# Permutations and Combinations Formula Tricks and Solved Examples

## 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:
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
There are mainly two types of categories of question frequently asked in competitive exams. They are
(I) How many ways the Non Repeated letters of word can be arranged?
(II) How many ways the Repeated letters of word can be arranged?

### 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 T T 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 JUDGE 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 wordANIMATION 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 consonants always come together?

Answer: The word CORPORATION 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?

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

_M_C_H_N_(AIE)
=5! (5 blanks) x 3!(3 vowels in 3 ways)
=> 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
be arranged that vowels never comes together?
The word SISTER consisting 6 letters with 2 vowels(IE) and 1 repeated letter (S).
As earlier method is used to find out 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

In this model we need to use formula  ⁿPᵣ = n!/(n-r)!
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;
ⁿPᵣ = n!/(n-r)!
⁴P₂ = 4!/ (4-2)!
= 24/2
=12 ways.
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?
The word BANKER has two vowels only AE, so the two vowels never comes together according to the question.
Step 1: Place the blanks between the letters by separating AE as a single group,
_B_N_K_R_(AE)
Here 4 consonants BNKR can be written in two 4 ways(4!) that no 2 vowels never together.
According to Permutation and Combination Formula
Number o f blanks (n) = 5 (greater value taken as n)
and number of vowels can arranged (r) = 2 (out of 5 places vowels can placed any two places only that never comes together, hence smaller value taken as r)
ⁿPᵣ = n!/(n-r)!
⁵P₂ = 5!/ (5-2)!
=120/6
=20
Finally that consonants (NBKR) can be arranged 4! (4 ways) that no 2 vowels comes together
=20 x 4!
=20 x 24
=480 ways.
(or)
Simply we can find the answer as ⁵P₂ x 4!
Example 2: The word “SISTER” is in how many ways can be arranged that no two vowels never comes together?
The word SISTER has two vowels only IE, so the two vowels never comes together according to the question.
Step 1: Place the blanks between the letters by separating IE as a single group,
_S_S_T_R_(IE)
Here 4 consonants SSTR can be written in two 4 ways(4!) that no 2 vowels never together.
According to Permutation and Combination Formula
Number o f blanks (n) = 5 (greater value taken as n)
and number of vowels can arranged (r) = 2 (out of 5 places vowels can placed any two places only that never comes together, hence smaller value taken as r)
ⁿPᵣ = n!/(n-r)!
⁵P₂ = 5!/ (5-2)!
=120/6
=20
Finally that consonants (SSTR) can be arranged 4! (4 ways) that no 2 vowels comes together and with 2 repeated “s” letters (2!).
=20 x 4!/2!
=20 x 24/2
=240 ways.
(or)
Simply we can find the answer as ⁵P₂ x 4!/2!

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?
See the word “DETAIL” has 6 letters and among them 1,3,5 positions are only odd along with 3 vowels in letter EAI.
Number of odd places (n) = 3
Number of vowels (r) = 3
Remaining consonants = 3 as (3!)
According to Permutation and Combination Formula
ⁿPᵣ = n!/(n-r)!
³P₃ x 3!
= 3!/(3-3)! x 3!
=3!/0! x 3!
= 6 x 6
= 36 ways.
Example 2: The word “POUNDING” is in how many ways can be arranged that vowels comes only in odd position?
The word “POUNDING” has 8 letters, 4 odd positions, 3 vowels, 5 consonants and 2 repeated letters (N).
Number of odd places (n) = 4
Number of vowels (r) = 3
Remaining consonants = 5 as (3!)
Repeated letters (S) = 2 as (2!)
According to Permutation and Combination Formula
ⁿPᵣ = n!/(n-r)!
⁴P₃ x 5!
2!
= 4!/(4-3)! x 5!➗ 2!
=24/1 x 120 ➗ 2
= 1440 ways.

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
ForRepetition 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
For 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
According to Permutation and Combination Formula for condition of repetition allowed
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
According to Permutation and Combination Formulafor condition of repetition allowed
nⁿ
= 7⁷
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
and repeated letter (O) = 2!
Permutation and Combination Formula for condition of repetition allowed
=n!/ repetition factorial
=4! / 2!
=24/2
=12.
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
and repeated letter (O) = 2!
According to Permutation and Combination Formula for condition of no repetition allowed
nⁿ/ repetition factorial
= 4⁴ / 2!
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”.

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?
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?
The 4 digit number has no “0” in its digits.
Hence the answer will be n!
here number of digits (n) = 4
n!= 4!
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
=4!/2!
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
3 digit numbers can be formed (r) = 3
ⁿPr= n!/ (n-r)!
⁵P₃= 5!/ (5-3)!
=5!/2!

(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?
The 4 digit number has “0” in its digits.
Hence the answer will be n- 1!
here number of digits (n-1!) = 3!
(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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