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}!)

## So far 45 Comments posted

thanku sir…awesum explaination….

U made it Simple… Thanks..:)

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Hi Sir

I am unable to solve the following question

Q: How many words beginning with vowels can be formed with the letters of the word EQUATION (source : BSC)

Akanshka,

EQUATION is an eight letter word and all the vowels of alphabets are in the word itself (a,e,i,o&u)

A vowel should be in first position so 5 vowels for one position can be selected in 5 ways

in rest of the positions any of the 7 letters can be filled so 7!

5×7!=25,200

Correct me if I am wrong.

It a eight letters words..

8!=8*7*6*5*4*3*2*1=40320

Yes,miss priya is right…because it should begin with vowel tht was the question only .

I am not sure but mosy probably the answer shud be 4C1 * 7!

Becoz there are 4 vowels and any vowel can be selcetd in 4C1 ways and after that there are 7 vacant spaces and 7 letters which can be arranged among themselves in 7! ways.

Thank u CS … now i got it …. bt vowels here are 5 (A,E,I,O,U)

so I guess it should be 5C1 * 7!

But in question it has not been mentioned that it should a 8 letter word only. So in my opinion it can be even 2 letter or 3 or 4 or 5 or 6 or 7 letter word also..

This is not a committee problem question….diffrent consonant make diffrent words so you should not divide here.

i agree..pleaase any one clear this doubt

Very helpful, thank you very much.

Sir Please check case 3 “RECUPERATE”… i think you unintentionally ignored the letter “T” for the calculation.

or i think need to change just the letter “E” with “T”..

how can i calculate 10*9*8*7*6*5*4*3*2*1=?

in less time?

Dear Friend U cannot get any shortcuts / tricks to solve continuous multiplication of numbers which are in decreasing order. U have to multiply it fully. Solving a problem may have some shortcuts but the basic numerical operations do not have…….

?

calculate once and memorise it and remember it for life long.. lol…. its a silly question.. if u want use tyra book for speed calculations..

EXCELLENTLY EXPLAINED…..NOW ITS AN INFINITESIMAL PIECE OF CAKE

in how many ways the letters of the word ” successful” can be organised so that all ‘ s ‘ remains together u?

plz help to solve this question

in how many ways the letters of the word ” successful” can be organised so that all ‘ s ‘ remains together ?

plz help me to solve this question

count all s as 1 alphabet only so in total there will be 8 alphabets. Ways to arrange 8 alphabets is 8! now as there are 2 c so we need to divide by 2!

so ans is 8!/2!

there are 2 c and 2 u……..i think ans will be (8! / 2!x2!)

I feel it should be 8*7! ways, please let me know if this is not right answer.

How many ways can the letter 06 the word NINETEEN rearranged?

Yaar aap tho jadugar ho

two consecutive vowels are taken at random from english alphabets, the probability that they are both consonants is ?

in how many ways can two balls be selected from a bag containing 1 red, 2 identical white and 3 identical pink balls?

what ll be its ane. mrunal?

acc. to me it shld b 6C2, but in book its 5.

pliz correct me where i m wrong.

thankw

Answer 5 is corect .

Solve this by taking cases:

Red White Pink Total

1 1 0 2

1 0 1 2

0 2 0 2

0 0 2 2

0 1 1 2

This is the answer because the balls (white and pink) are idential.

sujit: can u deduce the answer iin terms of formula since it is not posiible to write all the combinations everytime

How many diffirent ways can 3 red,4 yellow, and 2 blue bulbs be arranged in a string of christmas tree light with 9 socket?

शब्द “SUCCESSFUL” के अक्षरों को कितने विभिन्न तरीके से व्यवस्थित कर सकते है कि सभी ‘S’ एक साथ रहे?

sir ,.. please give me solution…………//

the number of ways of HOTEL can rearrange so that vowels come together?

in word SUCCESS how many arrangements so words always started with S..

nice explanation…!! sir…

Hello, I have a question.

How many different words can be formed from letters of the word “MINIMUM” when all M’s NEVER come together ?

Hi,

The question asks about words beginning with vowels. So in this word there are five vowels A,E,I,O,U.

SO in the first position we will keep a vowel and so there would be a total of five cases as we have five vowels:

A_ _ _ _ _ _ _: now for this 7 blanks we have to find arrangements which will be 7*6*5*4*3*2*1 which is 7!.

Similarly for words beginning with other vowels namely E,I,O,U we would have same result so we would have 5*7!.

In how many ways letter of the word LATTER be arranged in which E never comes before A?

Find the number of arrangements that can be made by taking 4 letters from the word ENTRANCE?

How many four-letter arrangements are there of the letters in the word SUCCEED. The answer is 270 but I don’t know how to calculate and apply the methods? Please help! Thanks in advance

Sorry another one I am stuck on:

Q. A new Pizza parlour offers 5 toppings, 3 pizza sizes to choose from. The owner wants to name the restaurant Pizza2000 to indicate all the different type of pizza that can be made. She increases the number of options to include three different cheeses, of which you can choose one, two or three. If the owner wants to offer more then 2000 possible pizzas, how many toppings choices must she make available?

The answer is 7 but I don’t know what formula to apply. Please help, thanks in advance!

In how many ways can the letter in the word SURPRISE be arranged?