# 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
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

## Permutations and Combinations Video Tutorial

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