = 15)$$
   • Arrange 4 women:$$ (4! = 24)$$
     $$[\text{Total} = 120 \times 15 \times 24 = 43,200]$$

Level 4: Advanced / Mind-Bending

7. Question 7: How many numbers between 1000 and 9999 have all different digits?
   Solution hint:
   • First digit (1–9) → 9 choices
   • Second digit → 9 choices (0–9 except first digit)
   • Third digit → 8 choices
   • Fourth digit → 7 choices
     $$[9 \times 9 \times 8 \times 7 = 4,536]$$
8. Question 8 (JEE Previous Year Style): In how many ways can 6 students be divided into 2 groups of 3 each?
   Solution hint:
   • Total ways to choose 3 out of 6: $$( {}^6C_3 = 20 )$$
   • Since groups are identical, divide by 2: $$( \frac{20}{2} = 10 )$$

Permutation and Combination Aptitude Questions

1. Basic Level

Q1: How many ways can 5 students sit in a row?
Solution: This is a permutation problem because order matters.

$$[\text{Number of ways} = 5! = 5 × 4 × 3 × 2 × 1 = 120]$$

Q2: How many ways can you choose 3 books from a shelf of 7 books?
Solution: This is a combination problem because order doesn’t matter.

$$[\text{Number of ways} = ^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7 × 6 × 5}{3 × 2 × 1} = 35]$$

Q3: A coin is tossed 4 times. In how many ways can exactly 2 heads appear?
Solution: This is a combination problem because the order of heads and tails matters in counting positions.

$$[^4C_2 = \frac{4 × 3}{2 × 1} = 6]$$

2. Medium Level

Q4: In how many ways can the letters of the word “MATH” be arranged?

• Solution: The word has 4 distinct letters → permutation.

$$[4! = 24 \text{ ways}]$$

Q5: How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition is not allowed?

• Solution: This is a permutation problem:

$$[^5P_3 = \frac{5!}{(5-3)!} = \frac{5 × 4 × 3 × 2 × 1}{2 × 1} = 60]$$

Q6: From 8 boys and 6 girls, a team of 4 students is to be selected. How many ways can this team be formed if the team must have 2 boys and 2 girls?

• Solution: Combination for each group:

$$[^8C_2 × ^6C_2 = \frac{8 × 7}{2} × \frac{6 × 5}{2} = 28 × 15 = 420]$$

3. Tricky Level / Advanced

Q7: How many 5-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 if the number must be even and no repetition is allowed?

• Solution:
  • Even numbers end with 2, 4, or 6 → 3 choices for last digit.
  • Remaining 4 digits can be arranged in 4! = 24 ways.

$$[\text{Total} = 3 × 24 = 72]$$

Q8: How many ways can 3 boys and 3 girls be seated alternately in a row?

• Solution:
  • Arrangement starts with boy: B G B G B G
    • Boys: 3! ways
    • Girls: 3! ways
  • Total = $$3! × 3! = 6 × 6 = 36$$

Q9: A committee of 5 is to be formed from 7 men and 4 women. In how many ways can the committee be formed so that it has at least 3 women?

• Solution: We consider 3, 4, or 5 women cases:

1. 3 women + 2 men →$$ ^4C3 × ^7C2 = 4 × 21 = 84$$
2. 4 women + 1 man → $$^4C4 × ^7C1 = 1 × 7 = 7$$

$$[\text{Total} = 84 + 7 = 91]$$

Q10: How many 6-letter words can be formed from the letters of “BANANA”?

• Solution: Letters: B, A, N, A, N, A → repeated letters (A appears 3 times, N appears 2 times)

$$[\text{Permutations} = \frac{6!}{3! × 2!} = \frac{720}{6 × 2} = \frac{720}{12} = 60]$$

Important Formulas for Quick Revision

1. Algebra

• Quadratic Formula: $$(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a})$$
• Sum of n terms of an AP: $$(S_n = \frac{n}{2}[2a + (n-1)d])$$
• Sum of n terms of a GP: $$(S_n = a \frac{r^n – 1}{r – 1}) (if (r \neq 1))$$
• Binomial Theorem: $$((a+b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k}b^k)$$

2. Geometry

• Area of Triangle (Heron’s Formula): $$(A = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2})$$
• Circle: Circumference $$(= 2\pi r), Area (= \pi r^2)$$
• Sphere: Surface Area$$ (= 4\pi r^2), Volume (= \frac{4}{3}\pi r^3)$$
• Cylinder: Surface Area $$(= 2\pi r(h+r)), Volume (= \pi r^2 h)$$

3. Trigonometry

• Basic Identities:
  $$(\sin^2\theta + \cos^2\theta = 1)$$
  $$(1 + \tan^2\theta = \sec^2\theta)$$
  $$(1 + \cot^2\theta = \csc^2\theta)$$
• Angle Formulas:
  $$(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B)$$
  $$(\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B)$$

4. Calculus

• Derivative: $$(\frac{d}{dx}(x^n) = nx^{n-1})$$
• Integration: $$(\int x^n dx = \frac{x^{n+1}}{n+1} + C) (for (n \neq -1))$$
• Derivative of Trig Functions:
  $$(\frac{d}{dx}(\sin x) = \cos x), (\frac{d}{dx}(\cos x) = -\sin x)$$

5. Physics (Mechanics)

• Newton’s Second Law: (F = ma)
• Kinematic Equations:
  $$(v = u + at)$$
  $$(s = ut + \frac{1}{2}at^2)$$
  $$(v^2 = u^2 + 2as)$$
• Gravitational Force: $$(F = G\frac{m_1 m_2}{r^2})$$

6. Chemistry

• Ideal Gas Law: $$(PV = nRT)$$
• Molarity:$$ (M = \frac{\text{moles of solute}}{\text{litres of solution}})$$
• pH: $$(pH = -\log[H^+])$$

Common mistakes in Permutation and Combination 

1. Confusing Permutation with Combination

Mistake: Using the wrong formula.

• Permutation (P) = Arrangement matters.
  Formula: $$( nP_r = \frac{n!}{(n-r)!} )$$
• Combination (C) = Selection matters, order doesn’t.
  Formula: $$( nC_r = \frac{n!}{r!(n-r)!} )$$

Example Mistake:
Choosing 3 students out of 5 for a team but calculating (5P3) instead of (5C3).

Tip: Ask yourself: “Does order matter?” If yes → Permutation; If no → Combination.

2. Forgetting to Consider Repetition

Mistake: Not accounting for repeated elements.

• Distinct items: Use regular formula.
• Repeated items: Use multinomial formula:
  $$[frac{n!}{n_1! n_2! timess n_k!}]$$

Example Mistake:
Number of ways to arrange letters in “LEVEL”. Ignoring repeated L and E gives (5!), but correct is:
$$[\frac{5!}{2! \times 2!} = 30]$$

3. Ignoring Constraints

Mistake: Overlooking restrictions like “no two girls together” or “must start with a vowel.”

Example Mistake:
Seating 3 boys and 2 girls in a row without two girls together. Many assume just (5!).
Correct approach: Arrange boys first, then insert girls in allowed spaces.

Tip: Draw diagrams or slots for restricted arrangements.

4. Double Counting / Overcounting

Mistake: Counting the same case multiple times.

Example Mistake:
Arranging books where two are identical and treating them as distinct: (6!) instead of (\frac{6!}{2!}).

Tip: Identify identical items or overlapping cases; divide by symmetry or use the inclusion-exclusion principle.

5. Misapplying Factorial Rules

Mistake: Forgetting factorial base cases or using incorrect factorial manipulation.

Example Mistake:
((n-1)! \neq n!) or assuming (0! = 1) is invalid.

Tip: Memorize basic factorial facts:
$$[0! = 1, \quad n! = n \times (n-1)!]$$

6. Wrong Approach in Circular Permutations

Mistake: Treating circular arrangements like linear ones.

• Linear: (n!)
• Circular: ((n-1)!)
• If clockwise & counter-clockwise are same, divide by 2: $$(\frac{(n-1)!}{2})$$

Tip: Always check if rotation matters.

7. Neglecting Complementary Counting

Mistake: Trying to count complex arrangements directly instead of using complement principle.

Example:
At least 1 boy in a team of 3 from 5 boys and 4 girls. Instead of counting “1, 2, 3 boys” separately, use:
$$[\text{Total ways} – \text{All girls} = 9C3 – 4C3]$$

8. Mixing Up nPr and nCr in Multi-Step Problems

Mistake: Using the wrong formula in different stages of the problem.

Example Mistake:
Choosing 2 team captains from 5 (order matters) and then selecting 3 members from remaining 3 (order doesn’t). Using combination for both steps gives wrong answer.

Tip: Break problems into stages and check if order matters at each stage.

9. Ignoring “Identical Groups” in Division Problems

Mistake: Treating identical groups as distinct.

Example Mistake:
Dividing 6 students into 3 groups of 2.
Wrong: $$(\frac{6!}{2!2!2!} = 720)$$
Correct: Divide by (3!) to account for identical groups:
$$[\frac{6!}{2!2!2! \times 3!} = 60]$$

10. Overlooking Edge Cases

• Choosing 0 items or all items sometimes changes formulas.
• Example: $$(nC_0 = 1) and (nC_n = 1)$$

General Tips to Avoid Mistakes

1. Always ask: Does order matter?

2. Verify whether elements in the set have identical characteristics or if there are multiple occurrences of items.

3. You should assess the restrictions before you start using the formulas.

4. Diagrams and slots serve as visual tools to show different possible arrangements.

5. When solving difficult problems that require “at least” and “at most” solutions use complementary counting methods.

6. Perform the solution process in distinct phases while identifying each phase as a permutation or combination.

Tips to Master Permutation and Combination Questions

1. Strengthen Your Basics

Before attempting tricky problems, make sure you thoroughly understand:

• Factorials: $$( n! = n \times (n-1) \times … \times 1 )$$
• Permutation Formula: $$( P(n,r) = \frac{n!}{(n-r)!} )$$
• Combination Formula: $$( C(n,r) = \frac{n!}{r!(n-r)!} )$$
• Basic properties:
  • $$( C(n,r) = C(n,n-r) )$$
  • $$( P(n,r) = r! \times C(n,r) )$$
• Special cases: Like arrangements of identical items, circular arrangements, or restrictions.

2. Categorize Questions

P&C problems often fall into recognizable patterns:

• Simple permutations/combinations: Straight application of formulas.
• With restrictions: E.g., certain elements must or must not be together.
• Circular arrangements: People around a table (use ( (n-1)! ) for circular).
• Repeated elements: Words with repeating letters.
• Selection problems: Choosing teams, committees, or objects.
• Complex arrangements: Rows, grids, or layered restrictions.

Knowing the type helps you pick the right formula or method.

3. Learn to Break the Problem

Many mistakes happen because people try to solve everything at once. Use this strategy:

• Step 1: Identify if the problem is about arrangement (order matters) or selection (order doesn’t matter).
• Step 2: Check for restrictions or repetitions.
• Step 3: Split into smaller cases if necessary and add or multiply appropriately:
  • Addition principle: If event A or event B happens → add counts.
  • Multiplication principle: If event A and event B happen → multiply counts.

4. Use Complementary Counting

Sometimes it’s easier to count the opposite:

• Example: “At least 2 girls in a team of 3 from 5 boys and 4 girls.”
  Instead of counting 2 girls and 3 girls separately, do:
  Total ways – ways with 0 or 1 girl.

5. Practice “Special Formulas”

Certain classic arrangements appear often:

• Circular permutations: $$( (n-1)! )$$
• Identical objects: $$( \frac{n!}{p!q!r!} )$$ for repeating items
• Distribution problems: Using stars and bars for identical items into boxes

6. Draw & Visualize

• Use tables, trees, or diagrams for small problems to visualize choices.
• For seating or arranging problems, draw chairs or boxes to track possibilities.

7. Avoid Overcomplication

• Don’t memorize every problem type. Focus on principles and logic.
• Always check if the same case is counted twice—avoid double-counting.

8. Practice Strategically

• Start with easy problems to master basics.
• Move to medium/hard problems, focusing on:
  • Conditional arrangements
  • Repetition and identical objects
  • Circular and restricted arrangements
• After solving, analyze mistakes and note formulas/tricks.

9. Quick Mental Tricks

• $$( C(n,1) = n, , C(n,2) = \frac{n(n-1)}{2} )$$
• For repeated letters: divide by factorial of repeats
• For complementary counting: sometimes faster than direct counting

10. Recommended Mindset

• Treat P&C like puzzle-solving rather than memorization.
• Focus on logic first, formulas second.
• Don’t fear multi-step questions; break them into manageable cases.

Conclusion

Combinatorial mathematics relies on permutation and combination methods which provide systematic techniques to count and organize objects. Permutations deal with object arrangements that require specific order while combinations allow users to select objects without considering their sequence. The ability to master these concepts enables students to solve probability and algebra problems while developing logical reasoning skills which they can use in practical situations such as scheduling and decision making and probability evaluation. The understanding of principles and formulas together with their application methods represents the fundamental requirement for solving advanced counting challenges.

Good evening everyone. Please confirm if I’m audible. You’re here to attend the class, right? Can you please confirm? I’m still waiting for confirmation that everything is fine. We’ll start the PYQ class based on the P&C topic. Am I audible? Yes? Okay, let me share my screen.

Roughly, there are quite a number of questions — about 10 sets from the January 2025 session and 10 sets from the April 2024 session. Some of the questions are binomial-type. You know, questions involving nCr can fall under both permutations and binomial, so the number of questions appears slightly higher. There are 26 questions across 20 sets. In some papers, two questions may have come from the same concept. Some are set-based questions from the Sets, Relations, and Functions topic, which is why the count is higher. So on average, you can assume at least one question per paper from here.

This division is rough and based on my judgment. Everyone has a different level. What I call easy, you might find difficult. What I call difficult, someone with very strong math might find easy. But this is a teacher’s general judgment. Questions based on concepts already studied, and similar to class questions and CTS sheets, fall into the easy category. If you’ve studied the topic and done basic practice, those are easy.

Some questions need a little thinking, where mistakes are more likely. Integer-type questions are harder to score in compared to MCQs, because small mistakes are harder to detect. Difficult questions are those where even the idea is hard to think of, even after basic practice.

We’ll understand this better as we solve questions. Let’s start with the first one.

Question 1 (Dictionary Order – “KANPUR”)

This is a very typical and easy question. If someone has revised class notes, they should definitely be able to score here.

We are forming all meaningful or meaningless words using the letters of KANPUR. First, arrange letters alphabetically:

$$A, K, N, P, R, U$$

There are 6 letters → total permutations = $$6!$$

But we are not asked for the rank of a word. Instead, we are asked: Which word is at the 44th position in dictionary order?

We proceed section-wise:

Words starting with A: 5! = 120

Starting with K: 120

Starting with N: 120

So up to N → 360 words

Next section is P. We must go inside the P section.

Within P:

$$PA section → 4! = 24 → total 384$$

$$PK section → +24 → 408$$

$$PN section → +24 → 432$$

We need 44th overall → actually 440th? (Teacher correcting in class flow.) After reaching 432, we go deeper.

Fix PRA:

Remaining letters $$= 3 → 3! = 6 → total 438$$

Next words:

PRAKN → 439

PRAKU → 440

So the 440th word is PRAKU N (final arrangement). Correct answer is B, not C.

Concept is straightforward — just careful counting.

Next Question (Repeated Concept)

A similar dictionary-order question appeared again in another year (Nagpur instead of Kanpur). This shows PYQs repeat ideas.

Question with Repeated Letters (e.g., “BIBJOK” type)

Total arrangements$$ = 5! / 2! = 60$$ (since one letter repeats)

We again find the 50th word in dictionary order.

Arrange alphabetically and proceed section-wise, remembering to divide by factorial for repeated letters. Final answer comes out as OBBJH (example structure).

7-Digit Numbers with Sum of Digits = 11 using digits 1,2,3

We form 7-digit numbers where digits are only 1, 2, or 3, and sum is 11.

Important: It does NOT mean all three digits must be used. That confusion can happen.

We create cases:

$$Five 1s + one 3 + one 3$$

$$Four 1s + two 2s + one 3$$

$$Three 1s + four 2s$$

For each case, count permutations:

$$\frac{7!}{5!\,2!} + \frac{7!}{4!\,2!\,1!} + \frac{7!}{3!\,4!} = 161$$

Rated average because students may miss cases.

Binomial Coefficient Ratio Question

Given:

$$binom{n}{r} = 56, \quad binom{n}{r-1} = 28, \quad binom{n}{r+1} = 70$$

Using ratios:

$$ \frac{{n \choose r}}{{n \choose r-1}} = \frac{n-r+1}{r} \quad \frac{{n \choose r-1}}{{n \choose r}} = \frac{r}{n-r+1} \quad \frac{{n \choose r+1}}{{n \choose r}} = \frac{n-r}{r+1} \quad \frac{{n \choose r}}{{n \choose r+1}} = \frac{r+1}{n-r} $$

Solving gives:

n = 8, r = 3

Then combined with coordinate geometry (centroid locus). After algebra and squaring/adding trig identities:

$\alpha = 20$

5-Digit Numbers > 50000 with digit sum condition

Digits allowed: given 8 digits. Repetition allowed (since not restricted).

We fix first digit cases:

Starting with 5

Starting with 6

Starting with 7

Apply condition: first digit + last digit ≤ 8

Count middle digits freely (repetition allowed). Subtract 1 to remove 50000.

Final answer matches Option D.

Group Selection (Men/Women)

Group 1: 4 men, 5 women

Group 2: 5 men, 4 women

Select 4 from each group such that total = 4 men + 4 women.

Make cases table:

$$(4M,0W), (3M,1W), (2M,2W), (1M,3W), (0M,4W)$$

Multiply combinations for each case and sum:

$$Final answer = 5626$$

$$Word “DAUGHTER” – vowels never together$$

$$Total permutations = 8!$$

Subtract cases where all 3 vowels are together:

Treat vowels as one block:

$$6! × 3!$$

$$Answer = 720 × 50 = 36,000$$

Queue Arrangement – Girls together, Boys together, B1 & B2 not adjacent

3 girls together block → 3!

4 boys with B1 & B2 separated → choose positions among 3 gaps → permutations

Multiply both group arrangements and ×2 (order of blocks)

Final answer = 144

Alphabet Question – Middle letter is M

Choose 5 letters in alphabetical order, middle letter fixed as M.

So 2 letters must be chosen from A–L (12 letters)

2 letters from N–Z (13 letters)

Order fixed (alphabetical), so just combinations:

$${12 \choose 2} \times {13 \choose 2} = 5148$$ $${12 \choose 2} = \frac{12 \times 11}{2} = 66$$ $${13 \choose 2} = \frac{13 \times 12}{2} = 78$$ $$66 \times 78 = 5148$$

The lecture emphasizes:

PYQs repeat concepts

Many questions are direct applications of class problems

Confidence matters — even partially prepared students can score

Carelessness causes more mistakes than difficulty