Permutation

Q: When do we use (n-1)! instead of n! ?

Use (n-1)! for circular arrangements of n distinct objects, like people around a round table with unnumbered seats. Rotations produce the same relative order, so we fix one object and permute the rest.

Q: What is the value of 0! and why?

0! is defined as 1, so that formulas like nPr = n!/(n-r)! and nC0 = 1 work uniformly across all valid r.

Q: In how many ways can n people be seated in a row?

n! ways. Each seat can be filled by any of the remaining people, giving n times (n-1) times (n-2) and so on down to 1.

Q: How do we count arrangements when some letters are repeated?

Divide n! by the factorial of each repeated group. For n letters with p alike, q alike, r alike, use n!/(p! q! r!). BANANA gives 6!/(3! 2!) = 60.

Q: Is permutation important for JEE Mains?

Yes. JEE Mains typically asks 1 to 2 questions from Permutations and Combinations every year, often on arrangements with restrictions, circular permutations, ranks of words, or sums of digit-based numbers.

Q: Does the formula n^r count arrangements or selections?

It counts arrangements. Each of the r ordered positions is filled independently by one of n objects, and order matters.

MathsPermutations And CombinationsFor JEE aspirants

A permutation is an ordered arrangement of objects, where the sequence matters. If you select $r$ objects from $n$ distinct objects and arrange them in a line, the number of possible arrangements is given by the permutation formula $^{n} P_{r} = \frac{n !}{( n - r )!}$ . Permutation problems in JEE Mains build on two counting principles (multiplication and addition), extend to circular arrangements $((n - 1)!)$ , and cover cases with repeated or identical objects using $\frac{n !}{p ! q ! r !}$ .

Key Formulas - Quick Reference

Multiplication principle: If Job A can be done in $m$ ways and Job B in $n$ ways, then both A AND B can be done in $m \times n$ ways.
Addition principle: If Job A can be done in $m$ ways and Job B in $n$ ways, then A OR B (exactly one) can be done in $m + n$ ways.
Permutations without repetition: $^{n} P_{r} = \frac{n !}{( n - r )!}$ , where $0 \leq r \leq n$ .
All $n$ objects arranged: $^{n} P_{n} = n!$
With repetition ( $r$ positions, each fillable by any of $n$ objects): total $= n^{r}$ .
Identical objects: Arrangements of $n$ objects with $p$ alike of one kind, $q$ alike of another, $r$ alike of another $= \frac{n !}{p ! q ! r !}$ .
Circular permutation (clockwise $\neq =$ anticlockwise): $(n - 1)!$
Necklace / bracelet (clockwise $=$ anticlockwise): $\frac{( n - 1 )!}{2}$
Sum of all $n$ -digit numbers formed from $n$ distinct non-zero digits $= (n - 1)! \times (sum of digits) \times n ones 111 \dots 1$

1. Fundamental Principles of Counting

Every permutation problem reduces to counting arrangements. Two principles do all the heavy lifting.

1.1 Multiplication Principle (AND)

If an event A can occur in

m

ways and, independently, an event B can occur in

n

ways, then the pair (A, B) can occur in

m \times n

ways.

Use this when a task splits into sub-tasks that must all be completed in sequence.

Figure 1: Multiplication principle. 3 routes from A to B, each with 4 onward routes to C, gives

3 \times 4 = 12

total paths.

Solved Example 1

A person travels from station A to station C via station B. There are 3 routes from A to B and 4 routes from B to C. In how many ways can he travel from A to C?

Solution:

Choosing a route from A to B and a route from B to C are two independent sub-tasks. By the multiplication principle: $3 \times 4 = 12$ ways.

1.2 Addition Principle (OR)

If event A can occur in

m

ways and event B can occur in

n

ways, and only one of them is to be performed, then the total number of ways

= m + n

Solved Example 2

A college offers 7 morning courses and 5 evening courses. In how many ways can a student pick exactly one course (either morning or evening)?

Solution:

Only one course is chosen, and it must come from one of the two disjoint groups. By the addition principle: $7 + 5 = 12$ ways.

Rule of thumb: "AND" means multiply, "OR" (exclusive) means add. If tasks can happen together, use multiplication; if the choice is between mutually exclusive options, use addition.

2. Permutations Without Repetition

A permutation of $n$ objects taken $r$ at a time is an ordered selection of $r$ objects from a set of $n$ distinct objects. To count them, imagine filling $r$ boxes in order:

The 1st box has $n$ choices.
The 2nd box has $n - 1$ choices (one already used).
The 3rd box has $n - 2$ choices, and so on.
The $r$ -th box has $n - r + 1$ choices.

^{n} P_{r} = n (n - 1) (n - 2) \dots (n - r + 1) = \frac{n !}{( n - r )!}, 0 \leq r \leq n

When all

n

objects are arranged:

^{n} P_{n} = n!

Solved Example 3

How many (i) 5-digit and (ii) 3-digit numbers can be formed using the digits

1, 2, 3, 4, 5

without repetition?

Solution:

(i) Filling 5 positions with 5 distinct digits: $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$ numbers.

(ii) Filling 3 positions: $5 \times 4 \times 3 = 60$ numbers. Equivalently, $^{5} P_{3} = \frac{5 !}{2 !} = 60$ .

3. Permutations With Repetition

If each of the $r$ positions can independently be filled by any of the $n$ objects (repetition allowed), then each position has $n$ choices and the total number of arrangements is:

Number of permutations with repetition

= n^{r}

Solved Example 4

How many 3-digit numbers can be formed using the digits

0, 1, 2, 3, 4, 5

if (a) digits may not be repeated, (b) digits may be repeated?

Solution:

Let the number be XYZ. The hundreds digit X cannot be 0.

(a) No repetition: X has 5 choices (1-5). Y has 5 choices left (including 0, minus the one used for X). Z has 4 choices. Total $= 5 \times 5 \times 4 = 100$ .

(b) Repetition allowed: X has 5 choices (1-5). Y and Z each have 6 choices (any digit 0-5). Total $= 5 \times 6 \times 6 = 180$ .

4. Permutations of Identical Objects

n

objects contain

p

alike of one kind,

q

alike of another kind, and

r

alike of a third kind (the rest all distinct), the number of distinct arrangements is

\frac{n !}{p ! q ! r !}

Divide by the factorial of each repeated group because permuting identical letters among themselves does not produce a new arrangement.

Solved Example 5

Using all the letters of the word HINDUSTAN, find the number of arrangements (anagrams) such that:

All arrangements (total).
Begin and end with a vowel.
All the vowels come together.
No two vowels are together.
Vowels and consonants occupy the same relative positions as in HINDUSTAN.

Solution:

HINDUSTAN has 9 letters: vowels ${I, U, A}$ (3 distinct), consonants ${H, D, S, T, N, N}$ (6 with N repeated).

(a) Total arrangements $= \frac{9 !}{2 !} = 181440$ (divide by $2!$ for the two N's).

(b) First position: 3 vowel choices. Last position: 2 remaining vowel choices. Middle 7 positions filled by the remaining 7 letters (1 vowel + 6 consonants including 2 N's) in $\frac{7 !}{2 !}$ ways. Total $= 3 \times 2 \times \frac{7 !}{2 !} = 6 \times 2520 = 15120$ .

(c) Bundle the 3 vowels IUA as one super-letter. We now arrange 7 units (bundle + 6 consonants with 2 N's) in $\frac{7 !}{2 !}$ ways; the 3 vowels inside the bundle permute in $3! = 6$ ways. Total $= \frac{7 !}{2 !} \times 6 = 2520 \times 6 = 15120$ .

(d) Arrange the 6 consonants first (with 2 N's alike): $\frac{6 !}{2 !} = 360$ ways. This creates 7 gaps (including ends) for the 3 vowels; choose and arrange 3 vowels in 7 gaps: $^{7} P_{3} = 210$ . Total $= 360 \times 210 = 75600$ .

(e) Vowel positions (3 slots) and consonant positions (6 slots) are fixed. Vowels permute among their slots in $3! = 6$ ways; consonants permute among theirs in $\frac{6 !}{2 !} = 360$ ways. Total $= 6 \times 360 = 2160$ .

5. Circular Permutations

In a linear arrangement, sliding everyone one seat to the right produces a new arrangement. In a circular arrangement, this rotation gives the same seating pattern - only relative positions matter. So we fix one object and permute the remaining $(n - 1)$ :

Circular arrangement of

n

distinct objects (clockwise

\neq =

anticlockwise):

\frac{n !}{n} = (n - 1)!

Necklace / bracelet (clockwise

=

anticlockwise, i.e., flipping is allowed):

\frac{( n - 1 )!}{2}

Figure 2: Circular arrangements. When clockwise and anticlockwise orderings are distinct, count is

(n - 1)!

. When they are treated as identical (necklace), divide by 2.

When positions are numbered (like numbered chairs at a table), a circular arrangement behaves like a linear one; use

n!

, not

(n - 1)!

Solved Example 6

A necklace contains 23 different coloured beads. In how many ways can they be arranged so that 3 specific beads are always together?

Solution:

Treat the 3 specific beads as one bundle. We effectively arrange $23 - 3 + 1 = 21$ objects on a necklace. Circular arrangements of 21 objects on a necklace $= \frac{( 21 - 1 )!}{2} = \frac{20 !}{2}$ . The 3 beads inside the bundle permute in $3!$ ways. Total $= \frac{20 ! \cdot 3 !}{2}$ .

Solved Example 7

In how many ways can 10 boys and 5 girls sit around a circular table so that no two girls sit together?

Solution:

First seat the 10 boys around the circle: $(10 - 1)! = 9!$ ways. This creates 10 gaps between boys. Place the 5 girls in any 5 of these 10 gaps (order matters): $^{10} P_{5} = \frac{10 !}{5 !}$ . Total $= 9! \times \frac{10 !}{5 !}$ .

6. Advanced JEE Techniques

6.1 Gap Method (for "no two together")

When items of one type must not be adjacent, first arrange the other type, then insert the restricted items into the gaps.

Figure 3: Gap method. Arrange

k

objects first, then place restricted items into the

(k + 1)

gaps so no two are adjacent.

6.2 Sum of All Numbers Formed

Sum of all

n

-digit numbers formed using

n

distinct non-zero digits (each used exactly once)

= (n - 1)! \times (sum of digits) \times n ones 111 \dots 1

Why: Each digit appears in each place value exactly

(n - 1)!

times.

Solved Example 8

Find the sum of all 4-digit numbers formed using the digits

1, 3, 5, 7

without repetition.

Solution:

Number of digits $n = 4$ . Sum of digits $= 1 + 3 + 5 + 7 = 16$ . Repunit with 4 ones $= 1111$ .

Sum $= (4 - 1)! \times 16 \times 1111 = 6 \times 16 \times 1111 = 106656$ .

6.3 Rank of a Word in Dictionary Order

To find where a word sits when all letter permutations are listed alphabetically: for each letter of the word (left to right), count letters strictly smaller than it that have not been used yet, then multiply by the factorial of remaining positions. Sum these counts and add 1.

Solved Example 9

Find the rank of the word MOTHER when its letters are arranged alphabetically.

Solution:

Letters of MOTHER in alphabetical order: ${E, H, M, O, R, T}$ (6 distinct letters).

1st letter M: letters before M in the list = ${E, H}$ , so $2 \times 5! = 240$ .
2nd letter O (M used): letters before O still available = ${E, H}$ , so $2 \times 4! = 48$ .
3rd letter T (M, O used): letters before T available = ${E, H, R}$ , so $3 \times 3! = 18$ .
4th letter H (M, O, T used): letters before H available = ${E}$ , so $1 \times 2! = 2$ .
5th letter E (M, O, T, H used): letters before E available = none, so $0 \times 1! = 0$ .
6th letter R: last position, adds 0.

Rank $= 240 + 48 + 18 + 2 + 0 + 0 + 1 = 309$ .

Common Mistakes to Avoid

Watch out

Confusing "AND" with "OR": multiply for AND, add for OR. "Choose one course and one book" is multiplication; "choose one course or one book" is addition.
Forgetting the zero-in-first-place rule for digit problems: a 3-digit number cannot start with 0.
Using $n!$ for circular arrangements instead of $(n - 1)!$ . Only use $n!$ when positions are numbered.
Missing the divide-by-2 for necklaces: a necklace can be flipped, so clockwise and anticlockwise are the same arrangement.
Not dividing by factorials of repeated letters: the total arrangements of MISSISSIPPI is $\frac{11 !}{4 ! 4 ! 2 !}$ , not $11!$ .
Bundling errors: when items must be together, treat them as one super-item, but do not forget to multiply by the internal arrangements of the bundle.

Frequently Asked Questions

Q1. What is the difference between a permutation and a combination?

A permutation is an ordered arrangement, so ABC and BCA are counted separately. A combination is an unordered selection, so both are the same choice ${A, B, C}$ . Formulaically, $^{n} P_{r} = r! \cdot^{n} C_{r}$ .

Q2. When do we use $(n - 1)!$ instead of $n!$ ?

Use $(n - 1)!$ for circular arrangements of $n$ distinct objects (like people around a round table with unnumbered seats). The rotation of any arrangement produces the same relative order, so we fix one object and permute the rest.

Q3. How is a necklace different from a round-table seating?

A round table is fixed in space, so clockwise and anticlockwise orderings look different. A necklace can be flipped over, so the mirror-image order is the same arrangement. That is why we divide by 2 for necklaces: $\frac{( n - 1 )!}{2}$ .

Q4. What is the value of $0!$ and why?

$0! = 1$ . This is a definition chosen so that formulas like $^{n} P_{r} = \frac{n !}{( n - r )!}$ work when $r = n$ (giving $^{n} P_{n} = n!$ ) and combinations $^{n} C_{0} = 1$ hold uniformly.

Q5. In how many ways can $n$ people be seated in a row?

$n!$ ways. Each of the $n$ seats can be filled by any of the remaining people, giving $n \times (n - 1) \times (n - 2) \times \dots \times 1 = n!$ .

Q6. How do we count arrangements when some letters are repeated?

Divide $n!$ by the factorial of each repeated group. For $n$ letters with $p$ alike, $q$ alike, $r$ alike, use $\frac{n !}{p ! q ! r !}$ . Example: BANANA has 6 letters with 3 A's and 2 N's, giving $\frac{6 !}{3 ! 2 !} = 60$ arrangements.

Q7. Is permutation important for JEE Mains?

Yes. JEE Mains typically asks 1-2 questions from Permutations and Combinations every year, often on arrangements with restrictions, circular permutations, ranks of words, or sums of digit-based numbers. It is also a prerequisite for Probability.

Q8. How do I find the rank of a word in dictionary order?

Sort the letters alphabetically. For each letter of the target word (left to right), count how many unused letters come before it in the sorted list, and multiply by the factorial of the number of remaining positions. Sum all these products and add 1 for the word itself.

Q9. Does the formula $n^{r}$ (with repetition) count arrangements or selections?

It counts arrangements. Each of the $r$ ordered positions is filled independently by one of $n$ objects, and order matters. For example, placing 3 letters in 5 boxes (repetition allowed) gives $5^{3} = 125$ arrangements.

Previous year questions on Permutation

10 questions from past papers, each with a step-by-step solution.

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1. Fundamental Principles of Counting

1.1 Multiplication Principle (AND)

1.2 Addition Principle (OR)

2. Permutations Without Repetition

3. Permutations With Repetition

4. Permutations of Identical Objects

5. Circular Permutations

6. Advanced JEE Techniques

6.1 Gap Method (for "no two together")

6.2 Sum of All Numbers Formed

6.3 Rank of a Word in Dictionary Order

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between a permutation and a combination?

Q2. When do we use (n−1)! instead of n!?

Q3. How is a necklace different from a round-table seating?

Q4. What is the value of 0! and why?

Q5. In how many ways can n people be seated in a row?

Q6. How do we count arrangements when some letters are repeated?

Q7. Is permutation important for JEE Mains?

Q8. How do I find the rank of a word in dictionary order?

Q9. Does the formula nr (with repetition) count arrangements or selections?

Previous year questions on Permutation

Ready to master Permutations And Combinations?

Q2. When do we use $(n - 1)!$ instead of $n!$ ?

Q4. What is the value of $0!$ and why?

Q5. In how many ways can $n$ people be seated in a row?

Q9. Does the formula $n^{r}$ (with repetition) count arrangements or selections?