# Permutations and Combinations Formula Tricks and Solved Examples

## Permutation and Combination Formula Tricks

## and

## Solved Examples

Permutation and Combination Formula is a typical and most important concept for any competitive exams like GMAT, CAT, Bank PO, Bank Clerk exams.

Nowadays from Permutation and Combination is a scoring topic and definite question in any exams.

If you want to crack this concept of Permutation and Combination Formula, first of all, you should learn what are definitions of terminology used in this concept and need to learn formulas, then finally learn factorial calculation, which is the most important to get a result for the given problem.

Only the formula you need to know, that is

**Permutation and Combination Formula**

**ⁿPᵣ = n!/(n-r)!**

Here, the symbol “**!**” called as **“factorial”**.

We can describe above formula as ” permutations as ‘n’ distinct objects taken ‘r’ at a time.

n = the number of objects, from which the permutation is framed.

r = the number of objects used to frame permutation.

### Basics in Permutation and Combination Formula

#### What is Permutation?

**Permutation**

All the possibles selections from the given set of things known as Permutation.

(or)

The arrangement of all or part of the set of any objects with regard to the order.

Here permutation focuses on the arrangement of objects in considering the order (different orders, but same letters) in which they can be arranged.

Example:**Suppose we have set of P, Q, R marbles in a bag. In these how many ways we can arrange 2 marbles from the set? **

**Answer:** Here in the set each possible of two marbles is an example for** permutation**. All the possibles for PQR set are PQ, PR, QP, QR, RP, RQ. (same letters have different order)

**Explanation:** Here the permutation is framed with 3 letters (i.e P, Q, R). According to permutation formula n = 3; and the permutation comprising 2 letters in each possible, hence r =2;

### what does combination mean?

** ** Combination

Required possibles selections from the given set of things are known as Combination. Here combination focuses without regarding the order in which objects are selected.

Example:

Example:

**Suppose we have set of P, Q, R marbles in a bag. In these how many ways we can arrange 2 marbles from the set?**

**Answer:**Here in the set, each possible of two marbles is an example for the

**combination**. All the possibles for PQR set, are PQ, PR, QR. Here we note that PQ and QP are considered to be one combination. Here the order of selected letters not important.

The main difference between permutation and combination is only the order that in which objects are selected or arranged.

**Combination Vs Permutation**

**Factorial Table – Permutation and Combination Formula**

**0! = 1**

**1! = 1**

**2! = 2**

**3! = 6**

**4! = 24**

**5! = 120**

**6! = 720**

**7! = 5040**

**8! = 40320**

**9! = 362880**

**10! = 3628800**

Byheart the above factorial table and it is useful for instant calculation.

**Procedure for Factorial Calculation**

The value always for 0! and 1! is equal to 1.

For 2! = 2 x 1 = 2;

3! = 3 x 2 x 1 = 6;

4! = 4 x 3 x 2 x 1 = 24;

5! = 5 x 4 x 3 x 2 x 1 = 120;

6! = 6 x 5 x 4 x 3 x 2x 1 = 720;

7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040;

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320;

9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880;

10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800;

**Types of Questions asked based on Permutation and Combination Formula**

(I) How many ways the

**Non Repeated**letters of word can be arranged?

**Repeated**letters of word can be arranged?

### Models based on Permutation and Combination Formula

### Vs

### Permutations and combinations questions

**Model I – Basic Questions based on Non-Repeated and Repeated letters in the word**

**Type I: How many ways**

**the word with**

__Non-Repeated__letters can be arranged?**Example 1: The word “BAT” in how many ways can be arranged?**

**Answer: **The word “BAT” consisting **3 letters**,which are non-repeated, hence **3! **= 3 x 2 x 1 = 6 ways is the answer.

**Explanation: “6 ways”** of word rewritten as BAT, BTA, ATB, ABT, TAB, TBA.

**Type II: How many ways the**

**word with**

__Repeated__letters can be arranged?**Example 1: The word “SISTER” is in how many ways can ****be arranged****?**

**Answer: **The word “SISTER” consisting total 6 letters and 2 repeated letters “S” in **SISTER**.

According to Permutation and Combination Formula

= 6!/2!

= 6 x 5 x 4 x 3 x 2 x 1/ 2 x 1

=720/2

=360.

**Example 2: ****The word “LETTER” is in how many ways can ****be arranged****?**

**Answer: **The word “LETTER” consisting total 6 letters and 2 (2!) repeated letters “T” and another 2 (2!) repeated letters “E” in **L E T T E R**.

According to Permutation and Combination Formula

= 6!/2! x 2!

= 720/ 2 x 2

= 720/ 4

= 180.

As like above example you need to calculate factorial for every respective repeated letter in the word and multiply like in above example.

**Model II – Questions based on Non-Repeated and Repeated letters in the word that vowels always come together**

**Type I: How many ways the word with Non-Repeated letters can be arranged with the condition of vowels always comes together?**

**Example 1:**

**The word “JUDGE” is in how many ways can**

**be arranged that**

__vowels always come together__**?**

**Answer:**The word J

**U**DG

**E**has vowels U and E. Then the word JUDGE can be written as below.

=>

**Step 1:**Place the blanks as _J_D_G_ between letters by making the separate group of letters of vowels.

=>

**Step 2:**Vowels together UE can write in 2 ways in either UE or EU (2!) together can write in 4

blanks (4!) shows between letters of the word.)

=>

**Step 3:**Hence the total

**product**of factorials gives an answer.

= 4! (4 blanks) x 2! (2 ways)

= 24 x 2

= 48 ways.

**Type II: How many ways the word with**

__Repeated__letters can be arranged with the condition of__vowels always comes together__?

**Example 1: The word “ANIMATION” is in how many ways can****be arranged****that**__vowels always come together__**?****Answer: **The word**A****N****I**M**A**T**ION** has vowels A(2 times repeated), I(2 times repeated), O (no repetition) and consonant N (2 times repeated)

=>** Step 1: **Place the blanks as _N_M_T_N_(AIAIO)

=> **Step 2: **Make fraction form as

__total blanks factorials x ways of vowels together can write in factorials __product of repeated letters in factorials

= 5! x 5!/2! x 2! x 2!

=120 x120 / 2 x 2x 2

= 14400/8

=1800.

**Model III – Questions based on Non-Repeated and Repeated letters in the word that Consonants always come together**

**Type I:**How many ways the word with

__Non-Repeated__**letters can be arranged with the condition of that**

__?__

**consonants****always comes together****Example 1: ****The word “****BANKER”**** is in how many ways can ****be arrange that ****consonants**__ always come together__**?**

**Answer: **The word **BANKER **having **BNKR** letters as consonants and make this as one separate group.

=>** Step 1: **Place the blanks between remaining letters as _A_E_ between letters by making the separate group of letters of consonants.

=>** Step 2: **consonants together BNKR can write in 4 ways (4!) in 3 blanks (3!) shows between letters of the word.)

=> **Step 3: **Hence the total** product **of factorials gives an answer.

= 3! x 4!

= 6 x 24

=144.

**Type II: **How many ways the word with __Repeated__** letters can be arranged with the condition of that consonants always**__ comes together__?

**Example 1: The word “CORPORATION” is in how many ways can**

**be arranged**

**that**

__consonant__

__s always come together__**?**

**Answer: **The word** CORPORATI****ON** has consonants R(2 times repeated), O (3 times repetition).

=>** Step 1: **Place the blanks as _O_O_A_I_O_ and making consonants (CRPRIN) as separate group.

=> **Step 2: **Make fraction form as

__total blanks factorials x ways of consonants together can write in factorials__

product of repeated letters in factorials

= 6! x 6!/2! x 3!

=720 x 720 / 2 x 6

= 518400/12

=43200.

**Model IV – Questions based on Some Letters always come together**

**Example 1: The word “CRYSTAL” is in how many wasys can be arranged that letters “AY” comes together? **

**The word CRYSTAL consisting “AY” letters and make them as a separate group and fill the blanks between remaining letters.**

**Answer:****Step 1:**Place the blanks as _C_R_S_T_L_ and making (AY) as separate group.

**=>Step 2:**Letters together AY can write in 2 ways in either AY or YA (2!) together can write in 6

blanks (6!) shows between letters of the word.)

=> **Step 3: **Hence the total** product **of factorials gives an answer.

= 6! (6 blanks) x 2! (2 ways)

= 720 x 2

= 1440 ways.

**Model V – Questions based on Vowels never comes together**

**Note:** In this model that we need to **subtract **ways **that vowels comes together **from total** ways of word can be arranged **both in non-repeated and repeated letters type of questions.

**Type 1:Non Repeated**

Example 1: The word “MACHINE” is in how many ways can

Example 1: The word “MACHINE” is in how many ways can

**be arranged**

**that**

__vowels never comes together__**?**

**Answer:**The word MACHINE consisting 7 letters non repeated and in the word 3 vowels are present as AIE.

As earlier we know that method that is how to find that word can be arranged with vowels comes together.

=>** Step 1: **find out first ways of word can be arranged that vowels comes together as earlier. Make vowels as separate group and place blanks between remaining letters of word.

**Step 2:**Subtract above value from total ways of complete word can be arranged.

MACHINE word has total 7 letters and can be arranged in 7 ways (7!)

=7!-5! x 3!

= 5040 -120 x 6

= 5040- 720

=4320 ways.

**Type 2: Repeated**

Example 1: The word “SISTER” is in how many ways can

Example 1: The word “SISTER” is in how many ways can

**be arranged**

**that**

__vowels never comes together__**?**

**Answer:**

=>** Step 1: **First find out that ways of word can be arranged with vowels comes together including repetition of letters. As we know for this how to calculate earlier.

_S_S_T_R_(IE)

= __Total blanks between letters x ways of vowels together comes in factorial__

factorial of repeated letters

= 5! x 2!/ 2!

= 120 x 2 /2

= 240/2

= 120.

=>** Step 2: **Subtract above value from total ways of complete word can be arranged with repeated letters.

Total ways of word with repeated letter = 6!/2!

=720/2

=360.

=>** Step 3: **Required answer = 360-120=240 ways**.**

**Model VI – Questions based on No Two Vowels never comes together**

**Explanation:**Suppose we have 4 letters as A, B, C, D and these 4 letters and assume that we need to place in 2 places, In how many ways that we can arrange 4 letters in 2 places?

**Answer:**Then formula applied as here, no of letters (n) = 4 and places (r)=2;

**Note:**The value of “n” always more than ‘r’ {∴ Always

**n >r**)

**Example 1: The word “BANKER” is in how many ways can**

**be arranged**

**that**

__no two__

__vowels never comes together__**?**

**Answer:**

**Example 2: The word “SISTER” is in how many ways can**

**be arranged**

**that**

__no two__

__vowels never comes together__**?**

**Answer:**

**Model VII – Questions based on that Vowels/ Consonants comes in Odd position or Even position**

**Example 1:**

**How many ways the word DETAIL can be arranged in such a way that vowels comes only in odd position?**

Answer:

Answer:

**EAI**.

**Example 2: The word “POUNDING” is in how many ways can**

**be arranged**

**that**

**vowels comes only in odd position?**

**Answer:**

__⁴P₃ x 5!__

2!

= 4!/(4-3)! x 5!➗ 2!

**Model VIII – Questions based on Repetition allowed and Repetition not allowed**

**Rule 1:**If any word given and asked the ways of word can be arranged

**Repetition allowed**condition use formula as

**n!**

**Note:**In case repeated letter in the word, then formula

=

**Total ways in factorial/ Repeated letter in factorial =**

**n!/ repetition factorial**)

**Rule 2:**If any word given and asked the ways of word can be arranged

**No Repetition allowed**condition use formula as

**nⁿ**

**Note:**In case repeated letter in the word, then formula

=

**nⁿ**

**/ Repeated letter in factorial**)

**Example 1: The word “CRYSTAL” in how many ways can be arranged if**

*repetition allowed*?**Answer:**The total number of letters in CRYSTAL word (n) = 7

**n!**

**= 7!**

**Example 2: The word “CRYSTAL” in how many ways can be arranged if**

*no repetition allowed*?**Answer:**The total number of letters in CRYSTAL word (n) = 7

**nⁿ**

**Example 3: The word “COOK” in how many ways can be arranged if**

*repetition allowed*?**Answer:**

**Answer:**The total number of letters in COOK word (n) = 4

**n!/ repetition factorial**

**Example 4: The word “COOK” in how many ways can be arranged if**

*no repetition allowed*?**Answer:**

**Answer:**The total number of letters in COOK word (n) = 4

**nⁿ**

**/ repetition factorial**

**Model IX – permutations and combinations formula Questions based**

**on**

**Word with Miscellaneous Concepts**

**Example: How many ways the word “PLAYGROUND” can be arranged such that**

(I) If the word starts with letter “Y”.

(I) If the word starts with letter “Y”.

**Answer:**The total letters in the word = 10

if Y is in 1st place with starting of word remaining places are only = 9

= 1 x 9!

=9 !

so, if the 1st place occupied by “Y”, then 9! ways the word can be arranged.

(II) If the word starts with letter “Y” and ends with “G” letters?

**Answer:**The total letters in the word = 10

if Y is in 1st place with starting of word and ends with G.

The remaining places are only = 8

= 1 x 8!

=8 !

so, if the 1st place occupied by “Y” and ends with “G”, then 8! ways the word can be arranged.

**(III) If the word ends with vowel (AOU)?**

**Answer: **There are 3 (AOU)vowels, among them any one of the occupied last place only.

Remaining places 9 filled with 9 other letters.

= 9! x 3 ways.

**(IV) If the word starts with vowel****(AOU)**** and ends with consonant (PLYGRND)?**

**Answer: **There are 3 (vowels) and 7 consonants, among them any one vowel at 1st and any one of consonant occupied last of word.

Remaining places 8 filled with 8 other letters.

= 3 x 8! x 7 ways.

**(V) If the word has no 2 vowels comes together****?**

**Answer: **Here no two vowels comes together means all the vowels should be in different places and no two vowels are placed side by side in the word.

The total number of vowels (OAU) = 3

_P_L_Y_G_R_N_D_(OAU) (follow the earlier concept in above models)

Number of blanks (n) = 8 ways

Number of vowels (r) = 3 ways

Number of consonants = 7 ways

According to Permutation and Combination Formula

ⁿPᵣ = n!/(n-r)!

⁸P₃ x 7! ways.

**(VI) If the word has vowels in odd positions****?**

**Answer: Number of odd positions in word (n)= 5**

**Number of vowels to be placed in odd positions (r) = 3**

**Number of consonants = 7**

According to permutation and combination formula

ⁿPᵣ = n!/(n-r)!

⁵P₃ x 7! ways.

**Model X – Questions based on ways of number can be formed with Digits with Non – Zero and Zero**

**(I) Ways of Number formed with No Zero in number**

**If any number if it has no ‘0’ in its digits the number of ways that it is formed = the number of digits factorial.**

Example: 1287 is a 4 digit number in how many ways that it can be formed with 4 digit numbers?

Example: 1287 is a 4 digit number in how many ways that it can be formed with 4 digit numbers?

**Answer:**

**Note:**If repetition occurs of any digit then divided by repeated digit factorials

**Example:**2342 (here 2 repeated to times, hence n!/repeated digits factoriasl

**Example: 93741 is a 4 digit number in how many ways that it can be formed with 3 digit numbers?**

**Answer:**Number of digits (n) = 5

**(II) Ways of Number formed with Zero in number**

**If any number if it has ‘0’ in its digits the number of ways that it is formed = number of digits- zero containing places.**

**Example: 2049 is a 4 digit number in how many ways that it can be formed with 4 digit numbers?**

**Answer:**

(because 0 has no value in 1st digit place and can’t form 4 digit number by placing 0 in 1st digit place, so we can consider up to 3!)

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