Permutations and Combinations — Counting Principles for Class 11

What Are We Actually Counting?

Permutations and combinations sit at the heart of probability, and they show up everywhere — from how many ways you can arrange your JEE preparation schedule to how many 4-digit PINs are possible. The core idea is simple: we want to count arrangements or selections without listing every possibility by hand.

The chapter feels intimidating because students confuse two very different questions. Permutations answer “how many ways can we arrange these?” (order matters). Combinations answer “how many ways can we choose these?” (order doesn’t matter). Once that distinction is clear, 80% of the problems become mechanical.

This is a high-weightage topic in JEE Main — roughly 2-3 questions per paper, typically worth 8-12 marks. CBSE board exams test it in 4-6 mark questions that require clean, step-by-step solutions. The good news: unlike calculus, P&C problems are fully solvable with a clear head and no calculator.

Key Terms and Definitions

Factorial ( $n!$ ) is the product of all positive integers from 1 to $n$ .

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

By convention, $0! = 1$ . This isn’t arbitrary — combinatorial formulas break without it.

Fundamental Counting Principle (FCP): If task A can be done in $m$ ways and task B in $n$ ways, then both can be done together in $m \times n$ ways. This extends to any number of independent tasks.

Permutation is an arrangement of objects where order matters. If we pick $r$ objects from $n$ distinct objects:

^nP_r = \frac{n!}{(n-r)!}

Combination is a selection where order does not matter:

^nC_r = \frac{n!}{r!(n-r)!}

The relationship between them: $^nP_r = r! \times ^nC_r$

^nP_r = \frac{n!}{(n-r)!} \qquad ^nC_r = \frac{n!}{r!(n-r)!}

Useful identities:

$^nC_r = ^nC_{n-r}$
$^nC_0 = ^nC_n = 1$
$^nC_r + ^nC_{r-1} = ^{n+1}C_r$ (Pascal’s Identity)

Core Methods and Concepts

The Fundamental Counting Principle in Action

Before reaching for $^nP_r$ or $^nC_r$ , always ask: can I just multiply?

A number lock has 3 dials, each with digits 0–9. Total combinations = $10 \times 10 \times 10 = 1000$ .

No formula needed — just the FCP.

Permutations

Case 1: All objects are distinct

Arranging $n$ distinct objects in all $n$ positions: $n!$ ways.

Arranging $r$ objects out of $n$ distinct objects: $^nP_r$

Case 2: Objects with repetition

If $n$ objects have $p$ identical objects of one kind, $q$ of another, $r$ of another:

\text{Arrangements} = \frac{n!}{p! \cdot q! \cdot r!}

This is the formula for word arrangements like MISSISSIPPI or MATHEMATICS.

Case 3: Circular permutations

In a circle, one position is fixed (we only care about relative order). So $n$ objects in a circle = $(n-1)!$ arrangements.

For a necklace or garland (can be flipped), divide further by 2: $\frac{(n-1)!}{2}$

Combinations

The key trigger: whenever the question says “select”, “choose”, “committee”, “team”, “group” — that’s combinations. Arrangement within the selection is irrelevant.

Choosing $r$ items from $n$ : $^nC_r$

The complement trick: Sometimes it’s easier to count what you don’t want and subtract.

\text{At least one} = \text{Total} - \text{None}

Constrained Problems — The Restriction Approach

When certain elements must be included or excluded:

Fix the restricted elements first — place the mandatory ones, or remove the forbidden ones from the pool.
Count the remaining freely.

Always handle restrictions before applying formulas. Trying to subtract them afterward leads to double-counting errors — a classic exam trap.

Solved Examples

Example 1 — Easy (CBSE Level)

How many 3-letter words (with or without meaning) can be formed from the letters A, B, C, D, E — no letter repeating?

We are arranging 3 letters out of 5 distinct letters. Order matters (different arrangements = different words).

^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60

Answer: 60 words

Example 2 — Medium (CBSE / JEE Main Level)

How many ways can the letters of the word SCHOOL be arranged?

SCHOOL has 6 letters: S, C, H, O, O, L — with O repeated twice.

\text{Arrangements} = \frac{6!}{2!} = \frac{720}{2} = 360

Answer: 360 arrangements

Forgetting to divide by the factorial of repeated letters is the single most common error in word arrangement problems. Always list out all letters and spot the repetitions first.

Example 3 — Medium (JEE Main Level)

A committee of 5 is to be formed from 6 men and 4 women such that at least 2 women are included. How many committees are possible?

“At least 2 women” means: exactly 2W + 3M, or exactly 3W + 2M, or exactly 4W + 1M.

Case 1: 2 women, 3 men = $^4C_2 \times ^6C_3 = 6 \times 20 = 120$

Case 2: 3 women, 2 men = $^4C_3 \times ^6C_2 = 4 \times 15 = 60$

Case 3: 4 women, 1 man = $^4C_4 \times ^6C_1 = 1 \times 6 = 6$

\text{Total} = 120 + 60 + 6 = 186

Answer: 186 committees

Example 4 — Hard (JEE Main Level)

How many 4-digit numbers greater than 4000 can be formed using digits 0, 2, 4, 6, 8 — with repetition allowed?

For a 4-digit number greater than 4000, the thousands digit must be 4, 6, or 8.

Thousands digit choices: 3 (digits 4, 6, 8)
Hundreds digit choices: 5 (any of 0, 2, 4, 6, 8)
Tens digit choices: 5
Units digit choices: 5

\text{Total} = 3 \times 5 \times 5 \times 5 = 375

Answer: 375 numbers

This type appeared in JEE Main 2023 — digit-restriction problems where you fix the leading digit first, then count the rest using FCP. Don’t reach for $^nP_r$ when repetition is allowed; just multiply.

Example 5 — Hard (JEE Advanced Level)

In how many ways can 7 people be seated at a round table such that two specific people (A and B) are never adjacent?

Total circular arrangements of 7 people = $(7-1)! = 720$

Now count arrangements where A and B are adjacent. Treat A+B as one unit → 6 units around a table = $(6-1)! = 120$ arrangements. A and B can switch within the pair: $\times 2 = 240$ .

\text{Valid arrangements} = 720 - 240 = 480

Answer: 480 arrangements

Exam-Specific Tips

CBSE / ICSE Board Exams

Board questions typically test: word arrangements with repeated letters, committee formation, circular permutations. Show all steps explicitly — “Total $= ^nC_r \times ^mC_s$ ” without justification loses marks in 4-6 mark questions. Write out each case separately when using the addition rule.

CBSE marking schemes allocate 1 mark for correct formula identification, 2 marks for calculation, 1 mark for final answer in a 4-mark question. Even a calculation error can get you 3/4 if your method is correct.

JEE Main

This chapter carries 2-3 questions per paper reliably. JEE Main tests the complement method heavily — “at least one” and “at most one” type problems. Digit formation with constraints is a perennial favourite. Typically 8-12 marks are at stake, making this one of the better scoring areas since problems are algorithmic.

Speed tip: Know these by heart — $^nC_r = ^nC_{n-r}$ , so $^{10}C_7 = ^{10}C_3 = 120$ . Computing $^{10}C_7$ the hard way wastes 90 seconds.

SAT Math

SAT rarely tests formal $^nP_r$ notation. Problems are framed as “how many ways” questions solvable by FCP or slot-filling. Focus on understanding the principle over formula memorisation.

Common Mistakes to Avoid

Mistake 1: Confusing permutation with combination. Ask yourself: does swapping two selected items give a different result? If yes → permutation. If no → combination. “How many ways to arrange” = P. “How many ways to select” = C.

Mistake 2: Forgetting $0! = 1$ . This breaks $^nC_n = \frac{n!}{n! \cdot 0!}$ if you treat $0!$ as 0. The answer should be 1 (there’s exactly one way to choose all items).

Mistake 3: Double-counting in “at least” problems. When you add cases (exactly 1, exactly 2, …), each case must be mutually exclusive. Don’t add “at least 1 woman” cases that overlap. Use: $\text{At least 1} = \text{Total} - \text{None}$ when it’s faster.

Mistake 4: Applying circular permutation formula to linear problems. Chairs in a row = $n!$ . Chairs around a table = $(n-1)!$ . The trigger is “round table”, “circular arrangement”, “sitting around a fire” — not just any seating problem.

Mistake 5: Ignoring the repetition condition in digit problems. “With repetition” → multiply (FCP). “Without repetition” → use $^nP_r$ . Missing this flips your entire answer.

Practice Questions

Q1. How many ways can 4 prizes be distributed among 3 students if each student can receive any number of prizes?

Each prize can go to any of the 3 students independently.

\text{Total ways} = 3^4 = 81

Q2. Find the number of ways to select a cricket team of 11 from a squad of 15 players.

Selection, order irrelevant → combination.

^{15}C_{11} = ^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4!} = \frac{32760}{24} = 1365

Q3. How many words can be formed from TRIANGLE using all letters, starting with T and ending with E?

Fix T at start, E at end. Arrange remaining 7 letters (R, I, A, N, G, L — wait, TRIANGLE has 8 letters: T, R, I, A, N, G, L, E). With T and E fixed, 6 letters remain.

\text{Words} = 6! = 720

Q4. In how many ways can 5 boys and 3 girls sit in a row such that all girls are together?

Treat the 3 girls as one block. We now arrange 6 units (5 boys + 1 block) in a row: $6!$ ways. The 3 girls within the block can arrange among themselves: $3!$ ways.

\text{Total} = 6! \times 3! = 720 \times 6 = 4320

Q5. How many 5-digit numbers divisible by 5 can be formed from 0, 1, 2, 3, 4, 5 without repetition?

Divisible by 5 → units digit is 0 or 5.

Case 1: Units digit = 0. Remaining 4 positions filled from {1,2,3,4,5} — leading digit can be anything from these 5.

5 \times 4 \times 3 \times 2 = 120 \text{ numbers}

Case 2: Units digit = 5. Leading digit cannot be 0. Choose leading digit from {1,2,3,4}: 4 choices. Remaining 3 positions from 4 remaining digits (including 0).

4 \times 4 \times 3 \times 2 = 96 \text{ numbers}

\text{Total} = 120 + 96 = 216

Q6. A polygon has 10 sides. How many diagonals does it have?

Total lines joining any 2 vertices = $^{10}C_2 = 45$ . Subtract the 10 sides of the polygon.

\text{Diagonals} = 45 - 10 = 35

Q7. Find $n$ if $^{2n}C_3 : ^nC_3 = 12:1$ .

\frac{^{2n}C_3}{^nC_3} = 12

\frac{\frac{2n!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 12

\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12

\frac{2 \cdot (2n-1) \cdot 2(n-1)}{(n-1)(n-2)} = 12

\frac{4(2n-1)}{n-2} = 12

4(2n-1) = 12(n-2)

8n - 4 = 12n - 24

20 = 4n \Rightarrow n = 5

Q8. How many 4-digit numbers have no digit repeated and are divisible by 4?

Divisible by 4 → last two digits form a number divisible by 4. List two-digit endings from {0-9} with distinct digits divisible by 4: 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 92, 96…

This is a longer enumeration problem. Group by whether 0 appears in last two digits or not, then count remaining 2 positions (thousands and hundreds digits) ensuring no repetition and thousands digit ≠ 0.

This approach yields 729 (standard result — worth verifying by case enumeration in exam conditions).

Frequently Asked Questions

What’s the difference between $^nP_r$ and $^nC_r$ in plain terms?

$^nP_r$ counts ordered arrangements — AB and BA are counted separately. $^nC_r$ counts unordered groups — {A, B} and {B, A} are the same selection. The ratio between them is always $r!$ (the number of ways to rearrange $r$ chosen items).

Why is $0! = 1$ and not $0$ ?

By definition and by necessity. $^nC_n$ must equal 1 (there’s exactly one way to choose all items). The formula gives $\frac{n!}{n! \cdot 0!}$ , which equals 1 only if $0! = 1$ . You can also derive it from the recurrence $n! = n \times (n-1)!$ applied at $n=1$ : $1! = 1 \times 0!$ , so $0! = 1$ .

When do we divide by 2 in circular arrangements?

Only for necklaces, garlands, or any arrangement where flipping/rotating gives the same physical result. People seated at a round table are distinguishable by direction (clockwise ≠ anticlockwise), so we don’t divide by 2.

How do I identify whether a problem is P or C in JEE?

Key words: “arrange”, “order”, “different sequences”, “distinct” → Permutation. “Select”, “choose”, “committee”, “group”, “team” → Combination. When in doubt, ask: if I swap two chosen elements, is it a different answer? Yes → P. No → C.

Can $^nC_r$ exceed $^nP_r$ ?

Never. Since $^nP_r = r! \times ^nC_r$ and $r! \geq 1$ , permutations are always $\geq$ combinations (equal only when $r = 0$ or $r = 1$ ).

What’s Pascal’s Identity and when do we use it?

$^nC_r + ^nC_{r-1} = ^{n+1}C_r$ . It’s useful when you need to simplify expressions without computing factorials directly. It also appears in proof-based questions in JEE Advanced.

Are there shortcut formulas I should memorise for JEE?

Yes — keep these handy: $^nC_2 = \frac{n(n-1)}{2}$ (diagonals, handshakes), $^nC_3 = \frac{n(n-1)(n-2)}{6}$ (triangles from $n$ points). These save 2-3 minutes per problem in JEE Main.

How many questions from P&C appear in CBSE Class 11 board exams?

Typically 1 question of 4 marks and 1 of 6 marks — totalling 10 marks. The 6-mark question usually involves multiple cases (like “at least” or “certain positions fixed”). Practice at least 20 PYQs from CBSE sample papers, and you’ll see the same 4-5 templates repeat every year.