**Permutation and Combination Tricks**

The topic of Permutation and Combination is counted in one of the important parts of Reasoning section. **Permutation and Combination Tricks **are available here for preparation of SSC/Railways/Banks and other competitive examinations. Solve the Permutation and Combination Questions through Best Tips and Shortcuts given on this page. Individuals can check Permutation and Combination Formulae and Concepts by going through the below section of this page.

**Permutation and Combination Tricks**

**What Do You Mean By Term Permutation and Combination?**

*Permutation:*

Permutation is basically called as a arrangement where order does matters. Here we need to arrange the digits, numbers , alphabets, colors and letters taking some or all at a time. The different arrangements of a given number of things by taking some or all at a time, are called permutations. It is represented as ** ^{n}P_{r}**.

__Note__:

^{n}P_{r}= n! / (n-r)!- If from the total set of n numbers p is of one kind and q ,r are others respectively then
^{n}P_{r}= n! / p! × q! × r!. ^{n}P_{n}= n!

*Combination:*

Combination is basically called as a selection where order does not matters. Here we need to arrange the digits , numbers , alphabets, colors and letters taking some or all at a time. Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination. It is represented as ** ^{n}C_{r}**.

__Note__:

^{n}C_{r }= n!/ r! × (n-r)!^{n}C_{0}= 1^{n}C_{n}= 1^{n}C_{r}= nCn – r^{n}C_{a}=^{n}C_{b}=> a = b => a+b = n^{n}C_{0}+^{n}C_{1}+^{n}C_{2}+^{n}C_{3}+ ……………+^{n}C_{n}=_{2}n

Check Here: __Permutation Combination Questions__

**Permutations and Combinations Formulas**

1) ** Permutations Formula:** Permutations can also be termed as ordered choices or arrangements. Each of the arrangements that can be made by selecting ‘r’ things out of ‘n’ things can be termed as a permutation. In permutations, the order in which the items are arranged is significant.

Consider arranging r things selected from n things. There are n possibilities for the first choice, (n – 1) possibilities for the second choice, (n – 2) possibilities for the third choice and so on. In other words, the available choices reduce by 1 after every selection.

- Therefore, ways to arrange r things selected out of n things are
*n x (n-1) x (n-2) …….. r terms = n x (n-1) x (n-2)… (n-r+1) = n!/(n-r)!* - Hence the generalise formulas for permutations is

https://i1.faceprep.in/Companies-1/permutations-and-combinations-formulas-1.PNG

2)* Similar Items In Permutations: *While arranging ‘n’ things from which ‘p’ things are of one kind and ‘q’ things are of the second kind, with the rest of the things being distinct, the number of different arrangements will be

https://i1.faceprep.in/Companies-1/permutations-and-combinations-formulas-2.PNG

3) The total number of ways in which ‘n’ things can be arranged in ‘r’ ways with repetition allowed is equal to **n ^{r}** ways.

4) *Circular Arrangements:*

- If ‘n’ objects are arranged in a circular way and if the clockwise and anti-clockwise arrangement is different, then the formula is
**(n-1)!** - When there is no difference between clockwise and anticlockwise arrangements. In those cases, the total possible arrangements are half of the original ways of arrangements, i.e
**(n-1)!/2**

5) ** Combinations Formula:** Combinations can also be termed as selections. Hence, each of the selection of ‘r’ items made out of a set of ‘n’ items is called a combination. Generally,

**. That is**

*r things are selected from n things in*^{n}C_{r}wayshttps://i1.faceprep.in/Companies-1/permutations-and-combinations-formulas-3.PNG

**It can be inferred that **

6) If an event or a trial has ‘m’ outcomes and for every outcome of the previous event, a second event has ‘n’ outcomes, then the number of different ways by which both events can be conducted will be the product of m and n.

Get Here: __55+ Mixture and Alligation Practice Questions__

**Permutation and Combinations Tips and Shortcuts**

- Use permutations if a problem calls for the number of arrangements of objects and different orders are to be counted.
- Use combinations if a problem calls for the number of ways of selecting objects and the order of selection is not to be counted.

*Summary Of Formula To Use*

Order | Repetition | Formula |

Permutation | Yes | nr |

Permutation | No | npr |

Combination | Yes | r + n – 1Cr |

Combination | No | nCr |

Download PDF of: __Number Series Question__

**Quick Tricks To Solve Permutation And Combination Questions**

*Factorial n!:** *

It is the product of all positive integers less than or equal to n.

Example: 4! = 4 × 3 × 2 × 1 = 24

*Theorem of Counting:*

1) Rule of Addition: If a first task is performed in x ways and second task is performed in y ways, then *either* of the two operations can be performed in (x + y) ways

2) Rule of Multiplication: If a first task is performed in x ways and second task is performed in y ways, then *both* of the two operations can be performed in (x × y) ways

*Suppose n different cakes are done in a _{1} , a _{2} , a _{3} , … a _{n} different ways respectively, independent of each other then:*

1) Any one of them can be done in a _{1} + a _{2} + a _{3} + … + a n ways. *(a _{ 1} way or a _{2} way or a _{3} + … + a _{n} way)*

2) All of them can be done in a_{1} × a_{2} × a_{3} × … × a_{n} ways *(a _{1} way and a_{2} way and a_{3} + … + an way)*

*All About Permutation:*

1) **Condition 1:** Number of permutations of n things, taken r at a time is given as follows:

^{ }^{n}P_{r} = n (n – 1) (n – 2) (n – 3)……. (n – r + 1) = n!/ (n – r)!

2) **Condition 2:** If there are N balls and out of these B_{1} balls are alike, B_{2} balls are alike , B_{3} balls are alike and so on and Br are alike of rth kind, such that (B_{1} balls + B_{2} balls + B_{3 }balls —– B_{r} balls) = N balls.

In such condition,

Number of permutation of these N balls = N!/ (B_{1})! × (B_{2})! × (B_{3})! × – – – – – (B_{r})!

3) **Condition 3:** If number of permutations of n objects are all taken at a time, then

^{n}P_{r} = n!/ 0! = n!

* **Important Points To Remember:*

1) If N different objects are to be arranged, then they can be arranged in N! ways.

2) N number of objects can be arranged around a circle in (N – 1)! ways.

3) Sometimes we have to solve problems on permutation considering the condition of Repetition

Repetition: This condition is not used unless specified. (*Remember*)

Number of permutation of *N objects taken r at a time* when each selected object can be repeated any number of times is given as:

Number of permutations = **n ^{ r}**

** ^{ }**Solve Out:

__Discount Questions and Answers__

4) **Restricted Permutation:** The number of permutations of n objects taken r at a time in which if k particular objects are:

a) Never included: (n – k)P_{r }—- (k are the number of objects not included)

b) Always included: (n – k) C_{r–k} _{x r!} —- (k is the number of objects always included)

*All About Combinations:*

1) Number of combinations of n objects, taken r at a time is given as follows:

This example will surely clear the concept!

Hint: In the example discussed below, the confusion related to addition and multiplication of terms will also be cleared.

Example: Suppose there are 12 boys and 8 girls, and we have to select 5 volunteers for a particular task. So we have to find the number of possible selections we can make.

Total students are (12 + 8) = 20 and we have to select 5 volunteers.

The total number of selections can be made

The question may be asked in different ways. 2 different conditions are specified below:

1) Out of 5 volunteers, 3 boys and 2 girls must be present.

2) Boys should be in majority.

Condition 1: 3 boys 2 girls are needed | Condition 2: If boys are in majority |

Out of 12 boys 3 are selected and out of 8 girls 2 are selected. (Boys) Both girls and boys are needed, hence they are multiplied.
| 1) If only boys are selected as volunteers: ^{12}C_{5}2) 4 boys and 3 girls: ^{12}C_{4} × ^{8}C_{1}3) 3 boys and 2 girls: ^{12}C_{3} × ^{8}C_{2}These will be the 3 possibilities, where boys are in majority. ( ( |

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__Note__:

Candidates must solve out these Permutation and Combination Tricks and match up their solutions with the given answers. Individuals can remain in touch with us by bookmarking our web portal recruitmentresult.com for more similar updates.

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