Permutations and Combinations - Formulas, Problems and Video Tutorials

Permutations and Combinations are arrangements or selections of objects out of groups. For example, selecting a team of 11 players out of 20 players.

Permutations

The arrangements made by taking some or all elements out of a group in a particular manner are called permutations.
For example, in how many way can the letters word ENGLISH be arranged, so that vowels never come together?
The number of permutation of n thing taking r at a time is denoted by $^{n} P_{r}$ and it is defined under :
$^{n} P_{r}\: =\: \frac{n!}{(n-r)!}\: ;$ r ≤ n.

Important notations
n! (Read as n factorial)
Product of first n positive integers is called n factorial.
n! = 1, 2, 3, 4, 5…….n
n! = (n-1)! N
A special case 0 ! = 1

Various Types of Permutations
Case 1 When in a permutation of n thing taken r at a time, a particular thing always occurs.
The required numbers of permutations = $r(^{n-1}P_{r-1})$

Case 2 The number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement is $^{n-1}P_{r}$

Case 3 (Permutation of like things) The number of n things taken all at a time, given that $p_{1}$ things are $1^{st}$ alike, $p_{2}$ things are $2^{nd}$ alike, and $p_{r}$ things are $r^{th}$ alike is
$\frac{n!}{p_{1}!p_{2}!......p_{r}!}$

Case 4 (Permutation with repetitions) The number of n different things taken r at a time when each may be repeated any number of times in each arrangements is n.

Case 5 (Circular permutations) Circular permutations are the permutations of things along the circumstance of a circle. We have to consider the relative position of the different things in a circular arrangement.

For example, if there are five letters P, Q, R, S and T, two of the arrangements would be PQRST, TPQRS. These two arrangements are obviously different if the things are to be placed in a straight line. But if the arrangements are written along the circumstance of a circle, then the two arrangements PQRST and TPQRS are one and the same.
As the number of circular permutations depends on the relative position of the objects, we fix the position of one object and then arrange the remaining (n-1)! Ways. Thus the circular arrangements of five letters P, Q, R, S, T will be
(5-1)! = 4! = 4.3.2.1= 24 ways.

Some important result of permutations
1. $np_{n-1}\:=\: np_{n}$
2. $np_{n}\:=\: n$ !
3. $np_{r}\:=\:n\left (n-1 p_{r-1} \right )$
4. $np_{r}\:=\:\left ( n-r+1 \right )\times\: n p_{r-1}$
5. $np_{r}\:=\:n-1p_{r}+ r \left(n-1p _{r-1} \right )$

Combinations

The groups or selections made by taking some or all elements out of a number of things are called combinations.
For example, find the number of different poker hands in pack 52 playing cards. The number of combinations of n thing r at a time is denoted by image and it is defined as under
$nc_{r}\: =\:\frac{n!}{r!\left ( n-r \right )!}=$

Types of combinations
Case 1 To find the number of ways selecting of ways selecting one or more items out of n given items is
$nc_{1}+nc_{2}+nc_{3}+.....+nc_{n}\: =\: 2^{n}-1$

Case 2 The number of combinations of n items taken r at a time in which given p particular items will always occur is
$n-pc_{r-p}$

Case 3 The number of combinations of n items taken r at a time in which p particular items never occur is image
Some important results of combinations
$n-pc_{r}$

Important result of combination :
1. $nc_{0}\:=\:1\:=\: nc_{n}$
2. $np_{r}\:=\: r!(nc_{r})$
3. $nc_{r}+nc_{r-1}\:=\: n+1c_{r}$
4. $\frac{nc_{r}}{n-r+1}\:=\: \frac{nc_{n-r}}{r}$
5. $n\times n-1c_{r-1}\:=\:(n-r+1)\times nc _{r-1}$

Permutations and Combinations Examples

Question-1: If $nc_{10}\:=\: nc_{14}$ find the value of n

Solution: $nc_{10}\:=\: nc_{14}\: \Rightarrow n\: =\: (10+14)\: =\: 24$

Question-2: if $nc_{3}\: =\: 220$ then find the value of n.

Solution: $\therefore \frac{n!}{(n-3)!3!}\: =\: 220$

$\frac{n(n-1)(n-2)(n-3)!}{(n-3)!6}\: =\: 220$

n(n-1)(n-2) = 1320 or n(n-1)(n-2) = $11\times 12\times 21$ or n= 12

Question-3: if $nc_{r}+nc_{n+1}\: =\:n+1 c_{x}$ then find the value of x.

Solution: $nc_{r}+nc_{n+1}\: =\:n+1 c_{r+1}$ , formula

$nc_{r}+nc_{n+1}\: =\:n+1 c_{x}$ given

by comparing above two statement we can get , x = r+1

Question-4: In how many ways can letter of the word 'APPLE' be arranged?

Solution: there is in all 5 letter .out of these two are p,one is A one is L and one is E.

$\frac{5!}{2!.1!.1!.1!}\:=\: 60$

Permutations and Combinations Questions from Previous Year Exams

This part will cover the Permutations and Combinations Syllabus of Bank Clerk Exam

Question 1 of 14

1. Question
1 points
How many different words can be made with the letters of the word STOREKEEPER, taking all at a time?
- 83160
- 415800
- 8136
- 813600
- None of these
Correct

There are four E, two R and five other different letters.

.'. The number of different words = $\frac{11!}{4!\, 2!}$

= $\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 2\times 1}=11\times 5\times 9\times 4\times 7\times 6\times5=415800$

Incorrect

There are four E, two R and five other different letters.

.'. The number of different words = $\frac{11!}{4!\, 2!}$

= $\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 2\times 1}=11\times 5\times 9\times 4\times 7\times 6\times5=415800$
Question 2 of 14

2. Question
1 points
In how many different ways can the letters of the word 'CLOUD' be arranged?
- 24
- 240
- 120
- 60
- None of these
Correct

The number of arrangement= 5! = $5\times 4\times 3\times 2\times 1=120$

Incorrect

The number of arrangement= 5! = $5\times 4\times 3\times 2\times 1=120$
Question 3 of 14

3. Question
1 points
In how many different ways can the letters of the word 'ABILITY' be arranged?
- 412
- 420
- 440
- 450
- None of these
Correct

The number of arrangement = $\frac{7!}{2!}=\frac{7\times 5\times 4\times 3\times 2!}{2!}=7\times 5\times 4\times3=420$

Incorrect

The number of arrangement = $\frac{7!}{2!}=\frac{7\times 5\times 4\times 3\times 2!}{2!}=7\times 5\times 4\times3=420$
Question 4 of 14

4. Question
1 points
In how many ways I persons can sit around a round dining table in such a manner that all will not have the same neighbours in any two arrangement?
- 5040
- 20160
- 2130
- 2250
- None of these
Correct

The number of ways= $\frac{\left ( 9-1\right )!}{2}= \frac{8!}{2}=\frac{40320}{2}=20160$

Incorrect

The number of ways= $\frac{\left ( 9-1\right )!}{2}= \frac{8!}{2}=\frac{40320}{2}=20160$
Question 5 of 14

5. Question
1 points
How many different numbers of 3 digits can be formed with digits 1,2, 4,5,7 without repeating any of them in any number?
- 60
- 27
- 19
- 12
- None of these
Correct

[su_table]

Place Hundred's Ten's Unit's

Number of digits which can be taken 3 4 5

[/su_table]
.'. The number of 3 digitnumbers= $3\times 4\times 5= 60$

Incorrect

[su_table]

Place Hundred's Ten's Unit's

Number of digits which can be taken 3 4 5

[/su_table]
.'. The number of 3 digitnumbers= $3\times 4\times 5= 60$
Question 6 of 14

6. Question
1 points
How many five digit numbers can be formed with digits 1,2,3, 4 and 5 which are divisible by 2?
- 52
- 48
- 46
- 40
- None of these
Correct

Since, the numbers are divisible by 2, therefore at unit's place only 2 or 4 will be taken

[su_table]

Place Ten thousand Thousands Hundreds Tens Units

Number of digits, which can be taken 1 2 3 4 5

[/su_table]
.'. The number of five digits number= $1\times 2\times 3\times 4\times 2 = 48$

Incorrect

Since, the numbers are divisible by 2, therefore at unit's place only 2 or 4 will be taken

[su_table]

Place Ten thousand Thousands Hundreds Tens Units

Number of digits, which can be taken 1 2 3 4 5

[/su_table]
.'. The number of five digits number= $1\times 2\times 3\times 4\times 2 = 48$
Question 7 of 14

7. Question
1 points
A tearn of 16 cricketers is selected for the ODI series. It consists of 10 old players and 6 new players. As a principle, it was decided that the team playing any of the matches must contain 8 players. The number of ways the team can be selected, is
- 1250
- 900
- 850
- 450
- None of these
Correct

The number of ways= ¹⁰C₈ $\times$ ⁶C₃

= $\frac{10!}{8!\, 2!}\times \frac{6!}{3!\,3!}=\left (\frac{10\times 9\times 8!}{8!\times 2\times 1} \right)\times \left(\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1}\right)$

= $45\times 20$ = 900

Incorrect

The number of ways= ¹⁰C₈ $\times$ ⁶C₃

= $\frac{10!}{8!\, 2!}\times \frac{6!}{3!\,3!}=\left (\frac{10\times 9\times 8!}{8!\times 2\times 1} \right)\times \left(\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1}\right)$

= $45\times 20$ = 900
Question 8 of 14

8. Question
1 points
How many five digit numbers greater than 6000 can be formed with the digits 0, 3, 6, 7 and I repetition not being allowed?
- 35
- 45
- 57
- 72
- None of these
Correct

Since, the numbers are thousand's place only 6, greater than 6000, therefore at 7 or 9 will be taken.
[su_table]

Place Ten thousand Thousands Hundreds Tens Units

Number of digits, which can be taken 4 (0 will not taken at this place) 3 3 2 1

[/su_table]
.'. The number of five digit numbers
= $4\times 3\times 3\times 2\times 1 = 72$

Incorrect

Since, the numbers are thousand's place only 6, greater than 6000, therefore at 7 or 9 will be taken.
[su_table]

Place Ten thousand Thousands Hundreds Tens Units

Number of digits, which can be taken 4 (0 will not taken at this place) 3 3 2 1

[/su_table]
.'. The number of five digit numbers
= $4\times 3\times 3\times 2\times 1 = 72$
Question 9 of 14

9. Question
1 points
How many six letter word.s can be formed with the letters of the word POLICY such that the vowels occur only at even places?
- 6
- 21
- 24
- 144
- None of these
Correct

Incorrect

Number of ways in which 3 even places can be filled with two vowels = ³P₂=6

Number of four vacant places can be filled with four consonants =⁴P₄=24.

.'. Total number of words in which vowels occupy even places = $6\times 24=144$
Question 10 of 14

10. Question
1 points
How many different words can be formed with the letters of the word TOMORROW, taking all at a time, such that all vowels don't occur together?
- 360
- 3360
- 3300
- 3000
- None of these
Correct

Incorrect

There are eight letters in the word TOMORROW of which there are two Rs, three Os and three other different numbers.

.'. The number of words in which vowels (two Os) occurs

$together\, =\, \frac{8!}{3!\, 2!}-\frac{\left ( 8-3+1 \right )!}{2!}$

= 3360-360= 3000
Question 11 of 14

11. Question
1 points
In how many ways 6' different coloured beads can be arranged in a necklace?
- 60
- 80
- 120
- 180
- None of these
Correct

Incorrect

the number of ways= $\frac{\left ( 6-1\right )!}{2}= 60$
Question 12 of 14

12. Question
1 points
How many triangles can be formed with 8 non-collinear points lying in a plane?
- 56
- 50
- 35
- 28
- None of these
Correct

Incorrect

The number of trangles =⁸C₃

= $\frac{8!}{3! \, 5!}=\frac{8\times 7\times 6}{3\times 2}=56$
Question 13 of 14

13. Question
1 points
How many diagonals can be drawn in a heptagon?
- 14
- 21
- 28
- 34
- None of these
Correct

A heptagon has 7 sides.

.'. Total number of diagonals= $\frac{7(7-3)}{2}=14$

Incorrect

A heptagon has 7 sides.

.'. Total number of diagonals= $\frac{7(7-3)}{2}=14$
Question 14 of 14

14. Question
1 points
A eommittee of 6 mernbers is to be selected from a group of 8 rnen and 6 women in such a way that at least 3 men are there in the committee. In how many different ways can it be done
- 2506
- 2534
- 1120
- 1150
- None of these
Correct

Incorrect

The number of ways = number of ways to select [(3 men + 3 women) + (4 men + 2 women) + (5 men + 1 women) + (6 men)]

= ⁸C₃ $\times$ ⁶C₃+ ⁸C₄ $\times$ ⁶C₂+ ⁸C₅ $\times$ ⁶C₁+ ⁸C₅ $\times$ ⁶C₀

= $\frac{8!}{3!\, 5!}\times \frac{6!}{3!\, 3!}+\frac{8!}{4!\, 4!}\times \frac{6!}{4!2!}+\frac{8!}{5!3!}\times \frac{6!}{1! 5!}+\frac{8!}{6!2!}\times \frac{6!}{0!6!}$

= $\frac{8\times 7\times 6}{3\times 2}\times \frac{6\times 5\times 4}{3\times 2}+\frac{8\times 7\times 6\times 5}{4\times 3\times 2}\times \frac{6}{1}+\frac{8\times 7}{2}$

= 1120 + 1050 + 336 + 28 =2534