Permutation or Arrangement Concept

Permutation or arrangement is one of the most important and frequently tested topics under permutation and combination in quantitative aptitude. It is asked in almost every competitive exam including CAT, SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB and CSAT. A strong understanding of permutation or arrangement concept, formulas and tricks is essential for scoring well in these exams. This post covers number of ways of arranging n different objects, circular arrangement when clockwise and anticlockwise are different, formation of garland when clockwise and anticlockwise are same, word formation, number formation and rank of a word in dictionary — all explained with clear formulas and solved examples.

📚 What You Will Learn in This Post

1. Permutation or Arrangement

(i) Number of ways in which ‘n’ different objects can be arranged

3. Circular Arrangement

(i) When clockwise ↻ and anticlockwise ↺ arrangement are different then number of arrangement of n different things around a circle

(ii) When clockwise ↻ and anticlockwise ↺ arrangement are same: i.e. Formation of Garland then number of arrangement of n different things

5. Word/Number Formation

(i) Formation of Word

(ii) Formation of Number

7. Rank of a Word in Dictionary

Permutation or Arrangement

Number of ways in which ‘n’ different onjects can be arranged:
Total object = n
& n object will occupy n different places.

∴ total numbere of ways = n×(n-1)×(n-2)×(n-3)×…………×1
= n!
Hence

Example:- In how many ways 20 students can be arranged in a straight line?
Solution:-

20! Answer

Example:- In how many ways 20 similar objects can be arranged in a straight line?
Solution:-

Since objects are similar
∴ number of ways = 1 Answer

Example:- In how many ways 5 students can be arranged on 10 chairs in a straight line?
Solution:-

Always remember : ” First select & then arrange”
∴ 5 chair can be selected out of 10 chairs in ¹⁰C₅ ways.
now 5 students can be arranged in these 5 chairs in 5! ways
Hence total number of ways = ¹⁰C₅×5!
= ¹⁰P₅ Answer

Example:- There are 10 students in a class. If three particular students Suresh, Amit & Ankit are to always sit together then in how many ways these 10 students vcasnm be arranged in a straight line?
Solution:-

= 8!.3! ways Answer

Example:- In above question if three particular students never sit together.
Solution:-
(10! – 8!.3!) ways Answer
Example:- In how many ways 20 students can be arranged in a straight line if four students A, B, C, D are always together & B is always at last of A, B, C.
Solution:-

= 20 – 4 + 1
= 17
Hence total number of ways = 17!.3!.1
= 17!.3! ways Answer

Example:- How many 5 digit numbers can be formed out of digits 1, 2, 3, 7, 9. if repetition of digit is not allowed.
Solution:-

= 5×4×3×2×1
= 5! ways Answer

Circular arrangement

The main difference between circular & linear arrangement is the point of reference.
⇒ When clockwise ↻ & anticlockwise ↺ arrangement are different.

Let there are 10 students & 10 chairs in a circle.
For the first student there is only one way to sit as chairs are in circle not in straight line. So all places ( of chairs) will be equal for 1ˢᵗ student. now for 2ⁿᵈ & onward student this first student will work as point of reference.
⇒ now for 2ⁿᵈ student there are 9 options available to sit.
⇒ For 3ʳᵈ student there are 8 options available to sit.
⇒For 4th student there are 7 options available to sit.
⋮
⋮
⋮
⋮
⇒ For 10th student there is only 1 option available to sit.
Hence total number of ways
=1×9×8×7×6×……..×1
= 1×9!
= 9!
=(10 – 1)! ways Answer

Hence

(Q). In how many ways 15 students can sit in 15 chairs in a circle.
Solution:-
(15 – 1)!
= 14! ways Answer

(Q). In how many ways 15 student can sit in a circle if 3 students A, B, C are to always sit together.
Solution:-

⦿ When clockwise ↻ & anticlockwise ↺ arrangement are same: Formation of Garland

we are to arrange n beads in a circle to form a garland.
Here if we look at garland clockwise ↻ or anticlockwise ↺ both look will be same. i .e. ↻ & ↺ arrangement are not different.

(Q). Find the number of ways in which 10 different colour beads can be arranged to form a necklace.
Solution:-

\(\frac{{(10 – 1)!}}{2}\)

= \(\boldsymbol{\frac{{9!}}{2}}\) Answer

Word/Number Formation

Formation of word:-

The number of arrangement of n things taken all at a time when p of them are identical of one type, q of them are identical of another type, r of them are identical of another type & remaining ‘n – (p + q +r)’ are all of different types is:

(Q). How many words can be formed from the letter of the word INDIA.
Solution:-
Here n = 5
p = 2 (I, I)
∴ total words = \(\frac{{5!}}{{2!}}\)
= 5×4×3
= 60 Answer

(Q). How many words can be formed from the letters of the word MISSISSIPPI.
Solution:-

Here n = 11
p = 4 (I)
q = 4 (s)
r = 2 (p)
∴ total words = \(\frac{{11!}}{{4!4!2!}}\)
= 34650 Answer

(Q). In previous question if 4 ‘I’ are always together.
Solution:-

∴ total words = \(\frac{{8!}}{{4!2!}}\)
= 840 Answer

Formation of Number

(Q).How many 5 digit numbers are there?
Solution:-

= 9×10×10×10×10
= 90000 Answer

(Q). How many 4 digit numbers can be formed from the digit 0, 1, 3, 6, 7, 9
(i) if repetition of digit is not allowed
(ii) if repetition of digit is allowed
Solution:-

(Q). Find the number of 5 digit numbers that can be formed from the digits: 6 6 6 9 9.
Solution:- = \(\frac{{5!}}{{3!2!}}\)

= 10 Answer

(Q). Find the nuumber of 5 digit numbers that can be formed from the digit 1 1 6 6 6 such that the number is divisible by 4.
Solution:-

A number is divisible by 4 ⇒ if last two digits of the number is divisible by 4
Hence

= \(\frac{{3!}}{{2!}}\)
= 3 Answer

Rank of a Word in Dictionary

Rank of a word in dictionary:-

If ‘ABC’ is arranged in dictionary order. find the rank of the word ‘CAB’
ABC → 1ˢᵗ
ACB →2ⁿᵈ
BAC → 3ʳᵈ
BCA → 4ᵗʰ
CAB → 5ᵗʰ → Answer
CBA → 6ᵗʰ

(Q). Find the rank of the word MOTHER when its letters are arranged in dictionary order.
Solution:-

Method (1) word ⇒ MOTHER
First arrange all the alphabets of the word in alphabetic order

now
(i) Total words starting with E = 5!
(ii) Total words starting with H = 5!
(iii) Total words starting with ME = 4!
(iv) Total words starting with MH = 4!
(v) Total words starting with MOE = 3!
(vi) Total words starting with MOH = 3!
(vii) Total words starting with MOR = 3!
(viii) Total words starting with MOTE = 2!
(ix) Total words starting with MOTHER = 1 ⇨ Final word
∴ Rank = 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1
= 309 Answer

Method(2)

Add 1 in this value to find the rank of the final word = 308 + 1 = 309 Answer

(i) Assign numerical value according to order of alphabets
(ii) 2 numbers are less than 3 at its right side → (2, 1) ⇒ write 2 below M
(iii) 2 numbers are less than 4 at its right side → (2, 1) ⇒ write 2 below O
(iv) 3 numbers are less than 6 at its right side →  (2, 1, 5) ⇒ write 3 below T
(v) 1 number is less than 2 at its right side →  (1) ⇒ write 1 below H
(vi) 0 numbers are less than 1 at its right side ⇒  write 0 below E
(vii) There is no number after last number 5 ⇒ write 0 below R
Next Step ⇒ Starting from right side write 0!, 1!, 2!, …….
now multiply & add all these values & add 1 in this value to find the rank of desired final word.

Let’s take another example:-
(Q). Find the rank of the word ‘MEDICAL’ when arranged in dictionary.

Method(1)

word ⇒ MEDICAL
alphabetic order ⇒ ACDEILM
(i) Total words starting with A = 6!
(ii) Total words starting with C = 6!
(iii) Total words starting with D = 6!
(iv) Total words starting with E = 6!
(v) Total words starting with I = 6!
(vi) Total words starting with L = 6!
(vii) Total words starting with MA = 5!
(viii) Total words starting with MC = 5!
(ix) Total words starting with MD = 5!
(x) Total words starting with MEA = 4!
(xi) Total words starting with MEC = 4!
(xii) Total words starting with MEDA = 3!
(xiii) Total words starting with MEDC = 3!
(xiv) Total words starting with MEDIA = 2!
(xv) Total words starting with MEDICAL = 1 → Final word

∴ Rank = 6×6! + 3×5! + 2×4! + 2×3! + 2! + 1
=4743 Answer

Method (2)

now add 1 to this value to find the rank of the final word

∴ Rank = 4742 + 1 = 4743 Answer

Let’s take an Example with repetition
(Q). Find the rank of the word REPETITION
Solution:-
Method (1)

Hence rank of the final word = 323959 + 1
= 323960 Answer

Method (2)

REPETITION

EEIINOPRTT → 2E 2I 2T
(i) Total number of words starting with E = \(\frac{{9!}}{{2!2!}}\) = 90720
(ii) Total number of words starting with I = \(\frac{{9!}}{{2!2!}}\) = 90720
(iii) Total number of words starting with N = \(\frac{{9!}}{{2!2!2!}}\) = 45360
(iv) Total number of words starting with 0 = \(\frac{{9!}}{{2!2!2!}}\) = 45360
(v) Total number of words starting with P = \(\frac{{9!}}{{2!2!2!}}\) = 45360
(vi) Total number of words starting with REE = \(\frac{{7!}}{{2!2!}}\) = 1260
(vii) Total number of words starting with REI = \(\frac{{7!}}{{2!}}\) = 2520
(viii) Total number of words starting with REN = \(\frac{{7!}}{{2!2!}}\) = 1260
(ix) Total number of words starting with REO = \(\frac{{7!}}{{2!2!}}\) = 1260
(x) Total number of words starting with REPEI = \(\frac{{5!}}{{2!}}\) = 60
(xi) Total number of words starting with REPEN = \(\frac{{5!}}{{2!2!}}\) = 30
(xii) Total number of words starting with REPEO = \(\frac{{5!}}{{2!2!}}\) = 30
(xiii) Total number of words starting with REPETII = 3! = 6
(xiv) Total number of words starting with REPETIN = 3! = 6
(xv) Total number of words starting with REPETIO = 3! = 6
(xvi) Total number of words starting with REPETITIN = 1! = 1
(xvii) Total number of words starting with REPETITION = 1 ⇒ Final word

Add all above values = 32960
Hence Rank = 32960 Answer

❓ Frequently Asked Questions on Permutation or Arrangement

Q1. What is permutation or arrangement in mathematics?

Permutation or arrangement is the number of ways in which n different objects can be arranged in a definite order. The number of ways of arranging n different objects is n! (n factorial). For example 20 students can be arranged in a straight line in 20! ways. This is one of the most fundamental concepts in permutation and combination and is frequently tested in CAT, SSC CGL, Bank PO and Railway RRB exams.

Q2. How many ways can n different objects be arranged in a straight line?

n different objects can be arranged in a straight line in n! ways. This is because the first place can be filled in n ways, second in (n-1) ways, third in (n-2) ways and so on till the last place is filled in 1 way. Hence total arrangements = n × (n-1) × (n-2) × … × 1 = n!. For example 5 students can be arranged in a straight line in 5! = 120 ways. If objects are identical then number of ways = 1.

Q3. What is circular arrangement and what is its formula?

Circular arrangement is the arrangement of n different objects around a circle. When clockwise and anticlockwise arrangements are different then number of circular arrangements = (n-1)!. This is because in a circle one object is fixed as a reference point and remaining (n-1) objects are arranged around it. For example 15 students can sit in a circle in (15-1)! = 14! ways. This concept is frequently asked in CAT, SSC CGL and Bank PO exams.

Q4. What is the formula for formation of garland or necklace?

When clockwise and anticlockwise arrangements are the same such as in formation of a garland or necklace then the number of arrangements of n different things = (n-1)!/2. This is because in a garland looking clockwise and anticlockwise gives the same arrangement so we divide by 2. For example 10 different colour beads can be arranged to form a necklace in (10-1)!/2 = 9!/2 ways.

Q5. What is the formula for word formation when letters are repeated?

When n letters are arranged and p of them are identical of one type, q of them are identical of another type and r of them are identical of another type then total number of words = n! / (p! × q! × r!). For example number of words formed from letters of INDIA where n=5 and p=2 (two I’s) = 5!/2! = 60. For MISSISSIPPI where n=11, p=4(I), q=4(S), r=2(P) total words = 11!/(4!×4!×2!) = 34650.

Q6. How many 4 digit numbers can be formed from given digits?

When forming numbers the first digit (leftmost) cannot be zero. So if digits include 0 then first place has one less option. For example from digits 0, 1, 3, 6, 7, 9 without repetition — first place has 5 options (not zero), second has 5, third has 4, fourth has 3 giving 5×5×4×3 = 300 four digit numbers. With repetition — first place has 5 options, remaining three places have 6 options each giving 5×6×6×6 = 1080 four digit numbers.

Q7. How to find the rank of a word in dictionary?

To find the rank of a word first arrange all its letters in alphabetical order. Then count all words that come before the given word alphabetically. In Method 2 assign numerical values to each letter according to alphabetical order, count how many numbers smaller than each letter appear to its right, multiply by the corresponding factorial starting from 0! on the right, add all values and add 1 to get the rank. For example rank of MOTHER = 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1 = 309.

Q8. Where can I practice permutation or arrangement questions?

After understanding the concept you can practice on our Permutation or Arrangement Exercise page, Circular Arrangement Exercise page, Word and Number Formation Exercise page and Rank of a Word in Dictionary Exercise page. You can also check our Permutation and Combination Concept page for the complete topic overview.

Permutation or Arrangement Concept — Formulas and Tricks