When you form words using alphabets (or letters), the ‘order’ matters.

Three alphabets T,A,C

Now you form words: CAT is not same as ACT; both words have different meanings hence order matters. This is a permuation problem.

- Case 1: All letters are different
- Case 2: Word contains repeating alphabets
- Case 3: More than one letter reappears in Word
- Case 4: Multipicking Consonents and Vowels
- Readymade Formulas for Word-arrangement problems
- Recommended Booklist for Aptitude in CSAT,CMAT,CAT,IBPS
- Previous articles on Aptitude

# Case 1: All letters are different

- Question: How many words can you form with 3 letters: A, C, T? OR
- How many ways can the letters of word “ACT” be arranged? OR
- Rephrase: How many ways can three gentlemen Abdul, Champaklal and Tarak Mehta (A, C, T) be seated in three chairs?

If you want to visualize, OR Manually arrange them (Desi Jugaad) it’ll look like this:

Otherwise it is a mere extention of Fundamental counting principle

3 candidates for first seat AND then 2 guys remain for second seat AND only one guy remains for the last seat

“AND” means multiplication

=3 x 2 x 1 ; meaning 3! ways

= 6

**Readmady Formula**: When “n” different items or alphabets or men are to be arranged in a line, it can be done in n! factorial ways.

“ACT” has three alphabets so it can be done in three factorial (3!=3x2x1 ways)

This was easy because all three alphabets A,C,T were different so you can apply the readymade formula.

But what if there is repeatation? For example the word “Repeat” itself!

# Case 2: Word contains repeating alphabets

Q. How many ways can the letters of word “REPEAT” be arranged?

To prevent mistakes in such question, Better make a frequency chart as shown below

R…I

E…II

P…I

A…I

T…I

The letter “E” appears twice, hence although we 6 letters word, there are only 5 “different” letters or alphabets.

lets label these two “E”s as E_{1} and E_{2}.

Form a word

REEPAT

It can be formed in two ways

RE_{1}E_{2}AT

or

RE_{2}E_{1}AT

Although both have same meaning.

When we arrange letters and form words, according to Funda.counting or Permutation method, these two words will be counting as ‘two different’ words. We’ve to remove this overcounting.

So we divide the answer with overcounting. Just like how we proceeded in Combination question in the very first article of PnC.

## Step:1 Assume all alphabets to be different

We’ve 6 alphabets

Arrange these 6 gentlemen into 6 seats? (Order matters, Permutation problem)

=6 x 5 x 4 x 3 x 2 x 1 =6P6=6! ways.

Out of these two gentlemen are same. How many ways can your arrange two men in two seats?

=2 options for the first seat AND then 1 guy remains for the second seat

=2 x 1 ; AND means multiplication

=2! ways

## Final answer

=6! divided by overcounting

=6!/2!

=360.

Let us now make the situation even more complex: What if multiple alphabets of a word, are getting repeated?

# Case 3: More than one letter reappears in Word

Q. How many ways can the letters of word “RECUPERATE” be arranged? -(From Sarvesh Kumar’s book)

Make frequency character to prevent mistakes

R…II

E…III

C…I

U…I

P…I

A…I

E…I

Total 10 letters but R appears twice (II) and E appears thrice (III)

Principle is same like previous case

First make permutation of all letters, assuming that they’re different. And then divide by overcounting.

So Break it in tasks

- Task 1: Arrange 10 letters (10!)
- Task 2: Divide by overcounting of Two Rs (2!)
- Task 3: AND Again divide by overcounting of Three Es (3!)
- Task 4: keep repeating…if more letters are repeated

Start Solving

=10!/2! ;Task 1

=10!/2! ; Task 2 again divide by 3!

=10!/(2! x 3!) ; Task 3

=302400 ways letters of the word “RECUPERATE” can be arranged.

# CASE 4: Multipicking Consonents and Vowels

Question from Indiabix on request of a reader.

**Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?**

Photo for better visualization of this concept

Words of 3 consonants and 2 vowels= 3+2=5 letters.We’ve to pick up some members in the Committee and then arrange them in ‘seats’.

Break it like this

Task 1 AND Task 2 AND Task 3

(Pick 3 conso out of given 7) AND (pick 2 vowels out of given 4) AND (Arrange them in 5 seats)

=Combi AND Combi AND Permu

=Combi x Combi x Permu ; because “AND” means multiplication

==7C3 x 4C2 x 5!

## Task 1: Pick 3 conso out of given 7

Same as Committee problem.

Pick three men out of 7:

7 x 6 x 5

but in Committee, order doesn’t matter so Divide the overcounting

(arranging the selected 3 men in three seats)

3 x 2 x 1

So 7C3= (7 x 6 x 5) / (3 x 2 x 1)

## Task 2: Pick 2 Vowels out of given 4

Same way you do for 4C2

Thus we selected 3 + 2 = 5 letters. (alphabets)

## Task 3: Arrange 5 letters in a word

Suppose you’ve three letters A, C and T. You start forming words-

CAT= is not same as ACT.

Both words have different meanings so order matters, this is a permuation problem.

So How many ways can you form words using 5 letters or alphabets?

=How many ways can you make 5 gentlemen sit in 5 chairs (permutation problem, as seen in first article on PnC)

5 x 4 x 3 x 2 x 1 =5! ways.

Now gather everything together in one place

Task 1 AND Task 2 AND Task 3

7C3 x 4C2 x 5!

=25200

# Readymade Formulas for Word-arrangement problems

- When all letters of the word are different: Number of permutation =n! ; where “n” is the number of letters.

But why does it work? Because we are doing permutation of “n” Men in “n” chairs and nPn=n!

- When the word contains “n” letters, out of which P
_{1}are alike and are of one type, P_{2}are alike and of second type and P_{3}are alike and of third type and all the rest are different, then number of permutations

=n!/ (P_{1}! x P_{2}! x P_{3}!)