MathsCBSE Class 11 · Permutations and Combinations

Permutations: Arrangements with Conditions, Repeated Objects and Rank of a Word

Updated 17 October 2026 · Fact-checked

The number of arrangements of n distinct objects taken r at a time is nPr = n!/(n − r)!. If p objects are alike of one kind, q of another, the arrangements of all n are n!/(p! q! …). For "together" glue the objects into one block and multiply by the block's internal arrangements; for "never together" subtract this from the total. For digit problems fill the restricted places first.

Understand Permutations: Arrangements with Conditions and Repeated Objects

Basics. By the fundamental principle of counting, $r$ places filled in turn from $n$ distinct objects without repetition give
$ $^{n} P_{r} = n (n - 1) \dots (n - r + 1) = \frac{n !}{( n - r )!},^{n} P_{n} = n!, 0! = 1.$ $
With repetition allowed the answer is $n^{r}$ . Use permutations whenever order matters.

Repeated objects. The arrangements of $n$ objects in which $p$ are alike (of one kind), $q$ alike (of another), $r$ alike... number $\frac{n !}{p ! q ! r ! \dots}$ . For PROPORTION (10 letters; O three times, P twice, R twice) this gives $\frac{10 !}{3 ! 2 ! 2 !} = 151200$ .

Conditions. Four standard moves:
1. Together: glue the objects into one block, arrange the blocks, then multiply by the arrangements inside the block.
2. Never together: total arrangements $-$ arrangements with them together. (For "no two together" use the gap method: arrange the others first, then choose gaps for the restricted objects.)
3. Fixed positions or fixed ends: put the restricted objects in place first and arrange the rest.
4. Digits and numbers: fill the most restricted place first (the first digit cannot be $0$ , the last digit decides even or odd or divisibility by $5$ ). Where the choice for a later place depends on an earlier one (as in the digit $0$ case for even numbers), split into cases.

Gap method example. COMPUTER has 5 consonants and 3 vowels. Arrange the consonants in $5!$ ways; they leave $6$ gaps; place the three vowels in distinct gaps in $^{6} P_{3}$ ways: $120 \times 120 = 14400$ arrangements with no two vowels together. Compare with "the three vowels not all together": $8! - 6! \cdot 3! = 36000$ .

Rank of a word. List the letters alphabetically. For each position, count how many unused letters are alphabetically smaller and multiply by the number of arrangements of the remaining positions, then add 1 for the word itself. Handle repeated letters by dividing by the factorials of repeats.

Order of two objects. Of all arrangements in which two chosen objects are separate, exactly half have the first object before the second (symmetry).

Permutations: Arrangements with Conditions and Repeated Objects: formulas to remember

Permutations of distinct objects: $n P r = n! / (n - r)!$; Order matters.
All objects distinct: $n! a r r an g e m e n t so f n o bj ec t s$; nPn = n!.
Repeated objects: $n! / (p! q! r! \dots)$; p, q, r are the repeat counts.
Together: $(b l oc k s)! \times (in t er na l a r r an g e m e n t s)$; Glue into a block.
Never together: $T o t a l - t o g e t h er$; Or use the gap method.
Repetition allowed: $n^{r}$; Each place has n choices.

How to solve Permutations: Arrangements with Conditions and Repeated Objects questions

Use these steps for arrangements with conditions.

1List the objects and note any repeats (use the factorial division) and any restrictions.
2Decide whether the restriction is "together", "never together", "fixed position", or a digit rule.
3For "together", glue the group; for "never together", compute the total minus the together case, or use the gap method.
4For digit problems, fill the most restricted place first and split into cases when later choices depend on earlier ones.
5Multiply the independent stages and add the cases.
6Check with a smaller version of the problem when possible.

Permutations: Arrangements with Conditions and Repeated Objects in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

PINs and repetition

Suppose a bank wants a 4-digit PIN using the digits 0 to 9.

How many PINs are possible if repetition is allowed and how many if all four digits must be different?

Try it first, then show the solutionSolution

With repetition: each of the 4 places has 10 choices, so $1 0^{4} = 10000$ .
Without repetition: $^{10} P_{4} = 10 \times 9 \times 8 \times 7 = 5040$ .

Answer: 10000 with repetition; 5040 with all digits different.

Memory hook: With repetition the number of choices stays the same at each place.

Neighbours in a photograph

Suppose 5 friends stand in a row for a photograph, and two of them, Asha and Bela, insist on standing next to each other.

How many arrangements are possible, and in how many are they not adjacent?

Try it first, then show the solutionSolution

Glue Asha and Bela into one block: 4 units can be arranged in $4! = 24$ ways, and the block has $2! = 2$ internal orders: $24 \times 2 = 48$ .
All arrangements: $5! = 120$ .
Not adjacent: $120 - 48 = 72$ .

Answer: 48 with them together; 72 with them apart.

Memory hook: Together is easy to count, so find it first and subtract.

Tricks to remember Permutations: Arrangements with Conditions and Repeated Objects

Short memory aids. Say each one aloud once, then use it on a question the same day.

Most restricted place first
Fill the first digit or the end positions before the free places.
Use it when: Digit and number problems.
Glue and multiply
Together: treat as one object, then multiply by the internal arrangements.
Use it when: Neighbour conditions.
Total minus together
Never together: count the complement.
Use it when: Separation conditions.
Gaps method
Arrange the unrestricted objects first, then choose gaps for the restricted ones.
Use it when: No two of a kind adjacent.

Permutations: Arrangements with Conditions and Repeated Objects: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Solving for n · 2 marks · Very short answerBoard level

If $^{n} P_{4} = 20^{n} P_{2}$ , find $n$ .

Try it first, then show the solutionSolution

$^{n} P_{4} = n (n - 1) (n - 2) (n - 3)$ and $^{n} P_{2} = n (n - 1)$ ; cancel $n (n - 1)$ to get $(n - 2) (n - 3) = 20$ . (1 mark)
$n^{2} - 5 n - 14 = 0$ , so $(n - 7) (n + 2) = 0$ ; $n = 7$ (reject $- 2$ ). (1 mark)

Answer: $n = 7$

Example 2: Repeated letters · 3 marks · Short answerBoard level

How many different arrangements can be made from all the letters of the word PROPORTION?

Try it first, then show the solutionSolution

The word has 10 letters: O three times, P twice, R twice, and T, I, N once each. (1 mark)
Number of arrangements $= \frac{10 !}{3 ! 2 ! 2 !}$ . (1 mark)
$= \frac{3628800}{24} = 151200$ . (1 mark)

Answer: 151200

Example 3: Even numbers with a restriction · 3 marks · Short answerHarder board level

How many 4-digit even numbers can be formed from the digits 0, 1, 2, 3, 4, 5 if no digit is repeated?

Try it first, then show the solutionSolution

Case 1: last digit 0. The first three places take 3 of the other 5 digits: $5 \times 4 \times 3 = 60$ . (1 mark)
Case 2: last digit 2 or 4 (2 ways). First digit: not 0 and not the last digit, so 4 ways; second place 4 ways; third place 3 ways: $2 \times 4 \times 4 \times 3 = 96$ . (1 mark)
Total $= 60 + 96 = 156$ . (1 mark)

Answer: 156

Example 4: Vowels together and apart · 4 marks · Short answerHarder board level

Using the letters of COMPUTER (8 distinct letters, 3 vowels O, U, E), find the number of arrangements in which (a) the three vowels are not all together, (b) no two vowels are together.

Try it first, then show the solutionSolution

Total arrangements $= 8! = 40320$ . (½ mark)
(a) All three vowels together: glue them: $6! \times 3! = 720 \times 6 = 4320$ . Not all together: $40320 - 4320 = 36000$ . (1½ marks)
(b) Arrange the 5 consonants: $5! = 120$ . They create 6 gaps; place the 3 vowels in different gaps: $^{6} P_{3} = 120$ . (1 mark)
Number $= 120 \times 120 = 14400$ . (1 mark)

Answer: (a) 36000 (b) 14400

Example 5: Rank of a word · 5 marks · Long answerHOTS / competency

Letters of the word MOTHER are arranged in all possible ways and listed alphabetically as in a dictionary. Find the position of MOTHER.

Try it first, then show the solutionSolution

Alphabetical order of letters: E, H, M, O, R, T. (½ mark)
Words starting with E or H: $2 \times 5! = 240$ . (1 mark)
Starting with M, second letter smaller than O (E or H): $2 \times 4! = 48$ . (1 mark)
MO then third letter smaller than T (E, H, R): $3 \times 3! = 18$ . MOT then fourth letter smaller than H (E): $1 \times 2! = 2$ . (1 mark)
MOTH then E: no smaller letter left (0 words). Words before MOTHER $= 240 + 48 + 18 + 2 = 308$ , so its rank is $309$ . (1½ marks)

Answer: 309th

Where students lose marks in Permutations: Arrangements with Conditions and Repeated Objects

The slips examiners see most often, and what to write instead.

Mistake: Ignoring repeated letters
Identical letters make some arrangements the same.
Wrong working
PROPORTION: $10! = 3628800$ .
Right working
$\frac{10 !}{3 ! 2 ! 2 !} = 151200$ .
Do this: Underline repeated letters before counting.
Usually costs: The answer.
Mistake: Forgetting the internal arrangements of a block
The glued group can be ordered in several ways.
Wrong working
Two together among 5: $4! = 24$ .
Right working
$4! \times 2! = 48$ .
Do this: Write "blocks" and "inside" as two factors.
Usually costs: The answer.
Mistake: Allowing 0 in the first place
A number cannot begin with 0.
Wrong working
4-digit numbers from 0 to 5: first place has 6 choices.
Right working
First place has 5 choices (not 0).
Do this: Fill the first digit first and check the digit 0 explicitly.
Usually costs: 1 to 2 marks.
Mistake: Counting a case twice
Overlapping cases are added without checking.
Wrong working
Even numbers: "last digit 0, 2 or 4" all counted as 60 each.
Right working
Last digit 0 gives 60; last digit 2 or 4 gives 48 each (first digit not 0).
Do this: Treat the zero case separately.
Usually costs: The answer.

Exam tips

Write the structure of the count (for example "5! × 6P3") before multiplying out; partial credit depends on it.
For digits, underline the places and write the number of choices over each one.
Verify "never together" with total minus together, and "no two together" with the gap method.
In rank problems, show a small table of position, smaller letters and remaining arrangements.
Repetition allowed means $n^{r}$ , not $^{n} P_{r}$ ; read the wording carefully.

Practice questions on Permutations: Arrangements with Conditions and Repeated Objects

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $^{8} P_{3}$ is:
(a) 56
(b) 336
(c) 512
(d) 24
Hint
$^{8} P_{3} = 8 \times 7 \times 6$ .
Answer
(b). $8 \times 7 \times 6 = 336$ .
Q2 · 1 mark · MCQHarder board level
In how many ways can 4 men and 3 women be arranged in a row so that no two women are together?
(a) 1440
(b) 5040
(c) 720
(d) 144
Hint
Arrange the men, then place the women in the gaps.
Answer
(a). $4! = 24$ arrangements of men leave 5 gaps; women in gaps: $^{5} P_{3} = 60$ ; $24 \times 60 = 1440$ .
Q3 · 3 marks · Short answerHarder board level
How many 5-digit numbers greater than 40000 can be formed from the digits 1, 2, 3, 4, 5, 6 without repetition?
Hint
The first digit must be 4, 5 or 6.
Answer
First digit: 3 ways. The remaining 4 places from the other 5 digits: $^{5} P_{4} = 120$ . Total $3 \times 120 = 360$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The number of arrangements of the letters of the word LEVEL is 30.
Reason (R): L occurs twice and E occurs twice, so the number is $\frac{5 !}{2 ! 2 !}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count the repeated letters.
Answer
(a). $\frac{120}{4} = 30$ .
Q5 · 4 marks · Short answerHOTS / competency
Seven distinct books are arranged on a shelf. In how many arrangements are two particular books $A$ and $B$ not next to each other, and in how many of these does $A$ stand to the left of $B$ ?
Hint
Count the "together" case first; then use symmetry.
Answer
Total $7! = 5040$ . $A$ and $B$ together: $6! \times 2 = 1440$ . Not together: $5040 - 1440 = 3600$ . By symmetry half of these have $A$ left of $B$ : $1800$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is nPr?Show answer
n!/(n − r)!.
2.How many arrangements of n objects with p alike of one kind and q alike of another?Show answer
n!/(p! q!).
3.How do you count "never together"?Show answer
Total minus together.
4.What is 0!?Show answer
1.
5.With repetition allowed, how many r-letter arrangements from n letters?Show answer
n^r.

Permutations: Arrangements with Conditions and Repeated Objects: frequently asked questions

When do I use permutations rather than combinations?

Use permutations when order matters (positions, arrangements, codes) and combinations when it does not (selections, groups).

Why do repeated letters reduce the count?

Swapping two identical letters gives the same word, so each distinct word is counted p! times for p alike letters; divide by p!.

Why does 0 cause trouble in number problems?

A number cannot start with 0, so the first place has one fewer choice, and counting needs separate cases.

Are these questions only for JEE?

They are a standard part of the chapter. The harder arrangement problems on this page are calibrated to school and JEE-foundation level.